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] | Mathematical Statistics/04_sampling_distributions.rmd | bmoretz/Statistical-Computing | 3 | ---
title: ''
mainfont: Arial
fontsize: 12pt
documentclass: report
header-includes:
- \PassOptionsToPackage{table}{xcolor}
- \usepackage{caption}
- \usepackage{amssymb}
- \usepackage{booktabs}
- \usepackage{longtable}
- \usepackage{array}
- \usepackage{multirow}
- \usepackage{wrapfig}
- \usepackage{float}
- \usepackage{colortbl}
- \usepackage{pdflscape}
- \usepackage{tabu}
- \usepackage{threeparttable}
- \usepackage{threeparttablex}
- \usepackage[normalem]{ulem}
- \usepackage{makecell}
- \usepackage[table]{xcolor}
- \usepackage{fancyhdr}
- \usepackage{boldline}
- \usepackage{tipa}
\definecolor{headergrey}{HTML}{545454}
\definecolor{msdblue}{HTML}{1C93D1}
\pagestyle{fancy}
\setlength\headheight{30pt}
\rhead{\color{headergrey}\today}
\fancyhead[L]{\color{headergrey}Moretz, Brandon}
\fancyhead[C]{\Large\bfseries\color{headergrey}Sampling Distributions}
\rfoot{\color{headergrey}\thepage}
\lfoot{\color{headergrey}Chapter 4}
\fancyfoot[C]{\rmfamily\color{headergrey}Mathematical Statistics}
geometry: left = 1cm, right = 1cm, top = 2cm, bottom = 3cm
date: "`r format(Sys.time(), '%d %B, %Y')`"
output:
pdf_document:
fig_caption: yes
latex_engine: xelatex
editor_options:
chunk_output_type: console
---
```{r knitr_setup, include = FALSE}
# DO NOT ADD OR REVISE CODE HERE
knitr::opts_chunk$set(echo = FALSE, eval = TRUE, dev = 'png')
options(knitr.table.format = "latex")
```
```{r report_setup, message = FALSE, warning = FALSE, include = FALSE}
library(data.table, quietly = TRUE, warn.conflicts = FALSE)
assignInNamespace("cedta.pkgEvalsUserCode", c(data.table:::cedta.pkgEvalsUserCode, "rtvs"), "data.table")
library(ggplot2, quietly = TRUE, warn.conflicts = FALSE)
library(ggrepel, quietly = TRUE, warn.conflicts = FALSE)
library(ggthemes, quietly = TRUE, warn.conflicts = FALSE)
library(knitr, quietly = TRUE, warn.conflicts = FALSE)
library(kableExtra, quietly = TRUE, warn.conflicts = FALSE)
library(Rblpapi, quietly = TRUE, warn.conflicts = FALSE)
library(scales, quietly = TRUE, warn.conflicts = FALSE)
library(pander, quietly = TRUE, warn.conflicts = FALSE)
library(dplyr, quietly = TRUE, warn.conflicts = FALSE)
library(formattable, quietly = TRUE, warn.conflicts = FALSE)
library(grid, quietly = TRUE, warn.conflicts = FALSE)
library(gridExtra, quietly = TRUE, warn.conflicts = FALSE)
library(png, quietly = TRUE, warn.conflicts = FALSE)
library(extrafont, quietly = TRUE, warn.conflicts = FALSE)
library(tinytex, quietly = TRUE, warn.conflicts = FALSE)
library(stringr, quietly = TRUE, warn.conflicts = FALSE)
library(lubridate, quietly = TRUE, warn.conflicts = FALSE)
library(reshape2, quietly = TRUE, warn.conflicts = FALSE)
library(ggrepel, quietly = TRUE, warn.conflicts = FALSE)
library(mnormt, quietly = TRUE, warn.conflicts = FALSE)
library(Ecdat, quietly = TRUE, warn.conflicts = FALSE)
library(MASS, quietly = TRUE, warn.conflicts = FALSE)
library(copula, quietly = TRUE, warn.conflicts = FALSE)
library(fGarch, quietly = TRUE, warn.conflicts = FALSE)
library(forecast, quietly = TRUE, warn.conflicts = FALSE)
library(tseries, quietly = TRUE, warn.conflicts = FALSE)
library(gmodels, quietly = TRUE, warn.conflicts = FALSE)
library(rugarch, quietly = TRUE, warn.conflicts = FALSE)
library(quantmod, quietly = TRUE, warn.conflicts = FALSE)
library(gtools, quietly = TRUE, warn.conflicts = FALSE)
options(tinytex.verbose = TRUE)
suppressMessages(library("tidyverse"))
pretty_kable <- function(data, title, dig = 2) {
kable(data, caption = title, digits = dig) %>%
kable_styling(bootstrap_options = c("striped", "hover")) %>%
kableExtra::kable_styling(latex_options = "hold_position")
}
theme_set(theme_light())
# Theme Overrides
theme_update(axis.text.x = element_text(size = 10),
axis.text.y = element_text(size = 10),
plot.title = element_text(hjust = 0.5, size = 16, face = "bold", color = "darkgreen"),
axis.title = element_text(face = "bold", size = 12, colour = "steelblue4"),
plot.subtitle = element_text(face = "bold", size = 8, colour = "darkred"),
legend.title = element_text(size = 12, color = "darkred", face = "bold"),
legend.position = "right", legend.title.align=0.5,
panel.border = element_rect(linetype = "solid",
colour = "lightgray"),
plot.margin = unit(c( 0.1, 0.1, 0.1, 0.1), "inches"))
data.dir <- "D:/Projects/Statistical-Computing/datasets/"
setwd("D:/Projects/Statistical-Computing/RDS")
```
```{r pander_setup, include = FALSE}
knitr::opts_chunk$set(comment = NA)
panderOptions('table.alignment.default', function(df)
ifelse(sapply(df, is.numeric), 'right', 'left'))
panderOptions('table.split.table', Inf)
panderOptions('big.mark', ",")
panderOptions('keep.trailing.zeros', TRUE)
```
### 4.1
Consider the population {1, 2, 5, 6, 10, 12}.
Find (and plot) the sampling distribution of medians for samples of size 3 without replacement.
```{r, echo = T, fig.height=4.5, fig.width=8}
p <- c(1, 2, 5, 6, 10, 12)
c <- combinations(v = p, n = 6, r = 3)
t <- apply(c, 1, median)
ggplot(data.table(value = t), aes(value, fill = ..count..)) +
geom_histogram(bins = 30) +
labs(title = "Median Sampling Distribution of p")
```
Compare the median of the population to the mean of the medians.
Median of p = __`r median(p)`__. Mean of Medians of p = __`r mean(t)`__
\newpage
### 4.2
Consider the population {3, 6, 7, 9, 11, 14}.
For samples of size 3 without replacement, find (and plot) the sampling distribution for the minimum.
```{r, echo = T, fig.height=4.5, fig.width=8}
p <- c(3, 6, 7, 9, 11, 14)
c <- combinations(v = p, n = 6, r = 3)
t <- apply(c, 1, min)
ggplot(data.table(value = t), aes(value, fill = ..count..)) +
geom_histogram(bins = 30) +
labs(title = "Minimum Sampling Distribution of p")
```
What is the mean of the sampling distribution? __`r mean(t)`__
The statistic is an estimate of some parameter - what is the value of that parameter?
This is an estimation of the minimum, which is: __`r min(p)`__
\newpage
### 4.3
Let _A_ denote the population {1, 3, 4, 5} and _B_ the population {5, 7, 9}.
```{r, echo = T}
A <- c(1, 3, 4, 5)
B <- c(5, 7, 9)
```
Let _X_ be a random value from _A_, and _Y_ and random value from _B_.
a.) Find the sampling distribution of _X + Y_.
```{r, echo = T}
result = numeric(12)
index <- 1
for(j in 1:length(A))
{
for(k in 1:length(B))
{
result[index] <- A[j] + B[k]
index <- index + 1
}
}
sort(result)
```
b.) In this example, does the sampling distribution depend on whether you sample with or without replacement?
_No._
Why or why not?
_Because 5 in is both sets._
c.) Compute the mean of the values for each of _A_ and _B_ and the values in the sampling distribution of _X + Y_.
Mean of _A_: __`r mean(A)`__. Mean of _B_: __`r mean(B)`__.
Mean of _A + B_: __`r mean(result)`__
How are the means related?
mean(A) + mean(B) = mean(A + B).
d.) Suppose you draw a random value from _A_ and a random value from _B_.
```{r, echo = T}
prob <- sum(result >= 13) / length(result)
```
What is the probability that the sum is 13 or larger? __`r prob*100`%__
\newpage
### 4.4
Consider the population {3, 5, 6, 6, 8, 11, 13, 15, 19, 20}.
a.) Compute the mean and standard deviation and create a dot plot of its distribution.
```{r, echo = T, fig.height=3.5, fig.width=8}
p <- c(3, 5, 6, 6, 8, 11, 13, 15, 19, 20)
mu <- mean(p)
sigma <- sd(p)
ggplot(data.table(value = p)) +
geom_dotplot(aes(value, fill = ..count..), binwidth = 1) +
labs(title = "Population Dot Plot")
```
$\mu = `r mu`$, $\sigma = `r sigma`$
b.) Simulate the sampling distribution of $\bar{X}$ by taking random samples of size 4 and plot your results.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e2
results <- numeric(N)
for( i in 1:N)
{
index <- sample(length(p), size = 4, replace = F)
results[i] <- mean( p[index] )
}
ggplot(data.table(value = results)) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs(title = "Sample Means")
xbar <- mean(results)
se <- sd(results) / sqrt(N)
```
Compute the mean and standard error, and compare to the population mean and standard deviation.
mean: `r xbar`, standard error: `r se`
c.) Use the simulation to find $P(\bar{X} < 11)$.
```{r, echo = T}
prob <- mean(results < 11)
```
$P(\bar{X} < 11) = `r prob*100`$%
\newpage
### 4.5
Consider two populations A = {3, 5, 7, 9, 10, 16}, B = {8, 10, 11, 15, 18, 25, 28}.
```{r, echo = T}
A <- c(3, 5, 7, 9, 10, 16)
B <- c(8, 10, 11, 15, 18, 25, 28)
```
a.) Using R, draw random samples (without replacement) of size 3 from each population, and simulate the sampling distribution of the sum of their maximums.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e2
results <- numeric(N)
for(i in 1:N)
{
samp.a <- sample(A, 3, replace = F)
samp.b <- sample(B, 3, replace = F)
results[i] <- max(samp.a) + max(samp.b)
}
ggplot(data.table(value = results)[, index := .I]) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs(title = "Sampling Distribution: max(A) + max(B)")
```
b.) Use your simulation to estimate the probability that the sum of the maximums is less than 20.
```{r, echo = T}
prob <- mean(results < 20)
```
Probability: `r prob*100`%
c.) Draw random samples of size 3 from each population, and find the maximum of the union of these two sets.
Simulate the sampling distribution of the maximums of this union.
```{r, echo = T, fig.height=3.5, fig.width=8}
results <- numeric(N)
for(i in 1:N)
{
samp.a <- sample(A, 3, replace = F)
samp.b <- sample(B, 3, replace = F)
results[i] <- max(union(samp.a, samp.b))
}
ggplot(data.table(value = results)[, index := .I]) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs(title = "Sampling Distribution: max(union(A,B))")
```
d.) Use simulation to find the probability that the maximum of the union is less than 20.
```{r, echo = T}
prob <- mean(results < 20)
```
Probability: `r prob*100`%
\newpage
### 4.6
The data set _Recidivism_ contains the poopulation of all Iowa offenders convicted of either a felony or misdemeanor who were released in 2010 (case study in Section 1.4).
```{r, echo = T}
Recidivism <- data.table(read.csv(paste0(data.dir, "Recidivism.csv"),
header = T))
```
Of these, 31.6% recidivated and were sent back to prision.
Simulate the sampling distribution of $\hat{p}$, the sample proportion of offeneders who recidivated, for random samples of size 25.
```{r, echo = T}
mean(Recidivism$Recid == "Yes")
N <- 10e2
results <- numeric(N)
for(i in 1:N)
{
samp <- sample(Recidivism$Recid, 25)
results[i] <- mean(samp == "Yes")
}
```
a.) Create a histogram and describe the simulated sampling distribution of $\hat{p}$.
```{r, echo = T, fig.height=3.5, fig.width=8}
ggplot(data.table(value = results)) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs(title = "Recidivism Sampling Distribution")
```
Estimate the mean and standard error.
```{r, echo = T}
mu <- mean(results)
se <- sd(results) / sqrt(25)
```
$\mu = `r mu`$, $\sigma = `r se`$
b.) Compare your estimate of the standard error with the theoretical standard error (_Corollary 4.3.2_).
```{r, echo = T}
tse <- mu * ( 1 - mu ) / sqrt(25)
```
Theoretical: `r tse`
c.) Repeat the above using samples of size 250, and compare with the $n = 25$ case.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e2
results <- numeric(N)
for(i in 1:N)
{
samp <- sample(Recidivism$Recid, 250)
results[i] <- mean(samp == "Yes")
}
ggplot(data.table(value = results)) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs(title = "Recidivism Sampling Distribution")
mu <- mean(results)
se <- sd(results) / sqrt(250)
```
$\mu = `r mu`$, $\sigma = `r se`$
### 4.7
The data set _FlightDelays_ contains the population of all flight departures by United Airlines and American Airlines out of LGA during May and June 2009 (case study in Section 1.1).
```{r, echo = T}
Flights <- data.table(read.csv(paste0(data.dir, "FlightDelays.csv"),
header = T))
```
a.) Create a histogram of _Delay_ and describe the distribution.
```{r, echo = T, fig.height=3.5, fig.width=8}
ggplot(Flights, aes(Delay)) +
geom_histogram(aes(fill = ..count..), bins = 30) +
scale_y_continuous(labels = comma) +
labs(title = "Flight Delays")
```
Compute the mean and standard deviation.
```{r, echo = T}
mu <- mean(Flights$Delay)
sigma <- sd(Flights$Delay)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
b.) Simulate the sampling distribution of $\bar{x}$, the sample mean of the length of the flight delays (_Delay_), for sample size 25.
```{r, echo = T}
N <- 10e2
results <- numeric(N)
for(i in 1:N)
{
samp <- sample(Flights$Delay, 25, replace = F)
results[i] <- mean(samp)
}
```
Create a histogram and describe the simulated sampling distribution of $\bar{x}$.
```{r, echo = T, fig.height=3.5, fig.width=8}
ggplot(data.table(value = results)) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs(title = "Flight Delay Sampling Distribution")
```
Estimate the mean and standard error.
```{r, echo = T}
mu <- mean(results)
se <- sd(results) / sqrt(25)
```
$\mu = `r mu`$, $\Sigma = `r se`$
c.) Compare your estimate of the standard error with the theoretical standard error (_Corollary A.4.1_).
```{r, echo = T}
tse <- var(results) / 25
```
Theoretical: `r tse`
d.) Repeat with sample size 250.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e2
results <- numeric(N)
for(i in 1:N)
{
samp <- sample(Flights$Delay, 250, replace = F)
results[i] <- mean(samp)
}
ggplot(data.table(value = results)) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs(title = "Flight Delay Sampling Distribution")
mu <- mean(results)
se <- sd(results) / sqrt(250)
tse <- var(results) / 250
```
$\mu = `r mu`$, $\Sigma = `r se`$
Theoretical: `r tse`
### 4.8
Let $X_1, X_2, \ldots, X_{25}$ be a random sample from some distribution and $W = T(X_1, X_2, \ldots, X_n)$ be a statistic.
Suppose the _sampling distribution_ of W has a pdf given by $f(x) = \frac{2}{x^2}$, for 1 < x < 2.
Find $P(w < 1.5)$
__Solution__:
```{r, echo = T, fig.height=3.5, fig.width=8}
f <- function(x) 2 / x^2
x <- seq( from = 1.0001, to = 1.999, by = 0.0001)
y <- f(x)
ggplot(data.table(x, y)) +
geom_point(aes(x, y), color = "cornflowerblue", size = .6) +
labs(title = "pdf f(x) = 2/x^2")
a <- cumsum(y) / sum(y)
p <- round( a[x == 1.5], 4 ) * 100
d <- data.table(x, y = a)
ggplot(d) +
geom_point(aes(x, y), color = "cornflowerblue", size = .6) +
geom_area(aes(x, y), data = d[x < 1.5], fill = "cornflowerblue", alpha = .3) +
labs(title = paste("cdf f(x) = 2/x^2, A =", p ))
```
Numerical solution: `r p`%
Analytical Solution: ${\int_{1}^{1.5}}\frac{2}{x^2} = \frac{2}{3}$
### 4.9
Let $X_1, X_2, \ldots, X_{n}$ be a random sample from some distribution and $Y = T(X_1, X_2, \ldots, X_n)$ be a statistic.
Suppose the _sampling distribution_ of Y has pdf $f(y) = (3/8)y^2 \ for \ 0 \le y \le 2$.
Find $P(0 \le Y \le \frac{1}{5})$
__Solution__:
```{r, echo = T, fig.height=3.5, fig.width=8}
f <- function(x) (3/8)*x**2
x <- seq( from = 0, to = 2, by = 0.001)
y <- f(x)
ggplot(data.table(x, y)) +
geom_point(aes(x, y), color = "cornflowerblue", size = .6) +
labs(title = paste("pdf: ", paste0(deparse(f), collapse = " ")))
a <- cumsum(y) / sum(y)
p <- round( a[x == 1/5], 4 ) * 100
d <- data.table(x, y = a)
ggplot(d) +
geom_point(aes(x, y), color = "cornflowerblue", size = .6) +
geom_area(aes(x, y), data = d[x < 1/5], fill = "cornflowerblue", alpha = .3) +
labs(title = paste("cdf f(x) = 2/x^2, A =", p ))
```
Numerical Solution: `r p`%
Analytical Solution: ${\int_{0}^{\frac{1}{5}}}\frac{x^3}{8} = \frac{.008}{8} = .001 = .1$ %
### 4.10
Suppose the heights of boys in a certain large city follow a distribution with mean 48 in. and variance $9^2$.
Use the CLT approximation to estimate the probability that in a random sample of 30 boys, the mean height is more than 51 in.
```{r, echo = T}
z <- (51 - 48) / (9^2 / sqrt(30))
p <- pnorm(z, lower.tail = F)
```
Probability: __`r round(p, 4)*100`%__
### 4.11
Let $X_1, X_2, \ldots, X_{36} \sim Bern(.55)$ be independent, and let $\hat{p}$ denote the sample proportion.
Use the CLT approximation with continuity correction to find the probability that $\hat{p} \le 0.5$.
```{r, echo = T}
z <- ( .5 - .55 ) / sqrt(.55 * (1 - .55) / 36)
p <- pnorm(z, lower.tail = T)
```
Probability: `r round(p, 4)*100`%
### 4.12
A random sample of size $n = 20$ is drawn from a distribution with mean 6 and variance 10.
Use the CLT approximation to estimate $P(\bar{X} \le 4.6)$.
```{r, echo = T}
z <- ( 4.6 - 6 ) / ( 10 * sqrt(20) )
p <- pnorm(z, lower.tail = T)
```
Probability: `r round(p, 4)*100`%
### 4.13
A random sample of size $n = 244$ is drawn from a distribution with pdf $f(x) = (3/16)(x - 4)^2, 2 \le x \le 6$.
Use the CLT approximation to estimate $P(X \ge 4.2)$.
```{r, echo = T, fig.height=2.6, fig.width=8}
pdf <- function(x) (3/16)*(x - 4)^2
x <- seq(from = 2, to = 6, by = 0.001)
y <- pdf(x)
ggplot(data.table(x,y)) +
geom_point(aes(x, y), col = "cornflowerblue", lwd = .8) +
labs(title = paste("PDF: ", paste0(deparse(f), collapse = " ")))
cdf <- function(x) (3/8)*(x - 4)
y <- cumsum(y) / sum(y)
ev <- x[min(which(y > .5))]
ggplot(data.table(x,y)) +
geom_point(aes(x, y), col = "cornflowerblue", lwd = .8) +
geom_vline(xintercept = ev) +
labs(title = paste("CDF: ", paste0(deparse(f), collapse = " ")))
z <- ( 4.2 - ev ) / sqrt(244)
pnorm(z, lower.tail = F)
```
### 4.14
According to the 2000 census, 28.6% of the US adult population recieved a high school diploma.
In a random sample of 800 US adults, what is the probability that between 220 and 230 (inclusive) people have a high school deploma?
Use the CLT approximation with continuity correction, and compare with the exact probability.
__Solution__:
The sampling distribution of $\hat{p}$ is approximately normal with:
```{r, echo = T}
n <- 800
mu <- .286
ev <- 800 * mu
sigma <- sqrt(n*mu*(1-mu))
```
$\mathbb{E}[X] = `r ev`$ and $\sigma = \sqrt{800(.286)(1 - .286)} = `r round(sigma, 4)`$
```{r, echo = T}
l <- pnorm((ev - 219.5) / sigma)
h <- pnorm((ev - 230.5) / sigma)
p <- l - h
```
Probability: `r round(p, 4)`
### 4.15
If $X_1, \ldots, X_n$ are i.i.d. from Unif[0, 1], how large should n be so that $P(\bar{X} - \frac{1}{2} < 0.05) \ge 0.90$,
that is, is there at least a 90% chance that the sample mean is within 0.05 of $\frac{1}{2}$? Use the CLT approximation.
```{r, echo = T}
```
### 4.16
Maria claims that she has drawn a random sample of size 30 from the exponential distribution with $\lambda = 1/10$.
The mean of her sample is 12.
a.) What is the expected value of a sample mean?
$X \sim Exp(\frac{1}{10}), \mathbb{E}(x) = 10$
b.) Run a simulation by drawing 1000 random samples, each of size 30, from Exp(1/10), and compute the mean for each sample.
```{r, echo = T, fig.height=3.5, fig.width=12}
N <- 1000
result <- numeric(N)
for( i in 1:N)
{
samp <- rexp( n = 30, rate = 1/10)
result[i] <- mean(samp)
}
ggplot(data.table(result), aes(result)) +
geom_histogram(aes(y = ..density.., fill = ..count..), bins = 30) +
geom_vline(xintercept = 12, col = "darkorange") +
stat_density( kernel = "gaussian", fill = "darkorange", alpha = .3) +
labs(title = "Exp(1/10) Sampling Distribution")
p <- mean(result > 12)
```
What proportion of the sample means is as large or larger than 12? __`r p*100`%__
c.) Is a mean of 12 unusual for a sample of size 30 from Exp(1/10)?
__Yes__, only ~13% of the sample means have a value of 12 or higher.
### 4.17
Let $X \sim N(15, 3^2)$ and $Y \sim N(4, 2^2)$ be independent random variables.
a.) What is the exact sampling distribution of $W = X - 2Y$?
$W \sim N(7, 5^2)$
b.) Use R to simulate the sampling distribution of $W$ and plot your results.
```{r, echo = T, fig.height=3.5, fig.width=8}
X <- rnorm(10e3, 15, 3)
Y <- rnorm(10e3, 4, 2)
W <- X - 2*Y
ggplot(data.table(x = W)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
geom_vline(xintercept = 7, col = "darkorange", lwd = 1.5) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
Check that the simulated mean and standard error are close to the theoretical mean and standard error.
```{r, echo = T}
mu <- mean(W)
sigma <- sd(W)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
c.) Use the simulated sampling to estimate $P(W \le 10)$, and then check your estimate with an exact calculation.
```{r, echo = T}
phat <- mean(W <= 10)
p <- pnorm(10, mean = 7, sd = 5)
```
$\hat{p}= `r phat*100`$%
$P(W \le 10) = `r round(p, 4)*100`$%
### 4.18
Let $X \sim Pois(4)$, $Y \sim Pois(12)$, $U \sim Pois(3)$ be independent random variables.
a.) What is the exact sampling distribution of $W = X + Y + U$?
$W \sim Pois(19)$
b.) Use R to simulate the sampling distribution of $W$ and plot your results.
```{r, echo = T, fig.height=3.5, fig.width=8}
W <- rpois(10e3, lambda = 19)
ggplot(data.table(x = W)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
geom_vline(xintercept = 19, col = "darkorange", lwd = 1.5) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
Check that the simulated mean and standard error are close to the theoretical mean and standard error.
```{r, echo = T}
mu <- mean(W)
sigma <- sd(W)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
c.) Use the simulated sampling distribution to estimate $P(W \le 14)$ and then check your estimate with an exact calculation.
```{r, echo = T}
phat <- mean(W <= 14)
p <- ppois(14, lambda = 19)
```
$\hat{p} = `r phat*100`$%
$P(W \le 14) = `r round(p, 4)*100`$%
### 4.19
Let $X_1, X_2, \ldots, X_{10} \sim^{i.i.d} N(20, 8^2)$ and $Y_1, Y_2, \ldots, Y_{15} \sim^{i.i.d} N(16, 7^2)$.
Let $W = \bar{X} + \bar{Y}$
a.) Give the exact sampling distribution of W.
$\sigma = (10 + 15) / sqrt( 10 + 15 - 1) = 3.1$
$W \sim N(36, 3.1^2)$
b.) Simulate the sampling distribution in R and plot your results.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e3
result <- numeric(N)
for( i in 1:N)
{
X <- rnorm(10, 20, 8)
Y <- rnorm(15, 16, 7)
result[i] <- mean(X) + mean(Y)
}
ggplot(data.table(x = result)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
geom_vline(xintercept = 36, col = "darkorange", lwd = 1.5) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
Check that the simulated mean and standard error are close to the exact mean and standard error.
```{r, echo = T}
mu <- mean(result)
sigma <- sd(result)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
c.) Use your simulation to find $P(W < 40)$. Calculate an exact answer and compare.
```{r, echo = T}
phat <- mean(result <= 40)
p <- pnorm(40, 36, 3)
```
$\hat{p} = `r phat*100`$%
$P(W < 40) = `r round(p, 4)*100`$%
### 4.20
Let $X_1, X_2, \ldots, X_9 \sim^{i.i.d.} N(7, 3^2)$, and $Y_1, Y_2, \ldots, X_{12} \sim^{i.i.d.} N(10, 5^2)$.
Let $W = \bar{X} - \bar{Y}$.
a.) Give the sampling distribution of $W$.
$\sigma = (3 + 5) / sqrt( 9 + 12 - 1 ) = 1.79$
$W = N(-3, 1.79^2)$
b.) Simulate the sampling distribution of W in R, and plot your results.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e3
result <- numeric(N)
for(i in 1:N)
{
X <- rnorm(9, 7, 3)
Y <- rnorm(12, 10, 5)
result[i] <- mean(X) - mean(Y)
}
ggplot(data.table(x = result)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
geom_vline(xintercept = -3, col = "darkorange", lwd = 1.5) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
Check that the simulated mean and standard error are close to the theoretical mean and standard error.
```{r, echo = T}
mu <- mean(result)
sigma <- sd(result)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
c.) Use your simulation to find $P(W < -1.5)$.
```{r, echo = T}
phat <- mean(result <= -1.5)
p <- pnorm(-1.5, -3, 1.79)
```
$\hat{p} = `r phat*100`$%
Calculate an exact answer and compare.
$P(W < 1.5) = `r round(p, 4)*100`$%
### 4.21
Let $X_1, X_2, \ldots, X_N$ be a random sample from $N(0, 1)$. Let $W = X^2_1 + X^2_2 + \ldots + X^2_n$
What is the mean and variance of the sampling distribution of W?
$\mu = 0$, $\sigma = 1$
Repeat using N = 4, N = 5.
$N = 4, \sigma = 4 / sqrt(4 - 1) = 2.3$
$N = 5, \sigma = 5 / sqrt(5 - 1) = 2.5$
What observations or conjectures do you have for general __n__?
### 4.22
Let $X$ be a uniform random variable on the interval $[40, 60]$ and Y be a uniform random variable on $[45, 80]$.
Assume that X and Y are independent.
a.) Compute the expected value and variance of $X + Y$.
$\mu = 112.5$, $Var = 1/24*(140 - 85)^2 = 126.04$
b.) Simulate a sampling distribution of $X + Y$.
```{r, echo = T, fig.height=3.5, fig.width=8}
X <- runif(1000, 40, 60)
Y <- runif(1000, 45, 80)
total <- X + Y
ggplot(data.table(value = total)) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs("X + Y Sampling Distribution")
```
Describe the sampling distribution of $X + Y$. __Approximately Normal__
Compute the mean and variance of the sampling distribution and compare this with the theoretical mean and variance.
```{r, echo = T}
mu <- mean(total)
var <- var(total)
```
$\mu = `r mu`$, $Var = `r var`$
c.) Suppose the time (in minutes) Andy takes to complete his statistics homework is $Unif[40, 60]$ and the time Adam takes is $Unif[45, 80]$.
Assume they work independently.
One day they announce that their total time to finish an assignment was less than 90 minutes.
How likely is this?
```{r, echo = T, fig.height=3.5, fig.width=8}
p <- punif(90, 85, 140)
```
Probability: __`r round(p, 4)*100`%__
### 4.23
Let $X_1, X_2, \ldots, X_{20} \sim^{i.i.d.} Exp(2)$. Let $X = \sum^{20}_{i=1}X_i$.
a.) Simulate the sampling distribution of $X$ in R.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e3
result <- numeric(N)
for(i in 1:N)
{
samp <- rexp(n = 20, rate = 2)
result[i] <- sum(samp)
}
```
b.) From your simulation, find $\mathbb{E}[X]$ and $Var[X]$.
```{r, echo = T}
mu <- mean(result)
var <- var(result)
```
$\mu = `r mu`$, $Var = `r var`$
c.) From your simulation, find $P( X \le 10)$.
```{r, echo = T}
phat <- mean(result <= 10)
```
Probablity: _`r round(phat, 4)*100`%_
### 4.24
Let $X_1, X_2, \ldots, X_{30} \sim^{i.i.d.} Exp(1/3)$ and let $\bar{X}$ denote the sample mean.
a.) Simulate the sampling distribution of $\bar{X}$ in R.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e3
result <- numeric(N)
for(i in 1:N)
{
samp <- rexp(n = 30, rate = 1/3)
result[i] <- mean(samp)
}
```
b.) Find the mean and standard error of the sampling distribution, and compare with the theoretical results.
```{r, echo = T}
mu <- mean(result)
sigma <- sd(result)
```
Sample: $\mu = `r mu`$, $\sigma = `r sigma`$
Theoretical: $\mu = \frac{\lambda}{1} = 3$, $sigma = 3/sqrt(30 - 1) = .56$
c.) From your simulation, find $P(\bar{X} \le 3.5)$.
```{r, echo = T}
phat <- mean(result <= 3.5)
```
$\hat{p} = `r phat`$
d.) Estimate $P(\bar{X} \le 3.5)$ by assuming the CLT approximation holds.
Compare this result with the one in part __(c)__.
```{r, echo = T}
z <- (3.5 - 3) / sigma
p <- pnorm(z)
```
$p = `r round(p, 4)`$
### 4.25
Consider the exponential distribution with density $f(x) \frac{1}{20}e^{-x/20}$, with mean and standard deviation of 20.
a.) Calculate the median of this distribution.
$0.5 = \int_{m}^{\inf}f(x)dx = -e^{-x/20}$
$\ldots = M = 20*log(2)$
$\ldots = `r 20*log(2)`$
b.) Using R, draw a random sample of size 50 and graph the histogram.
```{r, echo = T, fig.height=3.5, fig.width=8}
samp <- rexp( n = 50, rate = 1/20)
ggplot(data.table(x = samp)) +
geom_histogram(aes(x, fill = ..count..)) +
labs(title = "RExp(50, 1/20)")
```
What are the mean and standard deviation of your sample?
```{r, echo = T}
mu <- mean(samp)
sigma <- sd(samp)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
c.) Run a simulation to find the (approximate) sampling distribution for the median of sample size 50 from the exponential distribution and describe it.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e3
result <- numeric(N)
for(i in 1:N)
{
samp <- rexp( n = 50, rate = 1/20)
result[i] <- median(samp)
}
ggplot(data.table(x = result)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
What is the mean and the standard error of this sampling distribution?
```{r, echo = T}
mu <- mean(result)
sigma <- sd(result) / sqrt(50)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
d.) Repeat the above but use sample sizes $n = 100, 500 \ and \ 1,000$.
```{r, echo = T, fig.height=3, fig.width=8}
N <- 10e3
result <- numeric(N)
for(i in 1:N)
{
samp <- rexp( n = 100, rate = 1/20)
result[i] <- median(samp)
}
ggplot(data.table(x = result)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
```{r, echo = T}
mu <- mean(result)
sigma <- sd(result) / sqrt(50)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
```{r, echo = T, fig.height=3, fig.width=8}
N <- 10e3
result <- numeric(N)
for(i in 1:N)
{
samp <- rexp( n = 500, rate = 1/20)
result[i] <- median(samp)
}
ggplot(data.table(x = result)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
```{r, echo = T}
mu <- mean(result)
sigma <- sd(result) / sqrt(50)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
```{r, echo = T, fig.height=3, fig.width=8}
N <- 10e3
result <- numeric(N)
for(i in 1:N)
{
samp <- rexp( n = 1000, rate = 1/20)
result[i] <- median(samp)
}
ggplot(data.table(x = result)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
```{r, echo = T}
mu <- mean(result)
sigma <- sd(result) / sqrt(50)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
How does sample size affect the sampling distribution?
_The sample mean and standard error converge to the analytical solution with increased sample size._
### 4.26
Prove theorem 4.2.1.
### 4.27
Let $X_1, X_2 \sim^{i.i.d.} F$ with corresponding pdf $f(x) = \frac{2}{x^2}$, $1 \le x \le 2$.
a.) Find the pdf of $X_{max}$.
$F_{max}(x) = 8(\frac{1}{x^2} - \frac{1}{x^3})$
b.) Find the expected value of $X_{max}$.
$\mathbb{E}{[F_{max}]} = 1.545$
### 4.28
Let $X_1, X_2, \ldots, X_N \sim^{i.i.d.}$ with corresponding pdf $f(x) = 3x^2, 0 \le x \le 1$.
a.) Find the pdf for $X_{min}$.
b.) Find the pdf for $X_{max}$.
c.) If $n = 10$, find the probability that the largest value, $X_{max}$, is greater than 0.92.
### 4.29
Compute the pdf of the sampling distribution of the maximum samples of size 10 from a population with an exponential distribtuion with $\lambda = 12$.
### 4.30
Let $X_1, X_2, \ldots, X_N \sim^{i.i.d.} Exp(\lambda)$ with pdf $f(x) = \lambda e^{-\lambda x}, \lambda > 0, x > 0$.
a.) Find the pdf $f_{min}(x)$ for the sample minimum $X_{min}$. Recognize this as the pdf of a known distribution.
b.) Simulate in R the sampling distribution of $X_{min}$ of samples of size $n = 25$ from the exponential distribution with $\lambda = 7$.
Compare the theoretical expected value of $X_{min}$ with the simulated expected value.
### 4.31
Let $X_1, X_2, \ldots, X_n \sim^{i.i.d.} Pois(3)$. Let $X = \sum^{10}_{i=1}X_i$.
Find the pdf for the sampling distribution of X.
### 4.32
Let $X_1$ and $X_2$ be independent exponential random variables, both with parameter $\lambda > 0$.
Find the cumulative distribution function for the sampling distribution of $X = X_1 + X_2$.
### 4.33
This simulation illustrates the CLT for a finite population.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 400
n <- 5
finpop <- rexp(N, 1/10)
ggplot(data.table(x = finpop)) +
geom_histogram(aes(x, fill = ..count..)) +
labs(title = "Exp(1/10)")
mean(finpop) #mean (mu) of your pop.
sd(finpop) # stdev (sigma) of your pop.
sd(finpop)/sqrt(n) # theoretical standard error of sampling distribution
sd(finpop)/sqrt(n) * sqrt((N-n)/(N-1)) # without replacement
Xbar <- numeric(1000)
for(i in 1:1000)
{
x <- sample(finpop, n) # Random sample of size n
# (w/o replacement)
Xbar[i] <- mean(x)
}
p1 <- ggplot(data.table(x = Xbar)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
labs(title = "Mean Sampling Distribution Histogram")
p2 <- ggplot(data.table(x = Xbar), aes(sample = x)) +
stat_qq() +
stat_qq_line() +
labs(title = "Mean Sampling Distribution QQ-Plot")
grid.arrange(p1, p2, nrow = 2)
mean(Xbar)
sd(Xbar) # estimated standard error of sampling distribution
```
a.) Does the sampling distribution of sample means appear approximately normal?
b.) Compare the mean and standard error of your simulated sampling distribution with the theoretical ones.
c.) Calculate $(\sigma/\sqrt(n))(\sqrt{(N-n)/(N-1)}$, where $\sigma$ is the standard deviation of the finite population and compare with the (estimated) standard error of the sampling distribution.
d.) Repeat for larger __n__, say __n__ = 20 and __n__ = 100.
### 4.34
Let $X_1, X_2, \ldots, X_n$ be independent random variables from $N(\mu, \sigma)$.
We are interested in the sampling distribution of the variance.
Run a simulation to draw random samples of size 20 from $N(25, 7^2)$ and calculate the variance for each sample.
```{r, echo = T, fig.height=3.5, fig.width=8}
W <- numeric(1000)
for(i in 1:1000)
{
x <- rnorm(20, 25, 7)
W[i] <- var(x)
}
mean(W)
var(W)
p1 <- ggplot(data.table(x = W)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
labs(title = "Variance Sampling Distribution")
p2 <-
ggplot(data.table(value = W), aes(sample = value)) +
stat_qq() +
stat_qq_line() +
labs(title = "Mean Sampling Distribution QQ-Plot")
grid.arrange(p1, p2, nrow = 2)
```
Does the sampling distribution appear to be normally distributed?
Repeat with n = 50 and n = 200.
### 4.35
A random sample of size $n = 100$ is drawn from a distribution with pdf $f(x) = 3(1- x)^2, 0 \le x \le 1$.
a.) Use the CLT approximation to estimate $P(\bar{X} \le 0.27)$.
b.) Use the expanded CLT to estimate the same probability (dnorm).
c.) If $X_1, X_2, X_3 \sim^{i.i.d.} Unif[0, 1]$, then the minimum has density __f__ given above.
Use simulation to estimate the probability. | 5325 | ---
title: ''
mainfont: Arial
fontsize: 12pt
documentclass: report
header-includes:
- \PassOptionsToPackage{table}{xcolor}
- \usepackage{caption}
- \usepackage{amssymb}
- \usepackage{booktabs}
- \usepackage{longtable}
- \usepackage{array}
- \usepackage{multirow}
- \usepackage{wrapfig}
- \usepackage{float}
- \usepackage{colortbl}
- \usepackage{pdflscape}
- \usepackage{tabu}
- \usepackage{threeparttable}
- \usepackage{threeparttablex}
- \usepackage[normalem]{ulem}
- \usepackage{makecell}
- \usepackage[table]{xcolor}
- \usepackage{fancyhdr}
- \usepackage{boldline}
- \usepackage{tipa}
\definecolor{headergrey}{HTML}{545454}
\definecolor{msdblue}{HTML}{1C93D1}
\pagestyle{fancy}
\setlength\headheight{30pt}
\rhead{\color{headergrey}\today}
\fancyhead[L]{\color{headergrey}<NAME>, <NAME>}
\fancyhead[C]{\Large\bfseries\color{headergrey}Sampling Distributions}
\rfoot{\color{headergrey}\thepage}
\lfoot{\color{headergrey}Chapter 4}
\fancyfoot[C]{\rmfamily\color{headergrey}Mathematical Statistics}
geometry: left = 1cm, right = 1cm, top = 2cm, bottom = 3cm
date: "`r format(Sys.time(), '%d %B, %Y')`"
output:
pdf_document:
fig_caption: yes
latex_engine: xelatex
editor_options:
chunk_output_type: console
---
```{r knitr_setup, include = FALSE}
# DO NOT ADD OR REVISE CODE HERE
knitr::opts_chunk$set(echo = FALSE, eval = TRUE, dev = 'png')
options(knitr.table.format = "latex")
```
```{r report_setup, message = FALSE, warning = FALSE, include = FALSE}
library(data.table, quietly = TRUE, warn.conflicts = FALSE)
assignInNamespace("cedta.pkgEvalsUserCode", c(data.table:::cedta.pkgEvalsUserCode, "rtvs"), "data.table")
library(ggplot2, quietly = TRUE, warn.conflicts = FALSE)
library(ggrepel, quietly = TRUE, warn.conflicts = FALSE)
library(ggthemes, quietly = TRUE, warn.conflicts = FALSE)
library(knitr, quietly = TRUE, warn.conflicts = FALSE)
library(kableExtra, quietly = TRUE, warn.conflicts = FALSE)
library(Rblpapi, quietly = TRUE, warn.conflicts = FALSE)
library(scales, quietly = TRUE, warn.conflicts = FALSE)
library(pander, quietly = TRUE, warn.conflicts = FALSE)
library(dplyr, quietly = TRUE, warn.conflicts = FALSE)
library(formattable, quietly = TRUE, warn.conflicts = FALSE)
library(grid, quietly = TRUE, warn.conflicts = FALSE)
library(gridExtra, quietly = TRUE, warn.conflicts = FALSE)
library(png, quietly = TRUE, warn.conflicts = FALSE)
library(extrafont, quietly = TRUE, warn.conflicts = FALSE)
library(tinytex, quietly = TRUE, warn.conflicts = FALSE)
library(stringr, quietly = TRUE, warn.conflicts = FALSE)
library(lubridate, quietly = TRUE, warn.conflicts = FALSE)
library(reshape2, quietly = TRUE, warn.conflicts = FALSE)
library(ggrepel, quietly = TRUE, warn.conflicts = FALSE)
library(mnormt, quietly = TRUE, warn.conflicts = FALSE)
library(Ecdat, quietly = TRUE, warn.conflicts = FALSE)
library(MASS, quietly = TRUE, warn.conflicts = FALSE)
library(copula, quietly = TRUE, warn.conflicts = FALSE)
library(fGarch, quietly = TRUE, warn.conflicts = FALSE)
library(forecast, quietly = TRUE, warn.conflicts = FALSE)
library(tseries, quietly = TRUE, warn.conflicts = FALSE)
library(gmodels, quietly = TRUE, warn.conflicts = FALSE)
library(rugarch, quietly = TRUE, warn.conflicts = FALSE)
library(quantmod, quietly = TRUE, warn.conflicts = FALSE)
library(gtools, quietly = TRUE, warn.conflicts = FALSE)
options(tinytex.verbose = TRUE)
suppressMessages(library("tidyverse"))
pretty_kable <- function(data, title, dig = 2) {
kable(data, caption = title, digits = dig) %>%
kable_styling(bootstrap_options = c("striped", "hover")) %>%
kableExtra::kable_styling(latex_options = "hold_position")
}
theme_set(theme_light())
# Theme Overrides
theme_update(axis.text.x = element_text(size = 10),
axis.text.y = element_text(size = 10),
plot.title = element_text(hjust = 0.5, size = 16, face = "bold", color = "darkgreen"),
axis.title = element_text(face = "bold", size = 12, colour = "steelblue4"),
plot.subtitle = element_text(face = "bold", size = 8, colour = "darkred"),
legend.title = element_text(size = 12, color = "darkred", face = "bold"),
legend.position = "right", legend.title.align=0.5,
panel.border = element_rect(linetype = "solid",
colour = "lightgray"),
plot.margin = unit(c( 0.1, 0.1, 0.1, 0.1), "inches"))
data.dir <- "D:/Projects/Statistical-Computing/datasets/"
setwd("D:/Projects/Statistical-Computing/RDS")
```
```{r pander_setup, include = FALSE}
knitr::opts_chunk$set(comment = NA)
panderOptions('table.alignment.default', function(df)
ifelse(sapply(df, is.numeric), 'right', 'left'))
panderOptions('table.split.table', Inf)
panderOptions('big.mark', ",")
panderOptions('keep.trailing.zeros', TRUE)
```
### 4.1
Consider the population {1, 2, 5, 6, 10, 12}.
Find (and plot) the sampling distribution of medians for samples of size 3 without replacement.
```{r, echo = T, fig.height=4.5, fig.width=8}
p <- c(1, 2, 5, 6, 10, 12)
c <- combinations(v = p, n = 6, r = 3)
t <- apply(c, 1, median)
ggplot(data.table(value = t), aes(value, fill = ..count..)) +
geom_histogram(bins = 30) +
labs(title = "Median Sampling Distribution of p")
```
Compare the median of the population to the mean of the medians.
Median of p = __`r median(p)`__. Mean of Medians of p = __`r mean(t)`__
\newpage
### 4.2
Consider the population {3, 6, 7, 9, 11, 14}.
For samples of size 3 without replacement, find (and plot) the sampling distribution for the minimum.
```{r, echo = T, fig.height=4.5, fig.width=8}
p <- c(3, 6, 7, 9, 11, 14)
c <- combinations(v = p, n = 6, r = 3)
t <- apply(c, 1, min)
ggplot(data.table(value = t), aes(value, fill = ..count..)) +
geom_histogram(bins = 30) +
labs(title = "Minimum Sampling Distribution of p")
```
What is the mean of the sampling distribution? __`r mean(t)`__
The statistic is an estimate of some parameter - what is the value of that parameter?
This is an estimation of the minimum, which is: __`r min(p)`__
\newpage
### 4.3
Let _A_ denote the population {1, 3, 4, 5} and _B_ the population {5, 7, 9}.
```{r, echo = T}
A <- c(1, 3, 4, 5)
B <- c(5, 7, 9)
```
Let _X_ be a random value from _A_, and _Y_ and random value from _B_.
a.) Find the sampling distribution of _X + Y_.
```{r, echo = T}
result = numeric(12)
index <- 1
for(j in 1:length(A))
{
for(k in 1:length(B))
{
result[index] <- A[j] + B[k]
index <- index + 1
}
}
sort(result)
```
b.) In this example, does the sampling distribution depend on whether you sample with or without replacement?
_No._
Why or why not?
_Because 5 in is both sets._
c.) Compute the mean of the values for each of _A_ and _B_ and the values in the sampling distribution of _X + Y_.
Mean of _A_: __`r mean(A)`__. Mean of _B_: __`r mean(B)`__.
Mean of _A + B_: __`r mean(result)`__
How are the means related?
mean(A) + mean(B) = mean(A + B).
d.) Suppose you draw a random value from _A_ and a random value from _B_.
```{r, echo = T}
prob <- sum(result >= 13) / length(result)
```
What is the probability that the sum is 13 or larger? __`r prob*100`%__
\newpage
### 4.4
Consider the population {3, 5, 6, 6, 8, 11, 13, 15, 19, 20}.
a.) Compute the mean and standard deviation and create a dot plot of its distribution.
```{r, echo = T, fig.height=3.5, fig.width=8}
p <- c(3, 5, 6, 6, 8, 11, 13, 15, 19, 20)
mu <- mean(p)
sigma <- sd(p)
ggplot(data.table(value = p)) +
geom_dotplot(aes(value, fill = ..count..), binwidth = 1) +
labs(title = "Population Dot Plot")
```
$\mu = `r mu`$, $\sigma = `r sigma`$
b.) Simulate the sampling distribution of $\bar{X}$ by taking random samples of size 4 and plot your results.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e2
results <- numeric(N)
for( i in 1:N)
{
index <- sample(length(p), size = 4, replace = F)
results[i] <- mean( p[index] )
}
ggplot(data.table(value = results)) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs(title = "Sample Means")
xbar <- mean(results)
se <- sd(results) / sqrt(N)
```
Compute the mean and standard error, and compare to the population mean and standard deviation.
mean: `r xbar`, standard error: `r se`
c.) Use the simulation to find $P(\bar{X} < 11)$.
```{r, echo = T}
prob <- mean(results < 11)
```
$P(\bar{X} < 11) = `r prob*100`$%
\newpage
### 4.5
Consider two populations A = {3, 5, 7, 9, 10, 16}, B = {8, 10, 11, 15, 18, 25, 28}.
```{r, echo = T}
A <- c(3, 5, 7, 9, 10, 16)
B <- c(8, 10, 11, 15, 18, 25, 28)
```
a.) Using R, draw random samples (without replacement) of size 3 from each population, and simulate the sampling distribution of the sum of their maximums.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e2
results <- numeric(N)
for(i in 1:N)
{
samp.a <- sample(A, 3, replace = F)
samp.b <- sample(B, 3, replace = F)
results[i] <- max(samp.a) + max(samp.b)
}
ggplot(data.table(value = results)[, index := .I]) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs(title = "Sampling Distribution: max(A) + max(B)")
```
b.) Use your simulation to estimate the probability that the sum of the maximums is less than 20.
```{r, echo = T}
prob <- mean(results < 20)
```
Probability: `r prob*100`%
c.) Draw random samples of size 3 from each population, and find the maximum of the union of these two sets.
Simulate the sampling distribution of the maximums of this union.
```{r, echo = T, fig.height=3.5, fig.width=8}
results <- numeric(N)
for(i in 1:N)
{
samp.a <- sample(A, 3, replace = F)
samp.b <- sample(B, 3, replace = F)
results[i] <- max(union(samp.a, samp.b))
}
ggplot(data.table(value = results)[, index := .I]) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs(title = "Sampling Distribution: max(union(A,B))")
```
d.) Use simulation to find the probability that the maximum of the union is less than 20.
```{r, echo = T}
prob <- mean(results < 20)
```
Probability: `r prob*100`%
\newpage
### 4.6
The data set _Recidivism_ contains the poopulation of all Iowa offenders convicted of either a felony or misdemeanor who were released in 2010 (case study in Section 1.4).
```{r, echo = T}
Recidivism <- data.table(read.csv(paste0(data.dir, "Recidivism.csv"),
header = T))
```
Of these, 31.6% recidivated and were sent back to prision.
Simulate the sampling distribution of $\hat{p}$, the sample proportion of offeneders who recidivated, for random samples of size 25.
```{r, echo = T}
mean(Recidivism$Recid == "Yes")
N <- 10e2
results <- numeric(N)
for(i in 1:N)
{
samp <- sample(Recidivism$Recid, 25)
results[i] <- mean(samp == "Yes")
}
```
a.) Create a histogram and describe the simulated sampling distribution of $\hat{p}$.
```{r, echo = T, fig.height=3.5, fig.width=8}
ggplot(data.table(value = results)) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs(title = "Recidivism Sampling Distribution")
```
Estimate the mean and standard error.
```{r, echo = T}
mu <- mean(results)
se <- sd(results) / sqrt(25)
```
$\mu = `r mu`$, $\sigma = `r se`$
b.) Compare your estimate of the standard error with the theoretical standard error (_Corollary 4.3.2_).
```{r, echo = T}
tse <- mu * ( 1 - mu ) / sqrt(25)
```
Theoretical: `r tse`
c.) Repeat the above using samples of size 250, and compare with the $n = 25$ case.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e2
results <- numeric(N)
for(i in 1:N)
{
samp <- sample(Recidivism$Recid, 250)
results[i] <- mean(samp == "Yes")
}
ggplot(data.table(value = results)) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs(title = "Recidivism Sampling Distribution")
mu <- mean(results)
se <- sd(results) / sqrt(250)
```
$\mu = `r mu`$, $\sigma = `r se`$
### 4.7
The data set _FlightDelays_ contains the population of all flight departures by United Airlines and American Airlines out of LGA during May and June 2009 (case study in Section 1.1).
```{r, echo = T}
Flights <- data.table(read.csv(paste0(data.dir, "FlightDelays.csv"),
header = T))
```
a.) Create a histogram of _Delay_ and describe the distribution.
```{r, echo = T, fig.height=3.5, fig.width=8}
ggplot(Flights, aes(Delay)) +
geom_histogram(aes(fill = ..count..), bins = 30) +
scale_y_continuous(labels = comma) +
labs(title = "Flight Delays")
```
Compute the mean and standard deviation.
```{r, echo = T}
mu <- mean(Flights$Delay)
sigma <- sd(Flights$Delay)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
b.) Simulate the sampling distribution of $\bar{x}$, the sample mean of the length of the flight delays (_Delay_), for sample size 25.
```{r, echo = T}
N <- 10e2
results <- numeric(N)
for(i in 1:N)
{
samp <- sample(Flights$Delay, 25, replace = F)
results[i] <- mean(samp)
}
```
Create a histogram and describe the simulated sampling distribution of $\bar{x}$.
```{r, echo = T, fig.height=3.5, fig.width=8}
ggplot(data.table(value = results)) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs(title = "Flight Delay Sampling Distribution")
```
Estimate the mean and standard error.
```{r, echo = T}
mu <- mean(results)
se <- sd(results) / sqrt(25)
```
$\mu = `r mu`$, $\Sigma = `r se`$
c.) Compare your estimate of the standard error with the theoretical standard error (_Corollary A.4.1_).
```{r, echo = T}
tse <- var(results) / 25
```
Theoretical: `r tse`
d.) Repeat with sample size 250.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e2
results <- numeric(N)
for(i in 1:N)
{
samp <- sample(Flights$Delay, 250, replace = F)
results[i] <- mean(samp)
}
ggplot(data.table(value = results)) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs(title = "Flight Delay Sampling Distribution")
mu <- mean(results)
se <- sd(results) / sqrt(250)
tse <- var(results) / 250
```
$\mu = `r mu`$, $\Sigma = `r se`$
Theoretical: `r tse`
### 4.8
Let $X_1, X_2, \ldots, X_{25}$ be a random sample from some distribution and $W = T(X_1, X_2, \ldots, X_n)$ be a statistic.
Suppose the _sampling distribution_ of W has a pdf given by $f(x) = \frac{2}{x^2}$, for 1 < x < 2.
Find $P(w < 1.5)$
__Solution__:
```{r, echo = T, fig.height=3.5, fig.width=8}
f <- function(x) 2 / x^2
x <- seq( from = 1.0001, to = 1.999, by = 0.0001)
y <- f(x)
ggplot(data.table(x, y)) +
geom_point(aes(x, y), color = "cornflowerblue", size = .6) +
labs(title = "pdf f(x) = 2/x^2")
a <- cumsum(y) / sum(y)
p <- round( a[x == 1.5], 4 ) * 100
d <- data.table(x, y = a)
ggplot(d) +
geom_point(aes(x, y), color = "cornflowerblue", size = .6) +
geom_area(aes(x, y), data = d[x < 1.5], fill = "cornflowerblue", alpha = .3) +
labs(title = paste("cdf f(x) = 2/x^2, A =", p ))
```
Numerical solution: `r p`%
Analytical Solution: ${\int_{1}^{1.5}}\frac{2}{x^2} = \frac{2}{3}$
### 4.9
Let $X_1, X_2, \ldots, X_{n}$ be a random sample from some distribution and $Y = T(X_1, X_2, \ldots, X_n)$ be a statistic.
Suppose the _sampling distribution_ of Y has pdf $f(y) = (3/8)y^2 \ for \ 0 \le y \le 2$.
Find $P(0 \le Y \le \frac{1}{5})$
__Solution__:
```{r, echo = T, fig.height=3.5, fig.width=8}
f <- function(x) (3/8)*x**2
x <- seq( from = 0, to = 2, by = 0.001)
y <- f(x)
ggplot(data.table(x, y)) +
geom_point(aes(x, y), color = "cornflowerblue", size = .6) +
labs(title = paste("pdf: ", paste0(deparse(f), collapse = " ")))
a <- cumsum(y) / sum(y)
p <- round( a[x == 1/5], 4 ) * 100
d <- data.table(x, y = a)
ggplot(d) +
geom_point(aes(x, y), color = "cornflowerblue", size = .6) +
geom_area(aes(x, y), data = d[x < 1/5], fill = "cornflowerblue", alpha = .3) +
labs(title = paste("cdf f(x) = 2/x^2, A =", p ))
```
Numerical Solution: `r p`%
Analytical Solution: ${\int_{0}^{\frac{1}{5}}}\frac{x^3}{8} = \frac{.008}{8} = .001 = .1$ %
### 4.10
Suppose the heights of boys in a certain large city follow a distribution with mean 48 in. and variance $9^2$.
Use the CLT approximation to estimate the probability that in a random sample of 30 boys, the mean height is more than 51 in.
```{r, echo = T}
z <- (51 - 48) / (9^2 / sqrt(30))
p <- pnorm(z, lower.tail = F)
```
Probability: __`r round(p, 4)*100`%__
### 4.11
Let $X_1, X_2, \ldots, X_{36} \sim Bern(.55)$ be independent, and let $\hat{p}$ denote the sample proportion.
Use the CLT approximation with continuity correction to find the probability that $\hat{p} \le 0.5$.
```{r, echo = T}
z <- ( .5 - .55 ) / sqrt(.55 * (1 - .55) / 36)
p <- pnorm(z, lower.tail = T)
```
Probability: `r round(p, 4)*100`%
### 4.12
A random sample of size $n = 20$ is drawn from a distribution with mean 6 and variance 10.
Use the CLT approximation to estimate $P(\bar{X} \le 4.6)$.
```{r, echo = T}
z <- ( 4.6 - 6 ) / ( 10 * sqrt(20) )
p <- pnorm(z, lower.tail = T)
```
Probability: `r round(p, 4)*100`%
### 4.13
A random sample of size $n = 244$ is drawn from a distribution with pdf $f(x) = (3/16)(x - 4)^2, 2 \le x \le 6$.
Use the CLT approximation to estimate $P(X \ge 4.2)$.
```{r, echo = T, fig.height=2.6, fig.width=8}
pdf <- function(x) (3/16)*(x - 4)^2
x <- seq(from = 2, to = 6, by = 0.001)
y <- pdf(x)
ggplot(data.table(x,y)) +
geom_point(aes(x, y), col = "cornflowerblue", lwd = .8) +
labs(title = paste("PDF: ", paste0(deparse(f), collapse = " ")))
cdf <- function(x) (3/8)*(x - 4)
y <- cumsum(y) / sum(y)
ev <- x[min(which(y > .5))]
ggplot(data.table(x,y)) +
geom_point(aes(x, y), col = "cornflowerblue", lwd = .8) +
geom_vline(xintercept = ev) +
labs(title = paste("CDF: ", paste0(deparse(f), collapse = " ")))
z <- ( 4.2 - ev ) / sqrt(244)
pnorm(z, lower.tail = F)
```
### 4.14
According to the 2000 census, 28.6% of the US adult population recieved a high school diploma.
In a random sample of 800 US adults, what is the probability that between 220 and 230 (inclusive) people have a high school deploma?
Use the CLT approximation with continuity correction, and compare with the exact probability.
__Solution__:
The sampling distribution of $\hat{p}$ is approximately normal with:
```{r, echo = T}
n <- 800
mu <- .286
ev <- 800 * mu
sigma <- sqrt(n*mu*(1-mu))
```
$\mathbb{E}[X] = `r ev`$ and $\sigma = \sqrt{800(.286)(1 - .286)} = `r round(sigma, 4)`$
```{r, echo = T}
l <- pnorm((ev - 219.5) / sigma)
h <- pnorm((ev - 230.5) / sigma)
p <- l - h
```
Probability: `r round(p, 4)`
### 4.15
If $X_1, \ldots, X_n$ are i.i.d. from Unif[0, 1], how large should n be so that $P(\bar{X} - \frac{1}{2} < 0.05) \ge 0.90$,
that is, is there at least a 90% chance that the sample mean is within 0.05 of $\frac{1}{2}$? Use the CLT approximation.
```{r, echo = T}
```
### 4.16
<NAME> claims that she has drawn a random sample of size 30 from the exponential distribution with $\lambda = 1/10$.
The mean of her sample is 12.
a.) What is the expected value of a sample mean?
$X \sim Exp(\frac{1}{10}), \mathbb{E}(x) = 10$
b.) Run a simulation by drawing 1000 random samples, each of size 30, from Exp(1/10), and compute the mean for each sample.
```{r, echo = T, fig.height=3.5, fig.width=12}
N <- 1000
result <- numeric(N)
for( i in 1:N)
{
samp <- rexp( n = 30, rate = 1/10)
result[i] <- mean(samp)
}
ggplot(data.table(result), aes(result)) +
geom_histogram(aes(y = ..density.., fill = ..count..), bins = 30) +
geom_vline(xintercept = 12, col = "darkorange") +
stat_density( kernel = "gaussian", fill = "darkorange", alpha = .3) +
labs(title = "Exp(1/10) Sampling Distribution")
p <- mean(result > 12)
```
What proportion of the sample means is as large or larger than 12? __`r p*100`%__
c.) Is a mean of 12 unusual for a sample of size 30 from Exp(1/10)?
__Yes__, only ~13% of the sample means have a value of 12 or higher.
### 4.17
Let $X \sim N(15, 3^2)$ and $Y \sim N(4, 2^2)$ be independent random variables.
a.) What is the exact sampling distribution of $W = X - 2Y$?
$W \sim N(7, 5^2)$
b.) Use R to simulate the sampling distribution of $W$ and plot your results.
```{r, echo = T, fig.height=3.5, fig.width=8}
X <- rnorm(10e3, 15, 3)
Y <- rnorm(10e3, 4, 2)
W <- X - 2*Y
ggplot(data.table(x = W)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
geom_vline(xintercept = 7, col = "darkorange", lwd = 1.5) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
Check that the simulated mean and standard error are close to the theoretical mean and standard error.
```{r, echo = T}
mu <- mean(W)
sigma <- sd(W)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
c.) Use the simulated sampling to estimate $P(W \le 10)$, and then check your estimate with an exact calculation.
```{r, echo = T}
phat <- mean(W <= 10)
p <- pnorm(10, mean = 7, sd = 5)
```
$\hat{p}= `r phat*100`$%
$P(W \le 10) = `r round(p, 4)*100`$%
### 4.18
Let $X \sim Pois(4)$, $Y \sim Pois(12)$, $U \sim Pois(3)$ be independent random variables.
a.) What is the exact sampling distribution of $W = X + Y + U$?
$W \sim Pois(19)$
b.) Use R to simulate the sampling distribution of $W$ and plot your results.
```{r, echo = T, fig.height=3.5, fig.width=8}
W <- rpois(10e3, lambda = 19)
ggplot(data.table(x = W)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
geom_vline(xintercept = 19, col = "darkorange", lwd = 1.5) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
Check that the simulated mean and standard error are close to the theoretical mean and standard error.
```{r, echo = T}
mu <- mean(W)
sigma <- sd(W)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
c.) Use the simulated sampling distribution to estimate $P(W \le 14)$ and then check your estimate with an exact calculation.
```{r, echo = T}
phat <- mean(W <= 14)
p <- ppois(14, lambda = 19)
```
$\hat{p} = `r phat*100`$%
$P(W \le 14) = `r round(p, 4)*100`$%
### 4.19
Let $X_1, X_2, \ldots, X_{10} \sim^{i.i.d} N(20, 8^2)$ and $Y_1, Y_2, \ldots, Y_{15} \sim^{i.i.d} N(16, 7^2)$.
Let $W = \bar{X} + \bar{Y}$
a.) Give the exact sampling distribution of W.
$\sigma = (10 + 15) / sqrt( 10 + 15 - 1) = 3.1$
$W \sim N(36, 3.1^2)$
b.) Simulate the sampling distribution in R and plot your results.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e3
result <- numeric(N)
for( i in 1:N)
{
X <- rnorm(10, 20, 8)
Y <- rnorm(15, 16, 7)
result[i] <- mean(X) + mean(Y)
}
ggplot(data.table(x = result)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
geom_vline(xintercept = 36, col = "darkorange", lwd = 1.5) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
Check that the simulated mean and standard error are close to the exact mean and standard error.
```{r, echo = T}
mu <- mean(result)
sigma <- sd(result)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
c.) Use your simulation to find $P(W < 40)$. Calculate an exact answer and compare.
```{r, echo = T}
phat <- mean(result <= 40)
p <- pnorm(40, 36, 3)
```
$\hat{p} = `r phat*100`$%
$P(W < 40) = `r round(p, 4)*100`$%
### 4.20
Let $X_1, X_2, \ldots, X_9 \sim^{i.i.d.} N(7, 3^2)$, and $Y_1, Y_2, \ldots, X_{12} \sim^{i.i.d.} N(10, 5^2)$.
Let $W = \bar{X} - \bar{Y}$.
a.) Give the sampling distribution of $W$.
$\sigma = (3 + 5) / sqrt( 9 + 12 - 1 ) = 1.79$
$W = N(-3, 1.79^2)$
b.) Simulate the sampling distribution of W in R, and plot your results.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e3
result <- numeric(N)
for(i in 1:N)
{
X <- rnorm(9, 7, 3)
Y <- rnorm(12, 10, 5)
result[i] <- mean(X) - mean(Y)
}
ggplot(data.table(x = result)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
geom_vline(xintercept = -3, col = "darkorange", lwd = 1.5) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
Check that the simulated mean and standard error are close to the theoretical mean and standard error.
```{r, echo = T}
mu <- mean(result)
sigma <- sd(result)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
c.) Use your simulation to find $P(W < -1.5)$.
```{r, echo = T}
phat <- mean(result <= -1.5)
p <- pnorm(-1.5, -3, 1.79)
```
$\hat{p} = `r phat*100`$%
Calculate an exact answer and compare.
$P(W < 1.5) = `r round(p, 4)*100`$%
### 4.21
Let $X_1, X_2, \ldots, X_N$ be a random sample from $N(0, 1)$. Let $W = X^2_1 + X^2_2 + \ldots + X^2_n$
What is the mean and variance of the sampling distribution of W?
$\mu = 0$, $\sigma = 1$
Repeat using N = 4, N = 5.
$N = 4, \sigma = 4 / sqrt(4 - 1) = 2.3$
$N = 5, \sigma = 5 / sqrt(5 - 1) = 2.5$
What observations or conjectures do you have for general __n__?
### 4.22
Let $X$ be a uniform random variable on the interval $[40, 60]$ and Y be a uniform random variable on $[45, 80]$.
Assume that X and Y are independent.
a.) Compute the expected value and variance of $X + Y$.
$\mu = 112.5$, $Var = 1/24*(140 - 85)^2 = 126.04$
b.) Simulate a sampling distribution of $X + Y$.
```{r, echo = T, fig.height=3.5, fig.width=8}
X <- runif(1000, 40, 60)
Y <- runif(1000, 45, 80)
total <- X + Y
ggplot(data.table(value = total)) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs("X + Y Sampling Distribution")
```
Describe the sampling distribution of $X + Y$. __Approximately Normal__
Compute the mean and variance of the sampling distribution and compare this with the theoretical mean and variance.
```{r, echo = T}
mu <- mean(total)
var <- var(total)
```
$\mu = `r mu`$, $Var = `r var`$
c.) Suppose the time (in minutes) <NAME> takes to complete his statistics homework is $Unif[40, 60]$ and the time <NAME> takes is $Unif[45, 80]$.
Assume they work independently.
One day they announce that their total time to finish an assignment was less than 90 minutes.
How likely is this?
```{r, echo = T, fig.height=3.5, fig.width=8}
p <- punif(90, 85, 140)
```
Probability: __`r round(p, 4)*100`%__
### 4.23
Let $X_1, X_2, \ldots, X_{20} \sim^{i.i.d.} Exp(2)$. Let $X = \sum^{20}_{i=1}X_i$.
a.) Simulate the sampling distribution of $X$ in R.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e3
result <- numeric(N)
for(i in 1:N)
{
samp <- rexp(n = 20, rate = 2)
result[i] <- sum(samp)
}
```
b.) From your simulation, find $\mathbb{E}[X]$ and $Var[X]$.
```{r, echo = T}
mu <- mean(result)
var <- var(result)
```
$\mu = `r mu`$, $Var = `r var`$
c.) From your simulation, find $P( X \le 10)$.
```{r, echo = T}
phat <- mean(result <= 10)
```
Probablity: _`r round(phat, 4)*100`%_
### 4.24
Let $X_1, X_2, \ldots, X_{30} \sim^{i.i.d.} Exp(1/3)$ and let $\bar{X}$ denote the sample mean.
a.) Simulate the sampling distribution of $\bar{X}$ in R.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e3
result <- numeric(N)
for(i in 1:N)
{
samp <- rexp(n = 30, rate = 1/3)
result[i] <- mean(samp)
}
```
b.) Find the mean and standard error of the sampling distribution, and compare with the theoretical results.
```{r, echo = T}
mu <- mean(result)
sigma <- sd(result)
```
Sample: $\mu = `r mu`$, $\sigma = `r sigma`$
Theoretical: $\mu = \frac{\lambda}{1} = 3$, $sigma = 3/sqrt(30 - 1) = .56$
c.) From your simulation, find $P(\bar{X} \le 3.5)$.
```{r, echo = T}
phat <- mean(result <= 3.5)
```
$\hat{p} = `r phat`$
d.) Estimate $P(\bar{X} \le 3.5)$ by assuming the CLT approximation holds.
Compare this result with the one in part __(c)__.
```{r, echo = T}
z <- (3.5 - 3) / sigma
p <- pnorm(z)
```
$p = `r round(p, 4)`$
### 4.25
Consider the exponential distribution with density $f(x) \frac{1}{20}e^{-x/20}$, with mean and standard deviation of 20.
a.) Calculate the median of this distribution.
$0.5 = \int_{m}^{\inf}f(x)dx = -e^{-x/20}$
$\ldots = M = 20*log(2)$
$\ldots = `r 20*log(2)`$
b.) Using R, draw a random sample of size 50 and graph the histogram.
```{r, echo = T, fig.height=3.5, fig.width=8}
samp <- rexp( n = 50, rate = 1/20)
ggplot(data.table(x = samp)) +
geom_histogram(aes(x, fill = ..count..)) +
labs(title = "RExp(50, 1/20)")
```
What are the mean and standard deviation of your sample?
```{r, echo = T}
mu <- mean(samp)
sigma <- sd(samp)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
c.) Run a simulation to find the (approximate) sampling distribution for the median of sample size 50 from the exponential distribution and describe it.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e3
result <- numeric(N)
for(i in 1:N)
{
samp <- rexp( n = 50, rate = 1/20)
result[i] <- median(samp)
}
ggplot(data.table(x = result)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
What is the mean and the standard error of this sampling distribution?
```{r, echo = T}
mu <- mean(result)
sigma <- sd(result) / sqrt(50)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
d.) Repeat the above but use sample sizes $n = 100, 500 \ and \ 1,000$.
```{r, echo = T, fig.height=3, fig.width=8}
N <- 10e3
result <- numeric(N)
for(i in 1:N)
{
samp <- rexp( n = 100, rate = 1/20)
result[i] <- median(samp)
}
ggplot(data.table(x = result)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
```{r, echo = T}
mu <- mean(result)
sigma <- sd(result) / sqrt(50)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
```{r, echo = T, fig.height=3, fig.width=8}
N <- 10e3
result <- numeric(N)
for(i in 1:N)
{
samp <- rexp( n = 500, rate = 1/20)
result[i] <- median(samp)
}
ggplot(data.table(x = result)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
```{r, echo = T}
mu <- mean(result)
sigma <- sd(result) / sqrt(50)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
```{r, echo = T, fig.height=3, fig.width=8}
N <- 10e3
result <- numeric(N)
for(i in 1:N)
{
samp <- rexp( n = 1000, rate = 1/20)
result[i] <- median(samp)
}
ggplot(data.table(x = result)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
```{r, echo = T}
mu <- mean(result)
sigma <- sd(result) / sqrt(50)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
How does sample size affect the sampling distribution?
_The sample mean and standard error converge to the analytical solution with increased sample size._
### 4.26
Prove theorem 4.2.1.
### 4.27
Let $X_1, X_2 \sim^{i.i.d.} F$ with corresponding pdf $f(x) = \frac{2}{x^2}$, $1 \le x \le 2$.
a.) Find the pdf of $X_{max}$.
$F_{max}(x) = 8(\frac{1}{x^2} - \frac{1}{x^3})$
b.) Find the expected value of $X_{max}$.
$\mathbb{E}{[F_{max}]} = 1.545$
### 4.28
Let $X_1, X_2, \ldots, X_N \sim^{i.i.d.}$ with corresponding pdf $f(x) = 3x^2, 0 \le x \le 1$.
a.) Find the pdf for $X_{min}$.
b.) Find the pdf for $X_{max}$.
c.) If $n = 10$, find the probability that the largest value, $X_{max}$, is greater than 0.92.
### 4.29
Compute the pdf of the sampling distribution of the maximum samples of size 10 from a population with an exponential distribtuion with $\lambda = 12$.
### 4.30
Let $X_1, X_2, \ldots, X_N \sim^{i.i.d.} Exp(\lambda)$ with pdf $f(x) = \lambda e^{-\lambda x}, \lambda > 0, x > 0$.
a.) Find the pdf $f_{min}(x)$ for the sample minimum $X_{min}$. Recognize this as the pdf of a known distribution.
b.) Simulate in R the sampling distribution of $X_{min}$ of samples of size $n = 25$ from the exponential distribution with $\lambda = 7$.
Compare the theoretical expected value of $X_{min}$ with the simulated expected value.
### 4.31
Let $X_1, X_2, \ldots, X_n \sim^{i.i.d.} Pois(3)$. Let $X = \sum^{10}_{i=1}X_i$.
Find the pdf for the sampling distribution of X.
### 4.32
Let $X_1$ and $X_2$ be independent exponential random variables, both with parameter $\lambda > 0$.
Find the cumulative distribution function for the sampling distribution of $X = X_1 + X_2$.
### 4.33
This simulation illustrates the CLT for a finite population.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 400
n <- 5
finpop <- rexp(N, 1/10)
ggplot(data.table(x = finpop)) +
geom_histogram(aes(x, fill = ..count..)) +
labs(title = "Exp(1/10)")
mean(finpop) #mean (mu) of your pop.
sd(finpop) # stdev (sigma) of your pop.
sd(finpop)/sqrt(n) # theoretical standard error of sampling distribution
sd(finpop)/sqrt(n) * sqrt((N-n)/(N-1)) # without replacement
Xbar <- numeric(1000)
for(i in 1:1000)
{
x <- sample(finpop, n) # Random sample of size n
# (w/o replacement)
Xbar[i] <- mean(x)
}
p1 <- ggplot(data.table(x = Xbar)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
labs(title = "Mean Sampling Distribution Histogram")
p2 <- ggplot(data.table(x = Xbar), aes(sample = x)) +
stat_qq() +
stat_qq_line() +
labs(title = "Mean Sampling Distribution QQ-Plot")
grid.arrange(p1, p2, nrow = 2)
mean(Xbar)
sd(Xbar) # estimated standard error of sampling distribution
```
a.) Does the sampling distribution of sample means appear approximately normal?
b.) Compare the mean and standard error of your simulated sampling distribution with the theoretical ones.
c.) Calculate $(\sigma/\sqrt(n))(\sqrt{(N-n)/(N-1)}$, where $\sigma$ is the standard deviation of the finite population and compare with the (estimated) standard error of the sampling distribution.
d.) Repeat for larger __n__, say __n__ = 20 and __n__ = 100.
### 4.34
Let $X_1, X_2, \ldots, X_n$ be independent random variables from $N(\mu, \sigma)$.
We are interested in the sampling distribution of the variance.
Run a simulation to draw random samples of size 20 from $N(25, 7^2)$ and calculate the variance for each sample.
```{r, echo = T, fig.height=3.5, fig.width=8}
W <- numeric(1000)
for(i in 1:1000)
{
x <- rnorm(20, 25, 7)
W[i] <- var(x)
}
mean(W)
var(W)
p1 <- ggplot(data.table(x = W)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
labs(title = "Variance Sampling Distribution")
p2 <-
ggplot(data.table(value = W), aes(sample = value)) +
stat_qq() +
stat_qq_line() +
labs(title = "Mean Sampling Distribution QQ-Plot")
grid.arrange(p1, p2, nrow = 2)
```
Does the sampling distribution appear to be normally distributed?
Repeat with n = 50 and n = 200.
### 4.35
A random sample of size $n = 100$ is drawn from a distribution with pdf $f(x) = 3(1- x)^2, 0 \le x \le 1$.
a.) Use the CLT approximation to estimate $P(\bar{X} \le 0.27)$.
b.) Use the expanded CLT to estimate the same probability (dnorm).
c.) If $X_1, X_2, X_3 \sim^{i.i.d.} Unif[0, 1]$, then the minimum has density __f__ given above.
Use simulation to estimate the probability. | true | ---
title: ''
mainfont: Arial
fontsize: 12pt
documentclass: report
header-includes:
- \PassOptionsToPackage{table}{xcolor}
- \usepackage{caption}
- \usepackage{amssymb}
- \usepackage{booktabs}
- \usepackage{longtable}
- \usepackage{array}
- \usepackage{multirow}
- \usepackage{wrapfig}
- \usepackage{float}
- \usepackage{colortbl}
- \usepackage{pdflscape}
- \usepackage{tabu}
- \usepackage{threeparttable}
- \usepackage{threeparttablex}
- \usepackage[normalem]{ulem}
- \usepackage{makecell}
- \usepackage[table]{xcolor}
- \usepackage{fancyhdr}
- \usepackage{boldline}
- \usepackage{tipa}
\definecolor{headergrey}{HTML}{545454}
\definecolor{msdblue}{HTML}{1C93D1}
\pagestyle{fancy}
\setlength\headheight{30pt}
\rhead{\color{headergrey}\today}
\fancyhead[L]{\color{headergrey}PI:NAME:<NAME>END_PI, PI:NAME:<NAME>END_PI}
\fancyhead[C]{\Large\bfseries\color{headergrey}Sampling Distributions}
\rfoot{\color{headergrey}\thepage}
\lfoot{\color{headergrey}Chapter 4}
\fancyfoot[C]{\rmfamily\color{headergrey}Mathematical Statistics}
geometry: left = 1cm, right = 1cm, top = 2cm, bottom = 3cm
date: "`r format(Sys.time(), '%d %B, %Y')`"
output:
pdf_document:
fig_caption: yes
latex_engine: xelatex
editor_options:
chunk_output_type: console
---
```{r knitr_setup, include = FALSE}
# DO NOT ADD OR REVISE CODE HERE
knitr::opts_chunk$set(echo = FALSE, eval = TRUE, dev = 'png')
options(knitr.table.format = "latex")
```
```{r report_setup, message = FALSE, warning = FALSE, include = FALSE}
library(data.table, quietly = TRUE, warn.conflicts = FALSE)
assignInNamespace("cedta.pkgEvalsUserCode", c(data.table:::cedta.pkgEvalsUserCode, "rtvs"), "data.table")
library(ggplot2, quietly = TRUE, warn.conflicts = FALSE)
library(ggrepel, quietly = TRUE, warn.conflicts = FALSE)
library(ggthemes, quietly = TRUE, warn.conflicts = FALSE)
library(knitr, quietly = TRUE, warn.conflicts = FALSE)
library(kableExtra, quietly = TRUE, warn.conflicts = FALSE)
library(Rblpapi, quietly = TRUE, warn.conflicts = FALSE)
library(scales, quietly = TRUE, warn.conflicts = FALSE)
library(pander, quietly = TRUE, warn.conflicts = FALSE)
library(dplyr, quietly = TRUE, warn.conflicts = FALSE)
library(formattable, quietly = TRUE, warn.conflicts = FALSE)
library(grid, quietly = TRUE, warn.conflicts = FALSE)
library(gridExtra, quietly = TRUE, warn.conflicts = FALSE)
library(png, quietly = TRUE, warn.conflicts = FALSE)
library(extrafont, quietly = TRUE, warn.conflicts = FALSE)
library(tinytex, quietly = TRUE, warn.conflicts = FALSE)
library(stringr, quietly = TRUE, warn.conflicts = FALSE)
library(lubridate, quietly = TRUE, warn.conflicts = FALSE)
library(reshape2, quietly = TRUE, warn.conflicts = FALSE)
library(ggrepel, quietly = TRUE, warn.conflicts = FALSE)
library(mnormt, quietly = TRUE, warn.conflicts = FALSE)
library(Ecdat, quietly = TRUE, warn.conflicts = FALSE)
library(MASS, quietly = TRUE, warn.conflicts = FALSE)
library(copula, quietly = TRUE, warn.conflicts = FALSE)
library(fGarch, quietly = TRUE, warn.conflicts = FALSE)
library(forecast, quietly = TRUE, warn.conflicts = FALSE)
library(tseries, quietly = TRUE, warn.conflicts = FALSE)
library(gmodels, quietly = TRUE, warn.conflicts = FALSE)
library(rugarch, quietly = TRUE, warn.conflicts = FALSE)
library(quantmod, quietly = TRUE, warn.conflicts = FALSE)
library(gtools, quietly = TRUE, warn.conflicts = FALSE)
options(tinytex.verbose = TRUE)
suppressMessages(library("tidyverse"))
pretty_kable <- function(data, title, dig = 2) {
kable(data, caption = title, digits = dig) %>%
kable_styling(bootstrap_options = c("striped", "hover")) %>%
kableExtra::kable_styling(latex_options = "hold_position")
}
theme_set(theme_light())
# Theme Overrides
theme_update(axis.text.x = element_text(size = 10),
axis.text.y = element_text(size = 10),
plot.title = element_text(hjust = 0.5, size = 16, face = "bold", color = "darkgreen"),
axis.title = element_text(face = "bold", size = 12, colour = "steelblue4"),
plot.subtitle = element_text(face = "bold", size = 8, colour = "darkred"),
legend.title = element_text(size = 12, color = "darkred", face = "bold"),
legend.position = "right", legend.title.align=0.5,
panel.border = element_rect(linetype = "solid",
colour = "lightgray"),
plot.margin = unit(c( 0.1, 0.1, 0.1, 0.1), "inches"))
data.dir <- "D:/Projects/Statistical-Computing/datasets/"
setwd("D:/Projects/Statistical-Computing/RDS")
```
```{r pander_setup, include = FALSE}
knitr::opts_chunk$set(comment = NA)
panderOptions('table.alignment.default', function(df)
ifelse(sapply(df, is.numeric), 'right', 'left'))
panderOptions('table.split.table', Inf)
panderOptions('big.mark', ",")
panderOptions('keep.trailing.zeros', TRUE)
```
### 4.1
Consider the population {1, 2, 5, 6, 10, 12}.
Find (and plot) the sampling distribution of medians for samples of size 3 without replacement.
```{r, echo = T, fig.height=4.5, fig.width=8}
p <- c(1, 2, 5, 6, 10, 12)
c <- combinations(v = p, n = 6, r = 3)
t <- apply(c, 1, median)
ggplot(data.table(value = t), aes(value, fill = ..count..)) +
geom_histogram(bins = 30) +
labs(title = "Median Sampling Distribution of p")
```
Compare the median of the population to the mean of the medians.
Median of p = __`r median(p)`__. Mean of Medians of p = __`r mean(t)`__
\newpage
### 4.2
Consider the population {3, 6, 7, 9, 11, 14}.
For samples of size 3 without replacement, find (and plot) the sampling distribution for the minimum.
```{r, echo = T, fig.height=4.5, fig.width=8}
p <- c(3, 6, 7, 9, 11, 14)
c <- combinations(v = p, n = 6, r = 3)
t <- apply(c, 1, min)
ggplot(data.table(value = t), aes(value, fill = ..count..)) +
geom_histogram(bins = 30) +
labs(title = "Minimum Sampling Distribution of p")
```
What is the mean of the sampling distribution? __`r mean(t)`__
The statistic is an estimate of some parameter - what is the value of that parameter?
This is an estimation of the minimum, which is: __`r min(p)`__
\newpage
### 4.3
Let _A_ denote the population {1, 3, 4, 5} and _B_ the population {5, 7, 9}.
```{r, echo = T}
A <- c(1, 3, 4, 5)
B <- c(5, 7, 9)
```
Let _X_ be a random value from _A_, and _Y_ and random value from _B_.
a.) Find the sampling distribution of _X + Y_.
```{r, echo = T}
result = numeric(12)
index <- 1
for(j in 1:length(A))
{
for(k in 1:length(B))
{
result[index] <- A[j] + B[k]
index <- index + 1
}
}
sort(result)
```
b.) In this example, does the sampling distribution depend on whether you sample with or without replacement?
_No._
Why or why not?
_Because 5 in is both sets._
c.) Compute the mean of the values for each of _A_ and _B_ and the values in the sampling distribution of _X + Y_.
Mean of _A_: __`r mean(A)`__. Mean of _B_: __`r mean(B)`__.
Mean of _A + B_: __`r mean(result)`__
How are the means related?
mean(A) + mean(B) = mean(A + B).
d.) Suppose you draw a random value from _A_ and a random value from _B_.
```{r, echo = T}
prob <- sum(result >= 13) / length(result)
```
What is the probability that the sum is 13 or larger? __`r prob*100`%__
\newpage
### 4.4
Consider the population {3, 5, 6, 6, 8, 11, 13, 15, 19, 20}.
a.) Compute the mean and standard deviation and create a dot plot of its distribution.
```{r, echo = T, fig.height=3.5, fig.width=8}
p <- c(3, 5, 6, 6, 8, 11, 13, 15, 19, 20)
mu <- mean(p)
sigma <- sd(p)
ggplot(data.table(value = p)) +
geom_dotplot(aes(value, fill = ..count..), binwidth = 1) +
labs(title = "Population Dot Plot")
```
$\mu = `r mu`$, $\sigma = `r sigma`$
b.) Simulate the sampling distribution of $\bar{X}$ by taking random samples of size 4 and plot your results.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e2
results <- numeric(N)
for( i in 1:N)
{
index <- sample(length(p), size = 4, replace = F)
results[i] <- mean( p[index] )
}
ggplot(data.table(value = results)) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs(title = "Sample Means")
xbar <- mean(results)
se <- sd(results) / sqrt(N)
```
Compute the mean and standard error, and compare to the population mean and standard deviation.
mean: `r xbar`, standard error: `r se`
c.) Use the simulation to find $P(\bar{X} < 11)$.
```{r, echo = T}
prob <- mean(results < 11)
```
$P(\bar{X} < 11) = `r prob*100`$%
\newpage
### 4.5
Consider two populations A = {3, 5, 7, 9, 10, 16}, B = {8, 10, 11, 15, 18, 25, 28}.
```{r, echo = T}
A <- c(3, 5, 7, 9, 10, 16)
B <- c(8, 10, 11, 15, 18, 25, 28)
```
a.) Using R, draw random samples (without replacement) of size 3 from each population, and simulate the sampling distribution of the sum of their maximums.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e2
results <- numeric(N)
for(i in 1:N)
{
samp.a <- sample(A, 3, replace = F)
samp.b <- sample(B, 3, replace = F)
results[i] <- max(samp.a) + max(samp.b)
}
ggplot(data.table(value = results)[, index := .I]) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs(title = "Sampling Distribution: max(A) + max(B)")
```
b.) Use your simulation to estimate the probability that the sum of the maximums is less than 20.
```{r, echo = T}
prob <- mean(results < 20)
```
Probability: `r prob*100`%
c.) Draw random samples of size 3 from each population, and find the maximum of the union of these two sets.
Simulate the sampling distribution of the maximums of this union.
```{r, echo = T, fig.height=3.5, fig.width=8}
results <- numeric(N)
for(i in 1:N)
{
samp.a <- sample(A, 3, replace = F)
samp.b <- sample(B, 3, replace = F)
results[i] <- max(union(samp.a, samp.b))
}
ggplot(data.table(value = results)[, index := .I]) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs(title = "Sampling Distribution: max(union(A,B))")
```
d.) Use simulation to find the probability that the maximum of the union is less than 20.
```{r, echo = T}
prob <- mean(results < 20)
```
Probability: `r prob*100`%
\newpage
### 4.6
The data set _Recidivism_ contains the poopulation of all Iowa offenders convicted of either a felony or misdemeanor who were released in 2010 (case study in Section 1.4).
```{r, echo = T}
Recidivism <- data.table(read.csv(paste0(data.dir, "Recidivism.csv"),
header = T))
```
Of these, 31.6% recidivated and were sent back to prision.
Simulate the sampling distribution of $\hat{p}$, the sample proportion of offeneders who recidivated, for random samples of size 25.
```{r, echo = T}
mean(Recidivism$Recid == "Yes")
N <- 10e2
results <- numeric(N)
for(i in 1:N)
{
samp <- sample(Recidivism$Recid, 25)
results[i] <- mean(samp == "Yes")
}
```
a.) Create a histogram and describe the simulated sampling distribution of $\hat{p}$.
```{r, echo = T, fig.height=3.5, fig.width=8}
ggplot(data.table(value = results)) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs(title = "Recidivism Sampling Distribution")
```
Estimate the mean and standard error.
```{r, echo = T}
mu <- mean(results)
se <- sd(results) / sqrt(25)
```
$\mu = `r mu`$, $\sigma = `r se`$
b.) Compare your estimate of the standard error with the theoretical standard error (_Corollary 4.3.2_).
```{r, echo = T}
tse <- mu * ( 1 - mu ) / sqrt(25)
```
Theoretical: `r tse`
c.) Repeat the above using samples of size 250, and compare with the $n = 25$ case.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e2
results <- numeric(N)
for(i in 1:N)
{
samp <- sample(Recidivism$Recid, 250)
results[i] <- mean(samp == "Yes")
}
ggplot(data.table(value = results)) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs(title = "Recidivism Sampling Distribution")
mu <- mean(results)
se <- sd(results) / sqrt(250)
```
$\mu = `r mu`$, $\sigma = `r se`$
### 4.7
The data set _FlightDelays_ contains the population of all flight departures by United Airlines and American Airlines out of LGA during May and June 2009 (case study in Section 1.1).
```{r, echo = T}
Flights <- data.table(read.csv(paste0(data.dir, "FlightDelays.csv"),
header = T))
```
a.) Create a histogram of _Delay_ and describe the distribution.
```{r, echo = T, fig.height=3.5, fig.width=8}
ggplot(Flights, aes(Delay)) +
geom_histogram(aes(fill = ..count..), bins = 30) +
scale_y_continuous(labels = comma) +
labs(title = "Flight Delays")
```
Compute the mean and standard deviation.
```{r, echo = T}
mu <- mean(Flights$Delay)
sigma <- sd(Flights$Delay)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
b.) Simulate the sampling distribution of $\bar{x}$, the sample mean of the length of the flight delays (_Delay_), for sample size 25.
```{r, echo = T}
N <- 10e2
results <- numeric(N)
for(i in 1:N)
{
samp <- sample(Flights$Delay, 25, replace = F)
results[i] <- mean(samp)
}
```
Create a histogram and describe the simulated sampling distribution of $\bar{x}$.
```{r, echo = T, fig.height=3.5, fig.width=8}
ggplot(data.table(value = results)) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs(title = "Flight Delay Sampling Distribution")
```
Estimate the mean and standard error.
```{r, echo = T}
mu <- mean(results)
se <- sd(results) / sqrt(25)
```
$\mu = `r mu`$, $\Sigma = `r se`$
c.) Compare your estimate of the standard error with the theoretical standard error (_Corollary A.4.1_).
```{r, echo = T}
tse <- var(results) / 25
```
Theoretical: `r tse`
d.) Repeat with sample size 250.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e2
results <- numeric(N)
for(i in 1:N)
{
samp <- sample(Flights$Delay, 250, replace = F)
results[i] <- mean(samp)
}
ggplot(data.table(value = results)) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs(title = "Flight Delay Sampling Distribution")
mu <- mean(results)
se <- sd(results) / sqrt(250)
tse <- var(results) / 250
```
$\mu = `r mu`$, $\Sigma = `r se`$
Theoretical: `r tse`
### 4.8
Let $X_1, X_2, \ldots, X_{25}$ be a random sample from some distribution and $W = T(X_1, X_2, \ldots, X_n)$ be a statistic.
Suppose the _sampling distribution_ of W has a pdf given by $f(x) = \frac{2}{x^2}$, for 1 < x < 2.
Find $P(w < 1.5)$
__Solution__:
```{r, echo = T, fig.height=3.5, fig.width=8}
f <- function(x) 2 / x^2
x <- seq( from = 1.0001, to = 1.999, by = 0.0001)
y <- f(x)
ggplot(data.table(x, y)) +
geom_point(aes(x, y), color = "cornflowerblue", size = .6) +
labs(title = "pdf f(x) = 2/x^2")
a <- cumsum(y) / sum(y)
p <- round( a[x == 1.5], 4 ) * 100
d <- data.table(x, y = a)
ggplot(d) +
geom_point(aes(x, y), color = "cornflowerblue", size = .6) +
geom_area(aes(x, y), data = d[x < 1.5], fill = "cornflowerblue", alpha = .3) +
labs(title = paste("cdf f(x) = 2/x^2, A =", p ))
```
Numerical solution: `r p`%
Analytical Solution: ${\int_{1}^{1.5}}\frac{2}{x^2} = \frac{2}{3}$
### 4.9
Let $X_1, X_2, \ldots, X_{n}$ be a random sample from some distribution and $Y = T(X_1, X_2, \ldots, X_n)$ be a statistic.
Suppose the _sampling distribution_ of Y has pdf $f(y) = (3/8)y^2 \ for \ 0 \le y \le 2$.
Find $P(0 \le Y \le \frac{1}{5})$
__Solution__:
```{r, echo = T, fig.height=3.5, fig.width=8}
f <- function(x) (3/8)*x**2
x <- seq( from = 0, to = 2, by = 0.001)
y <- f(x)
ggplot(data.table(x, y)) +
geom_point(aes(x, y), color = "cornflowerblue", size = .6) +
labs(title = paste("pdf: ", paste0(deparse(f), collapse = " ")))
a <- cumsum(y) / sum(y)
p <- round( a[x == 1/5], 4 ) * 100
d <- data.table(x, y = a)
ggplot(d) +
geom_point(aes(x, y), color = "cornflowerblue", size = .6) +
geom_area(aes(x, y), data = d[x < 1/5], fill = "cornflowerblue", alpha = .3) +
labs(title = paste("cdf f(x) = 2/x^2, A =", p ))
```
Numerical Solution: `r p`%
Analytical Solution: ${\int_{0}^{\frac{1}{5}}}\frac{x^3}{8} = \frac{.008}{8} = .001 = .1$ %
### 4.10
Suppose the heights of boys in a certain large city follow a distribution with mean 48 in. and variance $9^2$.
Use the CLT approximation to estimate the probability that in a random sample of 30 boys, the mean height is more than 51 in.
```{r, echo = T}
z <- (51 - 48) / (9^2 / sqrt(30))
p <- pnorm(z, lower.tail = F)
```
Probability: __`r round(p, 4)*100`%__
### 4.11
Let $X_1, X_2, \ldots, X_{36} \sim Bern(.55)$ be independent, and let $\hat{p}$ denote the sample proportion.
Use the CLT approximation with continuity correction to find the probability that $\hat{p} \le 0.5$.
```{r, echo = T}
z <- ( .5 - .55 ) / sqrt(.55 * (1 - .55) / 36)
p <- pnorm(z, lower.tail = T)
```
Probability: `r round(p, 4)*100`%
### 4.12
A random sample of size $n = 20$ is drawn from a distribution with mean 6 and variance 10.
Use the CLT approximation to estimate $P(\bar{X} \le 4.6)$.
```{r, echo = T}
z <- ( 4.6 - 6 ) / ( 10 * sqrt(20) )
p <- pnorm(z, lower.tail = T)
```
Probability: `r round(p, 4)*100`%
### 4.13
A random sample of size $n = 244$ is drawn from a distribution with pdf $f(x) = (3/16)(x - 4)^2, 2 \le x \le 6$.
Use the CLT approximation to estimate $P(X \ge 4.2)$.
```{r, echo = T, fig.height=2.6, fig.width=8}
pdf <- function(x) (3/16)*(x - 4)^2
x <- seq(from = 2, to = 6, by = 0.001)
y <- pdf(x)
ggplot(data.table(x,y)) +
geom_point(aes(x, y), col = "cornflowerblue", lwd = .8) +
labs(title = paste("PDF: ", paste0(deparse(f), collapse = " ")))
cdf <- function(x) (3/8)*(x - 4)
y <- cumsum(y) / sum(y)
ev <- x[min(which(y > .5))]
ggplot(data.table(x,y)) +
geom_point(aes(x, y), col = "cornflowerblue", lwd = .8) +
geom_vline(xintercept = ev) +
labs(title = paste("CDF: ", paste0(deparse(f), collapse = " ")))
z <- ( 4.2 - ev ) / sqrt(244)
pnorm(z, lower.tail = F)
```
### 4.14
According to the 2000 census, 28.6% of the US adult population recieved a high school diploma.
In a random sample of 800 US adults, what is the probability that between 220 and 230 (inclusive) people have a high school deploma?
Use the CLT approximation with continuity correction, and compare with the exact probability.
__Solution__:
The sampling distribution of $\hat{p}$ is approximately normal with:
```{r, echo = T}
n <- 800
mu <- .286
ev <- 800 * mu
sigma <- sqrt(n*mu*(1-mu))
```
$\mathbb{E}[X] = `r ev`$ and $\sigma = \sqrt{800(.286)(1 - .286)} = `r round(sigma, 4)`$
```{r, echo = T}
l <- pnorm((ev - 219.5) / sigma)
h <- pnorm((ev - 230.5) / sigma)
p <- l - h
```
Probability: `r round(p, 4)`
### 4.15
If $X_1, \ldots, X_n$ are i.i.d. from Unif[0, 1], how large should n be so that $P(\bar{X} - \frac{1}{2} < 0.05) \ge 0.90$,
that is, is there at least a 90% chance that the sample mean is within 0.05 of $\frac{1}{2}$? Use the CLT approximation.
```{r, echo = T}
```
### 4.16
PI:NAME:<NAME>END_PI claims that she has drawn a random sample of size 30 from the exponential distribution with $\lambda = 1/10$.
The mean of her sample is 12.
a.) What is the expected value of a sample mean?
$X \sim Exp(\frac{1}{10}), \mathbb{E}(x) = 10$
b.) Run a simulation by drawing 1000 random samples, each of size 30, from Exp(1/10), and compute the mean for each sample.
```{r, echo = T, fig.height=3.5, fig.width=12}
N <- 1000
result <- numeric(N)
for( i in 1:N)
{
samp <- rexp( n = 30, rate = 1/10)
result[i] <- mean(samp)
}
ggplot(data.table(result), aes(result)) +
geom_histogram(aes(y = ..density.., fill = ..count..), bins = 30) +
geom_vline(xintercept = 12, col = "darkorange") +
stat_density( kernel = "gaussian", fill = "darkorange", alpha = .3) +
labs(title = "Exp(1/10) Sampling Distribution")
p <- mean(result > 12)
```
What proportion of the sample means is as large or larger than 12? __`r p*100`%__
c.) Is a mean of 12 unusual for a sample of size 30 from Exp(1/10)?
__Yes__, only ~13% of the sample means have a value of 12 or higher.
### 4.17
Let $X \sim N(15, 3^2)$ and $Y \sim N(4, 2^2)$ be independent random variables.
a.) What is the exact sampling distribution of $W = X - 2Y$?
$W \sim N(7, 5^2)$
b.) Use R to simulate the sampling distribution of $W$ and plot your results.
```{r, echo = T, fig.height=3.5, fig.width=8}
X <- rnorm(10e3, 15, 3)
Y <- rnorm(10e3, 4, 2)
W <- X - 2*Y
ggplot(data.table(x = W)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
geom_vline(xintercept = 7, col = "darkorange", lwd = 1.5) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
Check that the simulated mean and standard error are close to the theoretical mean and standard error.
```{r, echo = T}
mu <- mean(W)
sigma <- sd(W)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
c.) Use the simulated sampling to estimate $P(W \le 10)$, and then check your estimate with an exact calculation.
```{r, echo = T}
phat <- mean(W <= 10)
p <- pnorm(10, mean = 7, sd = 5)
```
$\hat{p}= `r phat*100`$%
$P(W \le 10) = `r round(p, 4)*100`$%
### 4.18
Let $X \sim Pois(4)$, $Y \sim Pois(12)$, $U \sim Pois(3)$ be independent random variables.
a.) What is the exact sampling distribution of $W = X + Y + U$?
$W \sim Pois(19)$
b.) Use R to simulate the sampling distribution of $W$ and plot your results.
```{r, echo = T, fig.height=3.5, fig.width=8}
W <- rpois(10e3, lambda = 19)
ggplot(data.table(x = W)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
geom_vline(xintercept = 19, col = "darkorange", lwd = 1.5) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
Check that the simulated mean and standard error are close to the theoretical mean and standard error.
```{r, echo = T}
mu <- mean(W)
sigma <- sd(W)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
c.) Use the simulated sampling distribution to estimate $P(W \le 14)$ and then check your estimate with an exact calculation.
```{r, echo = T}
phat <- mean(W <= 14)
p <- ppois(14, lambda = 19)
```
$\hat{p} = `r phat*100`$%
$P(W \le 14) = `r round(p, 4)*100`$%
### 4.19
Let $X_1, X_2, \ldots, X_{10} \sim^{i.i.d} N(20, 8^2)$ and $Y_1, Y_2, \ldots, Y_{15} \sim^{i.i.d} N(16, 7^2)$.
Let $W = \bar{X} + \bar{Y}$
a.) Give the exact sampling distribution of W.
$\sigma = (10 + 15) / sqrt( 10 + 15 - 1) = 3.1$
$W \sim N(36, 3.1^2)$
b.) Simulate the sampling distribution in R and plot your results.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e3
result <- numeric(N)
for( i in 1:N)
{
X <- rnorm(10, 20, 8)
Y <- rnorm(15, 16, 7)
result[i] <- mean(X) + mean(Y)
}
ggplot(data.table(x = result)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
geom_vline(xintercept = 36, col = "darkorange", lwd = 1.5) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
Check that the simulated mean and standard error are close to the exact mean and standard error.
```{r, echo = T}
mu <- mean(result)
sigma <- sd(result)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
c.) Use your simulation to find $P(W < 40)$. Calculate an exact answer and compare.
```{r, echo = T}
phat <- mean(result <= 40)
p <- pnorm(40, 36, 3)
```
$\hat{p} = `r phat*100`$%
$P(W < 40) = `r round(p, 4)*100`$%
### 4.20
Let $X_1, X_2, \ldots, X_9 \sim^{i.i.d.} N(7, 3^2)$, and $Y_1, Y_2, \ldots, X_{12} \sim^{i.i.d.} N(10, 5^2)$.
Let $W = \bar{X} - \bar{Y}$.
a.) Give the sampling distribution of $W$.
$\sigma = (3 + 5) / sqrt( 9 + 12 - 1 ) = 1.79$
$W = N(-3, 1.79^2)$
b.) Simulate the sampling distribution of W in R, and plot your results.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e3
result <- numeric(N)
for(i in 1:N)
{
X <- rnorm(9, 7, 3)
Y <- rnorm(12, 10, 5)
result[i] <- mean(X) - mean(Y)
}
ggplot(data.table(x = result)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
geom_vline(xintercept = -3, col = "darkorange", lwd = 1.5) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
Check that the simulated mean and standard error are close to the theoretical mean and standard error.
```{r, echo = T}
mu <- mean(result)
sigma <- sd(result)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
c.) Use your simulation to find $P(W < -1.5)$.
```{r, echo = T}
phat <- mean(result <= -1.5)
p <- pnorm(-1.5, -3, 1.79)
```
$\hat{p} = `r phat*100`$%
Calculate an exact answer and compare.
$P(W < 1.5) = `r round(p, 4)*100`$%
### 4.21
Let $X_1, X_2, \ldots, X_N$ be a random sample from $N(0, 1)$. Let $W = X^2_1 + X^2_2 + \ldots + X^2_n$
What is the mean and variance of the sampling distribution of W?
$\mu = 0$, $\sigma = 1$
Repeat using N = 4, N = 5.
$N = 4, \sigma = 4 / sqrt(4 - 1) = 2.3$
$N = 5, \sigma = 5 / sqrt(5 - 1) = 2.5$
What observations or conjectures do you have for general __n__?
### 4.22
Let $X$ be a uniform random variable on the interval $[40, 60]$ and Y be a uniform random variable on $[45, 80]$.
Assume that X and Y are independent.
a.) Compute the expected value and variance of $X + Y$.
$\mu = 112.5$, $Var = 1/24*(140 - 85)^2 = 126.04$
b.) Simulate a sampling distribution of $X + Y$.
```{r, echo = T, fig.height=3.5, fig.width=8}
X <- runif(1000, 40, 60)
Y <- runif(1000, 45, 80)
total <- X + Y
ggplot(data.table(value = total)) +
geom_histogram(aes(value, fill = ..count..), bins = 30) +
labs("X + Y Sampling Distribution")
```
Describe the sampling distribution of $X + Y$. __Approximately Normal__
Compute the mean and variance of the sampling distribution and compare this with the theoretical mean and variance.
```{r, echo = T}
mu <- mean(total)
var <- var(total)
```
$\mu = `r mu`$, $Var = `r var`$
c.) Suppose the time (in minutes) PI:NAME:<NAME>END_PI takes to complete his statistics homework is $Unif[40, 60]$ and the time PI:NAME:<NAME>END_PI takes is $Unif[45, 80]$.
Assume they work independently.
One day they announce that their total time to finish an assignment was less than 90 minutes.
How likely is this?
```{r, echo = T, fig.height=3.5, fig.width=8}
p <- punif(90, 85, 140)
```
Probability: __`r round(p, 4)*100`%__
### 4.23
Let $X_1, X_2, \ldots, X_{20} \sim^{i.i.d.} Exp(2)$. Let $X = \sum^{20}_{i=1}X_i$.
a.) Simulate the sampling distribution of $X$ in R.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e3
result <- numeric(N)
for(i in 1:N)
{
samp <- rexp(n = 20, rate = 2)
result[i] <- sum(samp)
}
```
b.) From your simulation, find $\mathbb{E}[X]$ and $Var[X]$.
```{r, echo = T}
mu <- mean(result)
var <- var(result)
```
$\mu = `r mu`$, $Var = `r var`$
c.) From your simulation, find $P( X \le 10)$.
```{r, echo = T}
phat <- mean(result <= 10)
```
Probablity: _`r round(phat, 4)*100`%_
### 4.24
Let $X_1, X_2, \ldots, X_{30} \sim^{i.i.d.} Exp(1/3)$ and let $\bar{X}$ denote the sample mean.
a.) Simulate the sampling distribution of $\bar{X}$ in R.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e3
result <- numeric(N)
for(i in 1:N)
{
samp <- rexp(n = 30, rate = 1/3)
result[i] <- mean(samp)
}
```
b.) Find the mean and standard error of the sampling distribution, and compare with the theoretical results.
```{r, echo = T}
mu <- mean(result)
sigma <- sd(result)
```
Sample: $\mu = `r mu`$, $\sigma = `r sigma`$
Theoretical: $\mu = \frac{\lambda}{1} = 3$, $sigma = 3/sqrt(30 - 1) = .56$
c.) From your simulation, find $P(\bar{X} \le 3.5)$.
```{r, echo = T}
phat <- mean(result <= 3.5)
```
$\hat{p} = `r phat`$
d.) Estimate $P(\bar{X} \le 3.5)$ by assuming the CLT approximation holds.
Compare this result with the one in part __(c)__.
```{r, echo = T}
z <- (3.5 - 3) / sigma
p <- pnorm(z)
```
$p = `r round(p, 4)`$
### 4.25
Consider the exponential distribution with density $f(x) \frac{1}{20}e^{-x/20}$, with mean and standard deviation of 20.
a.) Calculate the median of this distribution.
$0.5 = \int_{m}^{\inf}f(x)dx = -e^{-x/20}$
$\ldots = M = 20*log(2)$
$\ldots = `r 20*log(2)`$
b.) Using R, draw a random sample of size 50 and graph the histogram.
```{r, echo = T, fig.height=3.5, fig.width=8}
samp <- rexp( n = 50, rate = 1/20)
ggplot(data.table(x = samp)) +
geom_histogram(aes(x, fill = ..count..)) +
labs(title = "RExp(50, 1/20)")
```
What are the mean and standard deviation of your sample?
```{r, echo = T}
mu <- mean(samp)
sigma <- sd(samp)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
c.) Run a simulation to find the (approximate) sampling distribution for the median of sample size 50 from the exponential distribution and describe it.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 10e3
result <- numeric(N)
for(i in 1:N)
{
samp <- rexp( n = 50, rate = 1/20)
result[i] <- median(samp)
}
ggplot(data.table(x = result)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
What is the mean and the standard error of this sampling distribution?
```{r, echo = T}
mu <- mean(result)
sigma <- sd(result) / sqrt(50)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
d.) Repeat the above but use sample sizes $n = 100, 500 \ and \ 1,000$.
```{r, echo = T, fig.height=3, fig.width=8}
N <- 10e3
result <- numeric(N)
for(i in 1:N)
{
samp <- rexp( n = 100, rate = 1/20)
result[i] <- median(samp)
}
ggplot(data.table(x = result)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
```{r, echo = T}
mu <- mean(result)
sigma <- sd(result) / sqrt(50)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
```{r, echo = T, fig.height=3, fig.width=8}
N <- 10e3
result <- numeric(N)
for(i in 1:N)
{
samp <- rexp( n = 500, rate = 1/20)
result[i] <- median(samp)
}
ggplot(data.table(x = result)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
```{r, echo = T}
mu <- mean(result)
sigma <- sd(result) / sqrt(50)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
```{r, echo = T, fig.height=3, fig.width=8}
N <- 10e3
result <- numeric(N)
for(i in 1:N)
{
samp <- rexp( n = 1000, rate = 1/20)
result[i] <- median(samp)
}
ggplot(data.table(x = result)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
scale_y_continuous(labels = comma) +
labs(title = "Sampling Distribution")
```
```{r, echo = T}
mu <- mean(result)
sigma <- sd(result) / sqrt(50)
```
$\mu = `r mu`$, $\sigma = `r sigma`$
How does sample size affect the sampling distribution?
_The sample mean and standard error converge to the analytical solution with increased sample size._
### 4.26
Prove theorem 4.2.1.
### 4.27
Let $X_1, X_2 \sim^{i.i.d.} F$ with corresponding pdf $f(x) = \frac{2}{x^2}$, $1 \le x \le 2$.
a.) Find the pdf of $X_{max}$.
$F_{max}(x) = 8(\frac{1}{x^2} - \frac{1}{x^3})$
b.) Find the expected value of $X_{max}$.
$\mathbb{E}{[F_{max}]} = 1.545$
### 4.28
Let $X_1, X_2, \ldots, X_N \sim^{i.i.d.}$ with corresponding pdf $f(x) = 3x^2, 0 \le x \le 1$.
a.) Find the pdf for $X_{min}$.
b.) Find the pdf for $X_{max}$.
c.) If $n = 10$, find the probability that the largest value, $X_{max}$, is greater than 0.92.
### 4.29
Compute the pdf of the sampling distribution of the maximum samples of size 10 from a population with an exponential distribtuion with $\lambda = 12$.
### 4.30
Let $X_1, X_2, \ldots, X_N \sim^{i.i.d.} Exp(\lambda)$ with pdf $f(x) = \lambda e^{-\lambda x}, \lambda > 0, x > 0$.
a.) Find the pdf $f_{min}(x)$ for the sample minimum $X_{min}$. Recognize this as the pdf of a known distribution.
b.) Simulate in R the sampling distribution of $X_{min}$ of samples of size $n = 25$ from the exponential distribution with $\lambda = 7$.
Compare the theoretical expected value of $X_{min}$ with the simulated expected value.
### 4.31
Let $X_1, X_2, \ldots, X_n \sim^{i.i.d.} Pois(3)$. Let $X = \sum^{10}_{i=1}X_i$.
Find the pdf for the sampling distribution of X.
### 4.32
Let $X_1$ and $X_2$ be independent exponential random variables, both with parameter $\lambda > 0$.
Find the cumulative distribution function for the sampling distribution of $X = X_1 + X_2$.
### 4.33
This simulation illustrates the CLT for a finite population.
```{r, echo = T, fig.height=3.5, fig.width=8}
N <- 400
n <- 5
finpop <- rexp(N, 1/10)
ggplot(data.table(x = finpop)) +
geom_histogram(aes(x, fill = ..count..)) +
labs(title = "Exp(1/10)")
mean(finpop) #mean (mu) of your pop.
sd(finpop) # stdev (sigma) of your pop.
sd(finpop)/sqrt(n) # theoretical standard error of sampling distribution
sd(finpop)/sqrt(n) * sqrt((N-n)/(N-1)) # without replacement
Xbar <- numeric(1000)
for(i in 1:1000)
{
x <- sample(finpop, n) # Random sample of size n
# (w/o replacement)
Xbar[i] <- mean(x)
}
p1 <- ggplot(data.table(x = Xbar)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
labs(title = "Mean Sampling Distribution Histogram")
p2 <- ggplot(data.table(x = Xbar), aes(sample = x)) +
stat_qq() +
stat_qq_line() +
labs(title = "Mean Sampling Distribution QQ-Plot")
grid.arrange(p1, p2, nrow = 2)
mean(Xbar)
sd(Xbar) # estimated standard error of sampling distribution
```
a.) Does the sampling distribution of sample means appear approximately normal?
b.) Compare the mean and standard error of your simulated sampling distribution with the theoretical ones.
c.) Calculate $(\sigma/\sqrt(n))(\sqrt{(N-n)/(N-1)}$, where $\sigma$ is the standard deviation of the finite population and compare with the (estimated) standard error of the sampling distribution.
d.) Repeat for larger __n__, say __n__ = 20 and __n__ = 100.
### 4.34
Let $X_1, X_2, \ldots, X_n$ be independent random variables from $N(\mu, \sigma)$.
We are interested in the sampling distribution of the variance.
Run a simulation to draw random samples of size 20 from $N(25, 7^2)$ and calculate the variance for each sample.
```{r, echo = T, fig.height=3.5, fig.width=8}
W <- numeric(1000)
for(i in 1:1000)
{
x <- rnorm(20, 25, 7)
W[i] <- var(x)
}
mean(W)
var(W)
p1 <- ggplot(data.table(x = W)) +
geom_histogram(aes(x, fill = ..count..), bins = 30) +
labs(title = "Variance Sampling Distribution")
p2 <-
ggplot(data.table(value = W), aes(sample = value)) +
stat_qq() +
stat_qq_line() +
labs(title = "Mean Sampling Distribution QQ-Plot")
grid.arrange(p1, p2, nrow = 2)
```
Does the sampling distribution appear to be normally distributed?
Repeat with n = 50 and n = 200.
### 4.35
A random sample of size $n = 100$ is drawn from a distribution with pdf $f(x) = 3(1- x)^2, 0 \le x \le 1$.
a.) Use the CLT approximation to estimate $P(\bar{X} \le 0.27)$.
b.) Use the expanded CLT to estimate the same probability (dnorm).
c.) If $X_1, X_2, X_3 \sim^{i.i.d.} Unif[0, 1]$, then the minimum has density __f__ given above.
Use simulation to estimate the probability. |
[
{
"context": "---\ntitle: \"Kostyantyn Hrytsyuk & Khrystyna Kubatska. Finances Project\"\noutput:\n ",
"end": 31,
"score": 0.9998738169670105,
"start": 12,
"tag": "NAME",
"value": "Kostyantyn Hrytsyuk"
},
{
"context": "---\ntitle: \"Kostyantyn Hrytsyuk & Khrystyna Kubatska. Finances Project\"\noutput:\n pdf_document: defaul",
"end": 52,
"score": 0.9995737075805664,
"start": 34,
"tag": "NAME",
"value": "Khrystyna Kubatska"
}
] | Fin_Garch_Code.rmd | kostyantynHrytsyuk/Fin-GARCH | 0 | ---
title: "Kostyantyn Hrytsyuk & Khrystyna Kubatska. Finances Project"
output:
pdf_document: default
html_notebook: default
html_document:
df_print: paged
---
```{r}
require(dplyr)
require(ggplot2)
require(xts)
require(rugarch)
require(PerformanceAnalytics)
require(quantmod)
library(skewt)
```
```{r}
# Configure propet path to the Funds.csv file here
# setwd('./')
```
```{r}
read_funds <- function(lf) {
dfs <- list()
browser()
for (f in 1:length(lf)) {
cond <- substr(lf[f], nchar(lf[f])-3, nchar(lf[f])) == '.csv'
if (cond) {
temp <- data.frame(read.csv(lf[f], stringsAsFactors = FALSE))
temp$Date <- as.Date(temp$Date)
temp$Close <- as.numeric(temp$Close)
dfs[[f]] <- temp[,c(1,5)]
}
}
return(dfs)
}
funds_names <- c("Vanguard", "Blackrock", "Statestreet",
"JPmorgan", "Bankmellon", "Allianz")
```
```{r}
# Forming set of parameters for different GARCH model
get_garch_specs <- function() {
# Standard GARCH with normal distribution of errors
norm_garch_spec <- ugarchspec(mean.model = list(armaOrder = c(0,0)),
variance.model = list(model = 'sGARCH'),
distribution.model = 'norm')
# GJR GARCH with normal distribution of errors
norm_gjr_spec <- ugarchspec(mean.model = list(armaOrder = c(0,0)),
variance.model = list(model = 'gjrGARCH'),
distribution.model = 'norm')
# Standard GARCH with skewed Student t distribution of errors
sstd_garch_spec <- ugarchspec(mean.model = list(armaOrder = c(0,0)),
variance.model = list(model = 'sGARCH'),
distribution.model = 'sstd')
# GJR GARCH with skewed Student t distribution of errors
sstd_gjr_spec <- ugarchspec(mean.model = list(armaOrder = c(0,0)),
variance.model = list(model = 'gjrGARCH'),
distribution.model = 'sstd')
garch_specs <- list(norm_garch_spec, norm_gjr_spec,
sstd_garch_spec, sstd_gjr_spec)
return(garch_specs)
}
```
```{r}
# Apply GARCH model to our data
get_garch_fits <- function(fund) {
garch_specs <- get_garch_specs()
garch_fits <- list()
for (s in 1:length(garch_specs)) {
suppressWarnings(garch_fits[[s]] <- ugarchfit(data = fund,
spec = garch_specs[[s]]))
}
return(garch_fits)
}
# Visualizing standardized residuals for models
visualize_residuals <- function(garch_fits) {
for (f in 1:length(garch_fits)) {
chart.Histogram(residuals(garch_fits[[f]], standardize = T),
methods = c('add.normal', 'add.density'),
main = paste('Standardized residuals of',names(garch_fits)[f]))
}
}
# Models validation
test_garch_models <- function(garch_fits) {
for (f in 1:length(garch_fits)) {
standard_residuals <- residuals(garch_fits[[f]], standardize = T)
p <- acf(abs(standard_residuals),22, plot = F)
plot(p, main = names(garch_fits)[f])
cat('\n', names(garch_fits)[f],'\n')
print(Box.test(abs(standard_residuals), 22, type = 'Ljung-Box'))
}
}
#Coefficients
print_garch_coefficients <- function(garch_fits) {
for (f in 1:length(garch_fits)) {
cat('\nCoefficients of', names(garch_fits)[f], '\n')
print(round(garch_fits[[f]]@fit$matcoef,10))
cat('\nRobust coefficients of', names(garch_fits)[f], '\n')
print(round(garch_fits[[f]]@fit$robust.matcoef,10))
}
}
# Models comparing
compare_garch_models <- function(garch_fits, short_model_names) {
model_comparison <- data.frame()
for (f in 1:length(garch_fits)) {
temp <- data.frame()
temp[1,1] <- likelihood(garch_fits[[f]])
inf_criterion <- infocriteria(garch_fits[[f]])
temp <- rbind(temp, inf_criterion)
model_comparison <- c(model_comparison, temp)
}
model_comparison <- as.data.frame(model_comparison)
rownames(model_comparison) <- c('Likelihood',rownames(inf_criterion))
colnames(model_comparison) <- short_model_names
print(model_comparison)
}
# Visualizing impact of negative previous return on variance
visualize_dependecy_ret_var <- function(garch_fits, short_model_names) {
p <- ggplot()
for (f in 1:length(garch_fits)) {
garch_news <- as.data.frame(newsimpact(garch_fits[[f]])[1:2])
model_name <- short_model_names[f]
model_name <- enquo(model_name)
p <- p + geom_line(data = garch_news,
aes(x = zx, y = zy, color = !!model_name))
}
p <- p + labs(x = 'Error', y = 'Variance',
title = 'Dependence of variance on errors in different models') +
theme(plot.title = element_text(hjust = 0.5))
print(p)
}
# Visualizing volatility
visualizing_volatility <- function(garch_fits, short_model_names, fund_vol) {
p <- ggplot()
garch_vol <- list()
for (f in 1:length(garch_fits)) {
garch_vol[[f]] <- sigma(garch_fits[[f]])
model_name <- short_model_names[f]
model_name <- enquo(model_name)
p <- p + geom_line(data = garch_vol[[f]], aes(x = index(garch_vol[[f]][,1]),
y = garch_vol[[f]][,1],
color = !!model_name), alpha = 0.2)
}
names(garch_vol) <- short_model_names
p <- p + geom_line(data = fund_vol, aes(y = fund_vol[,1], x = index(fund_vol[,1]),
color = 'Actual volatility')) +
labs(x = 'Date', y = 'Volatility',
title = 'Volatility constructed by different models') +
theme(plot.title = element_text(hjust = 0.5))
suppressMessages(suppressWarnings(print(p)))
return(garch_vol)
}
# Predicting volatility for n.ahead periods
predict_volatility <- function(garch_fits, garch_vol,
fund_vol, short_model_names) {
fund_tail_volatility <- tail(fund_vol,10)
predict_results <- data.frame(fund_tail_volatility)
garch_sst <- c()
for (f in 1:length(garch_fits)) {
garch_forecast <- ugarchforecast(fitORspec = garch_fits[[f]],
data = garch_vol[[f]], n.ahead = 10)
predict_results <- cbind(predict_results, sigma(garch_forecast))
garch_sst[[f]] <- sum(fund_tail_volatility - sigma(garch_forecast))
names(garch_sst)[f] <- paste('TES for',short_model_names[f])
}
names(predict_results)[2:ncol(predict_results)] <- short_model_names
# Total error sum for models
print(garch_sst)
#Comparing predicted volatility for models with actual one
print(predict_results)
}
```
```{r}
# Data loading
df <- read.csv('Funds.csv', stringsAsFactors = F)
df <- df[,-1]
df$Date <- as.Date(df$Date)
# Transforming data from df to xts
funds <- xts(df[,2:ncol(df)], order.by = df$Date)
funds <- na.omit(funds)
rm(df)
# Subseting original data for April of 2020 year
funds_red <- funds['/202004']
```
```{r}
get_volatiles <- function(funds, width = 22, time_scale = 1, funds_names) {
vol_df <- data.frame()
for (i in 1:ncol(funds)) {
fund <- funds[,i]
temp <- rollapply(data = fund, width = 22, FUN = 'sd.annualized', scale = time_scale)
if (nrow(vol_df) == 0) {
vol_df <- temp
} else {
vol_df <- cbind(vol_df, temp)
}
}
names(vol_df) <- funds_names
return(vol_df)
}
visualize_funds_lines <- function(funds, y_axis_label = 'Volatility') {
for (i in 1:ncol(funds)) {
temp <- funds[,i]
temp_mean <- mean(temp, na.rm = TRUE)
p <- ggplot(temp, aes(x = index(temp), y = temp)) +
geom_line(aes(color = 'Volatility')) +
geom_hline(aes(yintercept = temp_mean, color = 'Mean'),
size=.5, linetype='dashed') +
geom_text( aes( min(index(temp)) , temp_mean, label = round(temp_mean, 4), vjust = 2)) +
labs(x = 'Date', y = y_axis_label, title = names(funds)[i]) +
theme(plot.title = element_text(hjust = 0.5))
suppressMessages(suppressWarnings(print(p)))
}
}
visualize_funds_hist <- function(funds, x_axis_label = 'Return') {
for (i in 1:ncol(funds)) {
temp <- funds[,i]
title <- paste(funds_names[i], 'returns')
chart.Histogram(temp,
methods = c('add.normal', 'add.density'),
main = title)
temp <- (temp - mean(temp, na.rm = T))/sd(temp, na.rm = T)
title <- paste('Standardized', title)
chart.Histogram(temp,
methods = c('add.normal', 'add.density'),
main = title)
}
}
```
```{r}
evaluate_garch <- function(fund, fund_vol) {
# Models naming
model_names <- c('Standard GARCH with normal distribution of errors',
'GJR GARCH with normal distribution of errors',
'Standard GARCH with skewed Student t distribution of errors',
'GJR GARCH with skewed Student t distribution of errors')
short_model_names <- c('Normal GARCH', 'Normal GJR', 'Skewed t GARCH', 'Skewed t GJR')
garch_fits <- get_garch_fits(fund)
names(garch_fits) <- model_names
visualize_residuals(garch_fits)
test_garch_models(garch_fits)
print_garch_coefficients(garch_fits)
compare_garch_models(garch_fits, short_model_names)
visualize_dependecy_ret_var(garch_fits, short_model_names)
garch_vol <- visualizing_volatility(garch_fits, short_model_names, fund_vol)
predict_volatility(garch_fits = garch_fits,
garch_vol = garch_vol,
fund_vol = fund_vol,
short_model_names = short_model_names)
}
```
```{r}
vol_df <- get_volatiles(funds, funds_names = funds_names)
vol_df <- na.omit(vol_df)
visualize_funds_lines(vol_df)
visualize_funds_hist(funds)
fund_tail_volatility <- tail(vol_df$Vanguard, 10)
```
```{r}
cat('Data up to 2020-04-30\n')
for (i in 1:ncol(funds_red)) {
cat(colnames(vol_df)[i],'\n')
evaluate_garch(funds_red[,i], vol_df[,i])
cat('\n--------------------------------\n')
}
```
```{r}
cat('Data up to 2020-05-08\n')
for (i in 1:ncol(funds)) {
cat(colnames(vol_df)[i],'\n')
evaluate_garch(funds_red[,i], vol_df[,i])
cat('\n--------------------------------\n')
}
```
| 2422 | ---
title: "<NAME> & <NAME>. Finances Project"
output:
pdf_document: default
html_notebook: default
html_document:
df_print: paged
---
```{r}
require(dplyr)
require(ggplot2)
require(xts)
require(rugarch)
require(PerformanceAnalytics)
require(quantmod)
library(skewt)
```
```{r}
# Configure propet path to the Funds.csv file here
# setwd('./')
```
```{r}
read_funds <- function(lf) {
dfs <- list()
browser()
for (f in 1:length(lf)) {
cond <- substr(lf[f], nchar(lf[f])-3, nchar(lf[f])) == '.csv'
if (cond) {
temp <- data.frame(read.csv(lf[f], stringsAsFactors = FALSE))
temp$Date <- as.Date(temp$Date)
temp$Close <- as.numeric(temp$Close)
dfs[[f]] <- temp[,c(1,5)]
}
}
return(dfs)
}
funds_names <- c("Vanguard", "Blackrock", "Statestreet",
"JPmorgan", "Bankmellon", "Allianz")
```
```{r}
# Forming set of parameters for different GARCH model
get_garch_specs <- function() {
# Standard GARCH with normal distribution of errors
norm_garch_spec <- ugarchspec(mean.model = list(armaOrder = c(0,0)),
variance.model = list(model = 'sGARCH'),
distribution.model = 'norm')
# GJR GARCH with normal distribution of errors
norm_gjr_spec <- ugarchspec(mean.model = list(armaOrder = c(0,0)),
variance.model = list(model = 'gjrGARCH'),
distribution.model = 'norm')
# Standard GARCH with skewed Student t distribution of errors
sstd_garch_spec <- ugarchspec(mean.model = list(armaOrder = c(0,0)),
variance.model = list(model = 'sGARCH'),
distribution.model = 'sstd')
# GJR GARCH with skewed Student t distribution of errors
sstd_gjr_spec <- ugarchspec(mean.model = list(armaOrder = c(0,0)),
variance.model = list(model = 'gjrGARCH'),
distribution.model = 'sstd')
garch_specs <- list(norm_garch_spec, norm_gjr_spec,
sstd_garch_spec, sstd_gjr_spec)
return(garch_specs)
}
```
```{r}
# Apply GARCH model to our data
get_garch_fits <- function(fund) {
garch_specs <- get_garch_specs()
garch_fits <- list()
for (s in 1:length(garch_specs)) {
suppressWarnings(garch_fits[[s]] <- ugarchfit(data = fund,
spec = garch_specs[[s]]))
}
return(garch_fits)
}
# Visualizing standardized residuals for models
visualize_residuals <- function(garch_fits) {
for (f in 1:length(garch_fits)) {
chart.Histogram(residuals(garch_fits[[f]], standardize = T),
methods = c('add.normal', 'add.density'),
main = paste('Standardized residuals of',names(garch_fits)[f]))
}
}
# Models validation
test_garch_models <- function(garch_fits) {
for (f in 1:length(garch_fits)) {
standard_residuals <- residuals(garch_fits[[f]], standardize = T)
p <- acf(abs(standard_residuals),22, plot = F)
plot(p, main = names(garch_fits)[f])
cat('\n', names(garch_fits)[f],'\n')
print(Box.test(abs(standard_residuals), 22, type = 'Ljung-Box'))
}
}
#Coefficients
print_garch_coefficients <- function(garch_fits) {
for (f in 1:length(garch_fits)) {
cat('\nCoefficients of', names(garch_fits)[f], '\n')
print(round(garch_fits[[f]]@fit$matcoef,10))
cat('\nRobust coefficients of', names(garch_fits)[f], '\n')
print(round(garch_fits[[f]]@fit$robust.matcoef,10))
}
}
# Models comparing
compare_garch_models <- function(garch_fits, short_model_names) {
model_comparison <- data.frame()
for (f in 1:length(garch_fits)) {
temp <- data.frame()
temp[1,1] <- likelihood(garch_fits[[f]])
inf_criterion <- infocriteria(garch_fits[[f]])
temp <- rbind(temp, inf_criterion)
model_comparison <- c(model_comparison, temp)
}
model_comparison <- as.data.frame(model_comparison)
rownames(model_comparison) <- c('Likelihood',rownames(inf_criterion))
colnames(model_comparison) <- short_model_names
print(model_comparison)
}
# Visualizing impact of negative previous return on variance
visualize_dependecy_ret_var <- function(garch_fits, short_model_names) {
p <- ggplot()
for (f in 1:length(garch_fits)) {
garch_news <- as.data.frame(newsimpact(garch_fits[[f]])[1:2])
model_name <- short_model_names[f]
model_name <- enquo(model_name)
p <- p + geom_line(data = garch_news,
aes(x = zx, y = zy, color = !!model_name))
}
p <- p + labs(x = 'Error', y = 'Variance',
title = 'Dependence of variance on errors in different models') +
theme(plot.title = element_text(hjust = 0.5))
print(p)
}
# Visualizing volatility
visualizing_volatility <- function(garch_fits, short_model_names, fund_vol) {
p <- ggplot()
garch_vol <- list()
for (f in 1:length(garch_fits)) {
garch_vol[[f]] <- sigma(garch_fits[[f]])
model_name <- short_model_names[f]
model_name <- enquo(model_name)
p <- p + geom_line(data = garch_vol[[f]], aes(x = index(garch_vol[[f]][,1]),
y = garch_vol[[f]][,1],
color = !!model_name), alpha = 0.2)
}
names(garch_vol) <- short_model_names
p <- p + geom_line(data = fund_vol, aes(y = fund_vol[,1], x = index(fund_vol[,1]),
color = 'Actual volatility')) +
labs(x = 'Date', y = 'Volatility',
title = 'Volatility constructed by different models') +
theme(plot.title = element_text(hjust = 0.5))
suppressMessages(suppressWarnings(print(p)))
return(garch_vol)
}
# Predicting volatility for n.ahead periods
predict_volatility <- function(garch_fits, garch_vol,
fund_vol, short_model_names) {
fund_tail_volatility <- tail(fund_vol,10)
predict_results <- data.frame(fund_tail_volatility)
garch_sst <- c()
for (f in 1:length(garch_fits)) {
garch_forecast <- ugarchforecast(fitORspec = garch_fits[[f]],
data = garch_vol[[f]], n.ahead = 10)
predict_results <- cbind(predict_results, sigma(garch_forecast))
garch_sst[[f]] <- sum(fund_tail_volatility - sigma(garch_forecast))
names(garch_sst)[f] <- paste('TES for',short_model_names[f])
}
names(predict_results)[2:ncol(predict_results)] <- short_model_names
# Total error sum for models
print(garch_sst)
#Comparing predicted volatility for models with actual one
print(predict_results)
}
```
```{r}
# Data loading
df <- read.csv('Funds.csv', stringsAsFactors = F)
df <- df[,-1]
df$Date <- as.Date(df$Date)
# Transforming data from df to xts
funds <- xts(df[,2:ncol(df)], order.by = df$Date)
funds <- na.omit(funds)
rm(df)
# Subseting original data for April of 2020 year
funds_red <- funds['/202004']
```
```{r}
get_volatiles <- function(funds, width = 22, time_scale = 1, funds_names) {
vol_df <- data.frame()
for (i in 1:ncol(funds)) {
fund <- funds[,i]
temp <- rollapply(data = fund, width = 22, FUN = 'sd.annualized', scale = time_scale)
if (nrow(vol_df) == 0) {
vol_df <- temp
} else {
vol_df <- cbind(vol_df, temp)
}
}
names(vol_df) <- funds_names
return(vol_df)
}
visualize_funds_lines <- function(funds, y_axis_label = 'Volatility') {
for (i in 1:ncol(funds)) {
temp <- funds[,i]
temp_mean <- mean(temp, na.rm = TRUE)
p <- ggplot(temp, aes(x = index(temp), y = temp)) +
geom_line(aes(color = 'Volatility')) +
geom_hline(aes(yintercept = temp_mean, color = 'Mean'),
size=.5, linetype='dashed') +
geom_text( aes( min(index(temp)) , temp_mean, label = round(temp_mean, 4), vjust = 2)) +
labs(x = 'Date', y = y_axis_label, title = names(funds)[i]) +
theme(plot.title = element_text(hjust = 0.5))
suppressMessages(suppressWarnings(print(p)))
}
}
visualize_funds_hist <- function(funds, x_axis_label = 'Return') {
for (i in 1:ncol(funds)) {
temp <- funds[,i]
title <- paste(funds_names[i], 'returns')
chart.Histogram(temp,
methods = c('add.normal', 'add.density'),
main = title)
temp <- (temp - mean(temp, na.rm = T))/sd(temp, na.rm = T)
title <- paste('Standardized', title)
chart.Histogram(temp,
methods = c('add.normal', 'add.density'),
main = title)
}
}
```
```{r}
evaluate_garch <- function(fund, fund_vol) {
# Models naming
model_names <- c('Standard GARCH with normal distribution of errors',
'GJR GARCH with normal distribution of errors',
'Standard GARCH with skewed Student t distribution of errors',
'GJR GARCH with skewed Student t distribution of errors')
short_model_names <- c('Normal GARCH', 'Normal GJR', 'Skewed t GARCH', 'Skewed t GJR')
garch_fits <- get_garch_fits(fund)
names(garch_fits) <- model_names
visualize_residuals(garch_fits)
test_garch_models(garch_fits)
print_garch_coefficients(garch_fits)
compare_garch_models(garch_fits, short_model_names)
visualize_dependecy_ret_var(garch_fits, short_model_names)
garch_vol <- visualizing_volatility(garch_fits, short_model_names, fund_vol)
predict_volatility(garch_fits = garch_fits,
garch_vol = garch_vol,
fund_vol = fund_vol,
short_model_names = short_model_names)
}
```
```{r}
vol_df <- get_volatiles(funds, funds_names = funds_names)
vol_df <- na.omit(vol_df)
visualize_funds_lines(vol_df)
visualize_funds_hist(funds)
fund_tail_volatility <- tail(vol_df$Vanguard, 10)
```
```{r}
cat('Data up to 2020-04-30\n')
for (i in 1:ncol(funds_red)) {
cat(colnames(vol_df)[i],'\n')
evaluate_garch(funds_red[,i], vol_df[,i])
cat('\n--------------------------------\n')
}
```
```{r}
cat('Data up to 2020-05-08\n')
for (i in 1:ncol(funds)) {
cat(colnames(vol_df)[i],'\n')
evaluate_garch(funds_red[,i], vol_df[,i])
cat('\n--------------------------------\n')
}
```
| true | ---
title: "PI:NAME:<NAME>END_PI & PI:NAME:<NAME>END_PI. Finances Project"
output:
pdf_document: default
html_notebook: default
html_document:
df_print: paged
---
```{r}
require(dplyr)
require(ggplot2)
require(xts)
require(rugarch)
require(PerformanceAnalytics)
require(quantmod)
library(skewt)
```
```{r}
# Configure propet path to the Funds.csv file here
# setwd('./')
```
```{r}
read_funds <- function(lf) {
dfs <- list()
browser()
for (f in 1:length(lf)) {
cond <- substr(lf[f], nchar(lf[f])-3, nchar(lf[f])) == '.csv'
if (cond) {
temp <- data.frame(read.csv(lf[f], stringsAsFactors = FALSE))
temp$Date <- as.Date(temp$Date)
temp$Close <- as.numeric(temp$Close)
dfs[[f]] <- temp[,c(1,5)]
}
}
return(dfs)
}
funds_names <- c("Vanguard", "Blackrock", "Statestreet",
"JPmorgan", "Bankmellon", "Allianz")
```
```{r}
# Forming set of parameters for different GARCH model
get_garch_specs <- function() {
# Standard GARCH with normal distribution of errors
norm_garch_spec <- ugarchspec(mean.model = list(armaOrder = c(0,0)),
variance.model = list(model = 'sGARCH'),
distribution.model = 'norm')
# GJR GARCH with normal distribution of errors
norm_gjr_spec <- ugarchspec(mean.model = list(armaOrder = c(0,0)),
variance.model = list(model = 'gjrGARCH'),
distribution.model = 'norm')
# Standard GARCH with skewed Student t distribution of errors
sstd_garch_spec <- ugarchspec(mean.model = list(armaOrder = c(0,0)),
variance.model = list(model = 'sGARCH'),
distribution.model = 'sstd')
# GJR GARCH with skewed Student t distribution of errors
sstd_gjr_spec <- ugarchspec(mean.model = list(armaOrder = c(0,0)),
variance.model = list(model = 'gjrGARCH'),
distribution.model = 'sstd')
garch_specs <- list(norm_garch_spec, norm_gjr_spec,
sstd_garch_spec, sstd_gjr_spec)
return(garch_specs)
}
```
```{r}
# Apply GARCH model to our data
get_garch_fits <- function(fund) {
garch_specs <- get_garch_specs()
garch_fits <- list()
for (s in 1:length(garch_specs)) {
suppressWarnings(garch_fits[[s]] <- ugarchfit(data = fund,
spec = garch_specs[[s]]))
}
return(garch_fits)
}
# Visualizing standardized residuals for models
visualize_residuals <- function(garch_fits) {
for (f in 1:length(garch_fits)) {
chart.Histogram(residuals(garch_fits[[f]], standardize = T),
methods = c('add.normal', 'add.density'),
main = paste('Standardized residuals of',names(garch_fits)[f]))
}
}
# Models validation
test_garch_models <- function(garch_fits) {
for (f in 1:length(garch_fits)) {
standard_residuals <- residuals(garch_fits[[f]], standardize = T)
p <- acf(abs(standard_residuals),22, plot = F)
plot(p, main = names(garch_fits)[f])
cat('\n', names(garch_fits)[f],'\n')
print(Box.test(abs(standard_residuals), 22, type = 'Ljung-Box'))
}
}
#Coefficients
print_garch_coefficients <- function(garch_fits) {
for (f in 1:length(garch_fits)) {
cat('\nCoefficients of', names(garch_fits)[f], '\n')
print(round(garch_fits[[f]]@fit$matcoef,10))
cat('\nRobust coefficients of', names(garch_fits)[f], '\n')
print(round(garch_fits[[f]]@fit$robust.matcoef,10))
}
}
# Models comparing
compare_garch_models <- function(garch_fits, short_model_names) {
model_comparison <- data.frame()
for (f in 1:length(garch_fits)) {
temp <- data.frame()
temp[1,1] <- likelihood(garch_fits[[f]])
inf_criterion <- infocriteria(garch_fits[[f]])
temp <- rbind(temp, inf_criterion)
model_comparison <- c(model_comparison, temp)
}
model_comparison <- as.data.frame(model_comparison)
rownames(model_comparison) <- c('Likelihood',rownames(inf_criterion))
colnames(model_comparison) <- short_model_names
print(model_comparison)
}
# Visualizing impact of negative previous return on variance
visualize_dependecy_ret_var <- function(garch_fits, short_model_names) {
p <- ggplot()
for (f in 1:length(garch_fits)) {
garch_news <- as.data.frame(newsimpact(garch_fits[[f]])[1:2])
model_name <- short_model_names[f]
model_name <- enquo(model_name)
p <- p + geom_line(data = garch_news,
aes(x = zx, y = zy, color = !!model_name))
}
p <- p + labs(x = 'Error', y = 'Variance',
title = 'Dependence of variance on errors in different models') +
theme(plot.title = element_text(hjust = 0.5))
print(p)
}
# Visualizing volatility
visualizing_volatility <- function(garch_fits, short_model_names, fund_vol) {
p <- ggplot()
garch_vol <- list()
for (f in 1:length(garch_fits)) {
garch_vol[[f]] <- sigma(garch_fits[[f]])
model_name <- short_model_names[f]
model_name <- enquo(model_name)
p <- p + geom_line(data = garch_vol[[f]], aes(x = index(garch_vol[[f]][,1]),
y = garch_vol[[f]][,1],
color = !!model_name), alpha = 0.2)
}
names(garch_vol) <- short_model_names
p <- p + geom_line(data = fund_vol, aes(y = fund_vol[,1], x = index(fund_vol[,1]),
color = 'Actual volatility')) +
labs(x = 'Date', y = 'Volatility',
title = 'Volatility constructed by different models') +
theme(plot.title = element_text(hjust = 0.5))
suppressMessages(suppressWarnings(print(p)))
return(garch_vol)
}
# Predicting volatility for n.ahead periods
predict_volatility <- function(garch_fits, garch_vol,
fund_vol, short_model_names) {
fund_tail_volatility <- tail(fund_vol,10)
predict_results <- data.frame(fund_tail_volatility)
garch_sst <- c()
for (f in 1:length(garch_fits)) {
garch_forecast <- ugarchforecast(fitORspec = garch_fits[[f]],
data = garch_vol[[f]], n.ahead = 10)
predict_results <- cbind(predict_results, sigma(garch_forecast))
garch_sst[[f]] <- sum(fund_tail_volatility - sigma(garch_forecast))
names(garch_sst)[f] <- paste('TES for',short_model_names[f])
}
names(predict_results)[2:ncol(predict_results)] <- short_model_names
# Total error sum for models
print(garch_sst)
#Comparing predicted volatility for models with actual one
print(predict_results)
}
```
```{r}
# Data loading
df <- read.csv('Funds.csv', stringsAsFactors = F)
df <- df[,-1]
df$Date <- as.Date(df$Date)
# Transforming data from df to xts
funds <- xts(df[,2:ncol(df)], order.by = df$Date)
funds <- na.omit(funds)
rm(df)
# Subseting original data for April of 2020 year
funds_red <- funds['/202004']
```
```{r}
get_volatiles <- function(funds, width = 22, time_scale = 1, funds_names) {
vol_df <- data.frame()
for (i in 1:ncol(funds)) {
fund <- funds[,i]
temp <- rollapply(data = fund, width = 22, FUN = 'sd.annualized', scale = time_scale)
if (nrow(vol_df) == 0) {
vol_df <- temp
} else {
vol_df <- cbind(vol_df, temp)
}
}
names(vol_df) <- funds_names
return(vol_df)
}
visualize_funds_lines <- function(funds, y_axis_label = 'Volatility') {
for (i in 1:ncol(funds)) {
temp <- funds[,i]
temp_mean <- mean(temp, na.rm = TRUE)
p <- ggplot(temp, aes(x = index(temp), y = temp)) +
geom_line(aes(color = 'Volatility')) +
geom_hline(aes(yintercept = temp_mean, color = 'Mean'),
size=.5, linetype='dashed') +
geom_text( aes( min(index(temp)) , temp_mean, label = round(temp_mean, 4), vjust = 2)) +
labs(x = 'Date', y = y_axis_label, title = names(funds)[i]) +
theme(plot.title = element_text(hjust = 0.5))
suppressMessages(suppressWarnings(print(p)))
}
}
visualize_funds_hist <- function(funds, x_axis_label = 'Return') {
for (i in 1:ncol(funds)) {
temp <- funds[,i]
title <- paste(funds_names[i], 'returns')
chart.Histogram(temp,
methods = c('add.normal', 'add.density'),
main = title)
temp <- (temp - mean(temp, na.rm = T))/sd(temp, na.rm = T)
title <- paste('Standardized', title)
chart.Histogram(temp,
methods = c('add.normal', 'add.density'),
main = title)
}
}
```
```{r}
evaluate_garch <- function(fund, fund_vol) {
# Models naming
model_names <- c('Standard GARCH with normal distribution of errors',
'GJR GARCH with normal distribution of errors',
'Standard GARCH with skewed Student t distribution of errors',
'GJR GARCH with skewed Student t distribution of errors')
short_model_names <- c('Normal GARCH', 'Normal GJR', 'Skewed t GARCH', 'Skewed t GJR')
garch_fits <- get_garch_fits(fund)
names(garch_fits) <- model_names
visualize_residuals(garch_fits)
test_garch_models(garch_fits)
print_garch_coefficients(garch_fits)
compare_garch_models(garch_fits, short_model_names)
visualize_dependecy_ret_var(garch_fits, short_model_names)
garch_vol <- visualizing_volatility(garch_fits, short_model_names, fund_vol)
predict_volatility(garch_fits = garch_fits,
garch_vol = garch_vol,
fund_vol = fund_vol,
short_model_names = short_model_names)
}
```
```{r}
vol_df <- get_volatiles(funds, funds_names = funds_names)
vol_df <- na.omit(vol_df)
visualize_funds_lines(vol_df)
visualize_funds_hist(funds)
fund_tail_volatility <- tail(vol_df$Vanguard, 10)
```
```{r}
cat('Data up to 2020-04-30\n')
for (i in 1:ncol(funds_red)) {
cat(colnames(vol_df)[i],'\n')
evaluate_garch(funds_red[,i], vol_df[,i])
cat('\n--------------------------------\n')
}
```
```{r}
cat('Data up to 2020-05-08\n')
for (i in 1:ncol(funds)) {
cat(colnames(vol_df)[i],'\n')
evaluate_garch(funds_red[,i], vol_df[,i])
cat('\n--------------------------------\n')
}
```
|
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[{"context":"X scATAC multiome data - PBMC 3K unsort\"\nauthor: \"Cankun\"\ndate: \"XXX\"\noutput:\n(...TRUNCATED) | vignettes/regulon.rmd | Wang-Cankun/iris3api | 0 | "---\ntitle: \"IRIS3 - 10X scATAC multiome data - PBMC 3K unsort\"\nauthor: \"Cankun\"\ndate: \"XXX\(...TRUNCATED) | 2553 | "---\ntitle: \"IRIS3 - 10X scATAC multiome data - PBMC 3K unsort\"\nauthor: \"<NAME>\"\ndate: \"XXX\(...TRUNCATED) | true | "---\ntitle: \"IRIS3 - 10X scATAC multiome data - PBMC 3K unsort\"\nauthor: \"PI:NAME:<NAME>END_PI\"(...TRUNCATED) |
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