Id stringlengths 1 6 | PostTypeId stringclasses 7 values | AcceptedAnswerId stringlengths 1 6 ⌀ | ParentId stringlengths 1 6 ⌀ | Score stringlengths 1 4 | ViewCount stringlengths 1 7 ⌀ | Body stringlengths 0 38.7k | Title stringlengths 15 150 ⌀ | ContentLicense stringclasses 3 values | FavoriteCount stringclasses 3 values | CreationDate stringlengths 23 23 | LastActivityDate stringlengths 23 23 | LastEditDate stringlengths 23 23 ⌀ | LastEditorUserId stringlengths 1 6 ⌀ | OwnerUserId stringlengths 1 6 ⌀ | Tags list |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
12826 | 1 | 12989 | null | 6 | 3767 | I'm interested in an estimator of the standard deviation in a Poisson regression. So the variance is
$$Var(y)=\phi\cdot V(\mu)$$
where $\phi=1$ and $V(\mu)=\mu$. So the variance should be $Var(y)=V(\mu)=\mu$. (I'm just interested in how the variance should be, so if overdispersion occurs ($\widehat{\phi}\neq 1$), I don't care about it). Thus an estimator of the variance should be
$$\widehat{Var}(y)=V(\widehat{\mu})=\widehat{\mu}$$
and an estimator of the standard deviation should be
$$\sqrt{\widehat{Var}(y)}=\sqrt{V(\widehat{\mu})}=\sqrt{\widehat{\mu}}.$$
Is this correct? I haven't found a discussion about standard deviation in the context with Poisson regression yet, that's why I'm asking.
Example:
So here is an easy example (which makes no sense btw) of what I'm talking about.
```
data1 <- function(x) {x^(2)}
numberofdrugs <- data1(1:84)
data2 <- function(x) {x}
healthvalue <- data2(1:84)
plot(healthvalue, numberofdrugs)
test <- glm(numberofdrugs ~ healthvalue, family=poisson)
summary(test) #beta0=5.5 beta1=0.042
mu <- function(x) {exp(5.5+0.042*x)}
plot(healthvalue, numberofdrugs)
curve(mu, add=TRUE, col="purple", lwd=2)
# the purple curve is the estimator for mu and it's also
# the estimator of the variance,but if I'd like to plot
# the (not constant) standard deviation I just take the
# square root of the variance. So it is var(y)=mu=exp(Xb)
# and thus the standard deviation is sqrt(exp(Xb))
sd <- function(x) {sqrt(exp(5.5+0.042*x))}
curve(sd, col="green", lwd=2)
```
Is the the green curve the correct estimator of the standard deviation in a Poisson regression? It should be, no?
| Estimating standard deviation in Poisson regression | CC BY-SA 3.0 | 0 | 2011-07-09T08:51:15.067 | 2011-07-13T14:53:14.843 | 2011-07-13T14:29:03.643 | 919 | 4496 | [
"regression",
"standard-deviation",
"poisson-distribution",
"generalized-linear-model"
] |
12827 | 2 | null | 12825 | 5 | null | You can't just "create a distribution" with a given form for the mean-variance relationship. In this case and others where quasi-likelihood is used, there is no probability distribution with the properties you desire.
That's why it's called [quasi-likelihood](http://en.wikipedia.org/wiki/Quasilikelihood):
>
"a quasi-likelihood function is not the log-likelihood corresponding to any actual probability distribution".
| null | CC BY-SA 3.0 | null | 2011-07-09T09:15:20.097 | 2011-07-09T09:15:20.097 | null | null | 449 | null |
12828 | 1 | 19380 | null | 7 | 1489 | Suppose I've data for 100 individuals for 5 variables, say Var1, Var2,...Var5. I run the cluster analysis using these 5 variables on these 100 rows & got 3 clusters. Now, I want to differentiate these 3 clusters based on 5 variables. That is, which among these 5 variables has been loaded more for which cluster, in order to get a meaningful interpretation of the clusters. Here I don't want to do PCA or other factor analysis.
I've heard that I can do that using discriminant analysis. Can anybody suggest me the method to do it?
| Classifying clusters using discriminant analysis | CC BY-SA 3.0 | null | 2011-07-09T09:33:29.813 | 2011-12-05T18:18:20.130 | 2011-07-09T09:45:33.007 | null | 4278 | [
"r",
"clustering",
"discriminant-analysis"
] |
12830 | 2 | null | 12768 | 2 | null | Based on your dataset and the above comments as well as your other post on [run time problem with lmer](https://stats.stackexchange.com/questions/12814/problem-with-three-level-hierarchical-regression-using-lmer), you'll need to specify that `choicenum` and `ipnum` are factors or `lmer` will treat them as covariates. This is probably what was causing your error message that group1:group2 is an interaction. I ran the model on your dataset as described and it worked fine.
```
dataset$choicenum <- as.factor(dataset$choicenum)
dataset$ipnum <- as.factor(dataset$ipnum)
mymodel <- lmer(ene ~ videocond + choicenum + (1|ipnum/choicenum),data=dataset)
```
| null | CC BY-SA 3.0 | null | 2011-07-09T10:36:59.340 | 2011-07-09T22:19:52.830 | 2017-04-13T12:44:37.583 | -1 | 3774 | null |
12831 | 2 | null | 12822 | 1 | null | If you have chronological data i.e.time series data then there are "knowns" and waiting to be discovered are the "unknowns" . For example if you have a sequence of data points for 10 periods such as 1,9,1,9,1,5,1,9,1,9 then based upon this sample one can reasonably expect 1,9,1,9,... to arise in the future. What data analysis reveals is that there is an "unusual" reading at period 6 even though it is well within +-3 sigma limits suggesting that the DGF did not hold. Unmasking the Inlier/Outlier allows us to reveal things about the data. We also note that the Mean Value is not the Expected Value. This idea easily extends to detecting Mean Shifts and/or Local Time Trends that may have been unknown before the data was analyzed ( Hypothesis Generation ). Now it is quite possible that the next 10 readings are also 1,9,1,9,1,5,1,9,1,9 suggesting that the "5" is not necessarily untoward. If we observe an error process from a suitable model that exhibits provable non-constant variance we might be revealing one of the following states of nature: 1) the parameters might have changed at a particular point in time ; 2. There may be a need for Weighted Analysis (GLS) ; 3. There may be a need to transform the data via a power transform; 4. There may be a need to actually model the variance of the errors. If you have daily data good analysis might reveal that there is a window of response (lead,contemporaneous and lag structure) around each Holiday reflecting consistent/predictable behavior. You might also be able to reveal that certain days of the month have a significant effect or that Fridays before a Monday holiday have exceptional activity.
| null | CC BY-SA 3.0 | null | 2011-07-09T10:42:38.040 | 2011-07-09T10:42:38.040 | null | null | 3382 | null |
12833 | 2 | null | 12822 | 6 | null | There's a whole field of exploratory data analysis (EDA), and an excellent book on this subject called [Exploratory Data Analysis](http://rads.stackoverflow.com/amzn/click/0201076160), by John W. Tukey.
I like that you are using graphs - there are many other graphs that can be useful, depending on your data - how many variables? What nature are the variables (Categorical? Numeric? Continuous? Counts? Ordinal?)
One graph that is often useful for data with multiple variables is a scatterplot matrix.
You can look for various types of outliers, which are often interesting points.
But I don't think this whole process can be made really methodical and scientific - exploration is what comes BEFORE the methodical and scientific approaches can be brought in. Here, I think the key aspect is playfulness.
| null | CC BY-SA 3.0 | null | 2011-07-09T12:20:41.997 | 2016-08-21T15:48:10.120 | 2016-08-21T15:48:10.120 | 7290 | 686 | null |
12834 | 2 | null | 12824 | 4 | null | A paired t-test is for testing a difference in means between two groups that are somehow connected. For example, do husbands and wives have equal IQs? Are people right and left eyes equally acute? and so on.
You have 3 groups, not 2. You have no matching mechanism that I can see. So, while you haven't really given us enough information, I can't see how the answer to your question could be "yes".
| null | CC BY-SA 3.0 | null | 2011-07-09T12:25:56.310 | 2011-07-09T12:25:56.310 | null | null | 686 | null |
12836 | 2 | null | 1184 | 0 | null | Just for reference, an [implementation](http://factbased.blogspot.com/2011/07/index-decomposition-with-r_09.html) of two index decomposition methods is now available.
| null | CC BY-SA 3.0 | null | 2011-07-09T16:18:31.180 | 2011-07-09T16:47:49.890 | 2011-07-09T16:47:49.890 | 573 | 573 | null |
12837 | 2 | null | 12816 | 0 | null | A combination of "is.na" and "lag" should do what you want, but, as previous commentators have pointed out, this may not be the best method of imputation.
| null | CC BY-SA 3.0 | null | 2011-07-09T17:17:00.117 | 2011-07-09T17:17:00.117 | null | null | 2817 | null |
12839 | 1 | null | null | 3 | 496 | I want to do a meta-analysis with count data for treatement/control cases (abundance of sea polichaetos). My problem is that I need to use zero values (an informative zero value) for the mean and standard deviation for one of the treatement, but R has a problem: "Studies with zero values for sd.e or sd.c get no weight in meta-analysis". I can agroup the case by Family (`byvar=Family`).
There is an option in `metabin`, where you could add either a numerical value to each cell frequency for studies with a zero cell count or the character string "TA" which stands for treatment arm continuity correction, but I want to use `metacont` which don´t have this option.
Can you help me? Thanks!
```
> datos<-read.table(file="datos_totales.csv",head=TRUE,sep=",",dec=",")
> datos[10:20,]
Study ControlN ControlMean ControlSD TreatedN TreatedMean TreatedSD Family
10 Ali.Gmb.S1.Inv 3 893.3333 180.36999 3 1213.33333 410.52812 Capitellidae
11 Ali.Gmb.S2.Inv 3 640.0000 348.71192 3 666.66667 220.30282 Capitellidae
12 Ali.Gmb.S3.Inv 3 426.6667 184.75209 3 920.00000 628.64935 Capitellidae
13 Ali.Cul.S1.Ver 3 213.3333 115.47005 3 0.00000 0.00000 Cirratulidae
14 Ali.Cul.S2.Ver 3 160.0000 144.22205 3 40.00000 40.00000 Cirratulidae
15 Ali.Cul.S3.Ver 3 293.3333 234.37861 3 26.66667 46.18802 Cirratulidae
16 Ali.Cul.S1.Inv 3 653.3333 532.66625 3 13.33333 23.09401 Cirratulidae
17 Ali.Cul.S2.Inv 3 706.6667 335.45988 3 26.66667 23.09401 Cirratulidae
18 Ali.Cul.S3.In 3 666.6667 674.48746 3 80.00000 80.00000 Cirratulidae
19 Ali.Gmb.S1.Vr 3 280.0000 69.28203 3 0.00000 0.00000 Cirratulidae
20 Ali.Gmb.S2.Ver 3 226.6667 220.30282 3 0.00000 0.00000 Cirratulidae
> library(meta)
> metaanalisis2<- metacont(TreatedN, TreatedMean, TreatedSD, ControlN, ControlMean,
ControlSD, data=datos, byvar=Family, label.c="Control", print.byvar=FALSE,
label.e="Treatment", studlab=Study)
**Mensajes de aviso perdidos
In metacont(TreatedN, TreatedMean, TreatedSD, ControlN, ControlMean, :
Studies with zero values for sd.e or sd.c get no weight in meta-analysis**
```
| Meta-analysis with zero values for mean and sd | CC BY-SA 3.0 | null | 2011-07-09T17:45:39.880 | 2022-11-29T14:23:31.707 | 2022-11-29T14:23:31.707 | 11887 | 5301 | [
"r",
"meta-analysis",
"biostatistics"
] |
12840 | 2 | null | 12816 | 2 | null | To just technically answer your question
```
set.seed(5)
a <- rnorm(20);
a[sample(1:20,4)] <- NA;
```
`a` is:
```
[1] -0.84085548 1.38435934 -1.25549186 0.07014277 NA -0.60290798
[7] -0.47216639 -0.63537131 -0.28577363 NA 1.22763034 -0.80177945
[13] -1.08039260 -0.15753436 NA -0.13898614 NA -2.18396676
[19] 0.24081726 -0.25935541
```
To set each NA to the previous value:
```
NAs <- which(is.na(a))
a[NAs] <- a[NAs-1]
```
giving
```
[1] -0.84085548 1.38435934 -1.25549186 0.07014277 0.07014277 -0.60290798
[7] -0.47216639 -0.63537131 -0.28577363 -0.28577363 1.22763034 -0.80177945
[13] -1.08039260 -0.15753436 -0.15753436 -0.13898614 -0.13898614 -2.18396676
[19] 0.24081726 -0.25935541
```
Note that this fails if first value is missing
| null | CC BY-SA 3.0 | null | 2011-07-09T17:51:01.607 | 2011-07-09T17:51:01.607 | null | null | 582 | null |
12842 | 1 | 12844 | null | 76 | 190235 | I read from my textbook that $\text{cov}(X,Y)=0$ does not guarantee X and Y are independent. But if they are independent, their covariance must be 0. I could not think of any proper example yet; could someone provide one?
| Covariance and independence? | CC BY-SA 3.0 | null | 2011-07-09T19:47:37.230 | 2023-03-22T19:09:01.843 | 2011-07-11T22:28:23.007 | 930 | 3525 | [
"independence",
"covariance"
] |
12843 | 1 | 12851 | null | 19 | 41685 | I am trying to generate random samples from a custom pdf using R. My pdf is:
$$f_{X}(x) = \frac{3}{2} (1-x^2), 0 \le x \le 1$$
I generated uniform samples and then tried to transform it to my custom distribution. I did this by finding the cdf of my distribution ($F_{X}(x)$) and setting it to the uniform sample ($u$) and solving for $x$.
$$ F_{X}(x) = \Pr[X \le x] = \int_{0}^{x} \frac{3}{2} (1-y^2) dy = \frac{3}{2} (x - \frac{x^3}{3}) $$
To generate a random sample with the above distribution, get a uniform sample $u \in[0,1]$ and solve for $x$ in $$\frac{3}{2} (x - \frac{x^3}{3}) = u $$
I implemented it in `R` and I don't get the expected distribution. Can anyone point out the flaw in my understanding?
```
nsamples <- 1000;
x <- runif(nsamples);
f <- function(x, u) {
return(3/2*(x-x^3/3) - u);
}
z <- c();
for (i in 1:nsamples) {
# find the root within (0,1)
r <- uniroot(f, c(0,1), tol = 0.0001, u = x[i])$root;
z <- c(z, r);
}
```
| Generating random samples from a custom distribution | CC BY-SA 3.0 | null | 2011-07-09T20:24:10.803 | 2014-03-07T19:54:04.130 | 2011-07-09T21:56:55.660 | 2513 | 2513 | [
"r",
"sampling",
"uniform-distribution"
] |
12844 | 2 | null | 12842 | 73 | null | Easy example: Let $X$ be a random variable that is $-1$ or $+1$ with probability 0.5. Then let $Y$ be a random variable such that $Y=0$ if $X=-1$, and $Y$ is randomly $-1$ or $+1$ with probability 0.5 if $X=1$.
Clearly $X$ and $Y$ are highly dependent (since knowing $Y$ allows me to perfectly know $X$), but their covariance is zero: They both have zero mean, and
$$\eqalign{
\mathbb{E}[XY] &=&(-1) &\cdot &0 &\cdot &P(X=-1) \\
&+& 1 &\cdot &1 &\cdot &P(X=1,Y=1) \\
&+& 1 &\cdot &(-1)&\cdot &P(X=1,Y=-1) \\
&=&0.
}$$
Or more generally, take any distribution $P(X)$ and any $P(Y|X)$ such that $P(Y=a|X) = P(Y=-a|X)$ for all $X$ (i.e., a joint distribution that is symmetric around the $x$ axis), and you will always have zero covariance. But you will have non-independence whenever $P(Y|X) \neq P(Y)$; i.e., the conditionals are not all equal to the marginal. Or ditto for symmetry around the $y$ axis.
| null | CC BY-SA 3.0 | null | 2011-07-09T21:03:44.923 | 2011-07-11T18:05:57.753 | 2011-07-11T18:05:57.753 | 919 | 5289 | null |
12845 | 1 | 12850 | null | 1 | 54 | This will sound weird, but I'm trying to calculate something that I don't know the name.
I want to calculate how big/low a variable is in relation to the others. Like Google webmaster tools says "your site loads in 3s, this is slower than 59% of sites". How can I calculate this?
For example, the points a player has. He has 4440 points and he is ranked on #437 out of 16801 players. The average of points in 976 and the top player has 30076 points.
| Calculate how big a variable is in relation to others | CC BY-SA 3.0 | null | 2011-07-09T21:35:23.160 | 2011-07-11T17:17:05.473 | 2011-07-11T17:17:05.473 | 919 | 5341 | [
"distributions",
"descriptive-statistics",
"quantiles",
"ranking"
] |
12846 | 2 | null | 12805 | 13 | null | I think it's perfectly well justified. (In fact, I've use this convention in papers I've published; or you can call them "nats" if you prefer to stick with logarithms of base $e$).
The justification runs as follows: the log-likelihood of the fitted model can be viewed as a Monte Carlo estimate of the [KL divergence](http://en.wikipedia.org/wiki/Kullback%E2%80%93Leibler_divergence) between the "true" (unknown) data distribution and the distribution implied by the fitted model. Let $P(x)$ denote the "true" distribution of the data, and let $P_\theta(x)$ denote the distribution (i.e., the likelihood $P(x|\theta))$ provided by a model.
Maximum likelihood fitting involves maximizing
$L(\theta) = \frac{1}{N}\sum_i \log P_\theta(x_i) \approx \int P(x) \log P_\theta(x) dx$
The left hand side (the log-likelihood, scaled by the # datapoints $N$) is a Monte Carlo estimate for the right hand side, i.e., since the datapoints $x_i$ were drawn from $P(x)$. So we can rewrite
$L(\theta) \approx \int P(x) \log P_\theta(x) dx = \int P(x) \log \frac{P_\theta(x)}{P(x)} dx + \int P(x) \log P(x)dx$
$ = -D_{KL}(P,P_\theta) - H(x)$
So the log-likelihood normalized by the number of points is an estimate of the (negative) KL-divergence between $P$ and $P_\theta$ minus the (true) entropy of $x$. The KL divergence has units of "bits" (if we use log 2), and can be understood as the number of "extra bits" you would need to encode data from $P(x)$ using a codebook based on $P_\theta(x)$. (If $P = P_\theta$, you don't need any extra bits, so KL divergence is zero).
Now: when you take the log-likelihood ratio of two different models, it should be obvious that you end up with:
$\log \frac{P_{\theta_1(x)}}{P_{\theta_2}(x)} \approx D_{KL}(P,P_{\theta_2}) - D_{KL}(P,P_{\theta_1})$
The entropy $H(x)$ terms cancel. So the log-likelihood ratio (normalized by $N$) is an estimate of the difference between the KL divergence of the true distribution and the distribution provided by model 1, and the true distribution provided by model 2. It's therefore an estimate of the number of "extra bits" you need to code your data with model 2 compared to coding it with model 1. So I think the "bits" units are perfectly well justified.
One important caveat: when using this statistic for model-comparison, you should really use LLR computed on cross-validated data. The log-likelihood of training data is generally artificially high (favoring the model with more parameters) due to overfitting. That is, the model assigns this data higher probability than it would if it were fit to an infinite set of training data and then evaluated at the points $x_i \dots x_N$ in your dataset. So the procedure many people follow is to:
- train models 1 and 2 using training data;
- evaluate the log-likelihood ratio of a test dataset and report the resulting number in units of bits as a measure of the improved "code" provided by model 1 compared to model
The LLR evaluated on training data would generally give an unfair advantage to the model with more parameters / degrees of freedom.
| null | CC BY-SA 3.0 | null | 2011-07-09T21:38:42.010 | 2014-12-28T05:43:48.117 | 2014-12-28T05:43:48.117 | 62321 | 5289 | null |
12849 | 2 | null | 12823 | 5 | null | You want to test whether $$p_A-p_B>0,$$ where $$p_A, p_B$$ are the accuracies of the classifiers. To test this, you need an estimate of $$p_A-p_B$$ and $$Var(p_A-p_B)=Var(p_A)+Var(p_B)-2Cov(p_A,p_B)$$. Without knowing the samples that each classifier gets right/wrong, you won't be able to estimate the covariance, thus you can't statistically compare the classifiers.
| null | CC BY-SA 3.0 | null | 2011-07-09T23:34:54.450 | 2011-07-09T23:34:54.450 | null | null | 4797 | null |
12850 | 2 | null | 12845 | 3 | null | You can make various statements about relations, such as
- Your player has $\frac{4440}{976} \approx 4.5$ times the average number of points.
- He has $\frac{4440}{30076} \approx 15\%$ of the highest score.
- His points total is lower than that of $\frac{436}{16801} \approx 2.6\%$ of players.
| null | CC BY-SA 3.0 | null | 2011-07-09T23:38:06.087 | 2011-07-09T23:38:06.087 | null | null | 2958 | null |
12851 | 2 | null | 12843 | 11 | null | It looks like you figured out that your code works, but @Aniko pointed out that you could improve its efficiency. Your biggest speed gain would probably come from pre-allocating memory for `z` so that you're not growing it inside a loop. Something like `z <- rep(NA, nsamples)` should do the trick. You may get a small speed gain from using `vapply()` (which specifies the returned variable type) instead of an explicit loop (there's a great [SO question](https://stackoverflow.com/questions/2275896/is-rs-apply-family-more-than-syntactic-sugar) on the apply family).
```
> nsamples <- 1E5
> x <- runif(nsamples)
> f <- function(x, u) 1.5 * (x - (x^3) / 3) - u
> z <- c()
>
> # original version
> system.time({
+ for (i in 1:nsamples) {
+ # find the root within (0,1)
+ r <- uniroot(f, c(0,1), tol = 0.0001, u = x[i])$root
+ z <- c(z, r)
+ }
+ })
user system elapsed
49.88 0.00 50.54
>
> # original version with pre-allocation
> z.pre <- rep(NA, nsamples)
> system.time({
+ for (i in 1:nsamples) {
+ # find the root within (0,1)
+ z.pre[i] <- uniroot(f, c(0,1), tol = 0.0001, u = x[i])$root
+ }
+ })
user system elapsed
7.55 0.01 7.78
>
>
>
> # my version with sapply
> my.uniroot <- function(x) uniroot(f, c(0, 1), tol = 0.0001, u = x)$root
> system.time({
+ r <- vapply(x, my.uniroot, numeric(1))
+ })
user system elapsed
6.61 0.02 6.74
>
> # same results
> head(z)
[1] 0.7803198 0.2860108 0.5153724 0.2479611 0.3451658 0.4682738
> head(z.pre)
[1] 0.7803198 0.2860108 0.5153724 0.2479611 0.3451658 0.4682738
> head(r)
[1] 0.7803198 0.2860108 0.5153724 0.2479611 0.3451658 0.4682738
```
And you don't need the `;` at the end of each line (are you a MATLAB convert?).
| null | CC BY-SA 3.0 | null | 2011-07-09T23:59:35.713 | 2011-07-09T23:59:35.713 | 2017-05-23T12:39:27.620 | -1 | 1445 | null |
12852 | 2 | null | 12824 | 1 | null | Definitely not paired t-tests. Your samples (the classes) are independent. Assuming your dependant variable is a continuous score, you could run an Anova analysis, which is a generalization of the t-test for more than 2 groups. You could also run 2 or 3 distinct independant samples t-tests, but you'll need to take into account the risk of family-wise error. Look for "bonferroni adjustment" for more details.
| null | CC BY-SA 3.0 | null | 2011-07-10T01:23:16.527 | 2011-07-10T01:23:16.527 | null | null | 4754 | null |
12853 | 1 | 12876 | null | 17 | 7567 | I am trying to perform document-level clustering. I constructed the term-document frequency matrix and I am trying to cluster these high dimensional vectors using k-means. Instead of directly clustering, what I did was to first apply LSA's (Latent Semantic Analysis) singular vector decomposition to obtain the U,S,Vt matrices, selected a suitable threshold using the scree plot and applied clustering on the reduced matrices (specifically Vt because it gives me a concept-document information) which seemed to be giving me good results.
I've heard some people say SVD (singular vector decomposition) is clustering (by using cosine similarity measure etc.) and was not sure if I could apply k-means on the output of SVD. I thought it was logically correct because SVD is a dimensionality reduction technique, gives me a bunch of new vectors. k-means, on the other hand, will take the number of clusters as the input and divide these vectors into the specified number of clusters. Is this procedure flawed or are there ways in which this can be improved? Any suggestions?
| When do we combine dimensionality reduction with clustering? | CC BY-SA 3.0 | null | 2011-07-10T01:30:54.663 | 2018-06-19T06:28:26.880 | 2018-06-19T06:28:05.063 | 3277 | 2164 | [
"clustering",
"pca",
"dimensionality-reduction",
"text-mining",
"svd"
] |
12854 | 1 | null | null | 25 | 107752 | What is the limit to the number of independent variables one may enter in a multiple regression equation? I have 10 predictors that I would like to examine in terms of their relative contribution to the outcome variable. Should I use a bonferroni correction to adjust for multiple analyses?
| Maximum number of independent variables that can be entered into a multiple regression equation | CC BY-SA 3.0 | null | 2011-07-10T02:05:05.680 | 2011-07-11T12:39:08.500 | 2011-07-10T04:12:35.960 | 183 | 4716 | [
"regression",
"predictor",
"importance",
"bonferroni"
] |
12855 | 2 | null | 12854 | 29 | null | You need to think about what you mean by a "limit".
There are limits, such as when you have more predictors than cases, you run into issues in parameter estimation (see the little R simulation at the bottom of this answer).
However, I imagine you are talking more about soft limits related to statistical power and good statistical practice.
In this case the language of "limits" is not really appropriate.
Rather, bigger sample sizes tend to make it more reasonable to have more predictors and the threshold of how many predictors is reasonable arguably falls on a continuum of reasonableness.
You may find the [discussion of rules of thumb for sample size in multiple regression](https://stats.stackexchange.com/questions/10079/rules-of-thumb-for-minimum-sample-size-for-multiple-regression) relevant, as many such rules of thumb make reference to the number of predictors.
A few points
- If you are concerned more with overall prediction than with statistical significance of individual predictors, then it is probably reasonable to include more predictors than if you are concerned with statistical significance of individual predictors.
- If you are concerned more with testing a specific statistical model that relates to your research question (e.g., as is common in many social science applications), presumably you have reasons for including particular predictors.
However, you may also have opportunities to be selective in which predictors you include (e.g., if you have multiple variables that measure a similar construct, you might only include one of them).
When doing theory based model testing, there are a lot of choices, and the decision about which predictors to include involves close connection between your theory and research question.
- I don't often see researchers using bonferroni corrections being applied to significance tests of regression coefficients.
One reasonable reason for this might be that researchers are more interested in appraising the overall properties of the model.
- If you are interested in assessing relative importance of predictors, I find it useful to examine both the bivariate relationship between the predictor and the outcome, as well as the relationship between the predictor and outcome controlling for other predictors. If you include many predictors, it is often more likely that you include predictors that are highly intercorrelated. In such cases, interpretation of both the bivariate and model based importance indices can be useful, as a variable important in a bivariate sense might be hidden in a model by other correlated predictors (I elaborate more on this here with links).
---
## A little R simulation
I wrote this little simulation to highlight the relationship between sample size and parameter estimation in multiple regression.
```
set.seed(1)
fitmodel <- function(n, k) {
# n: sample size
# k: number of predictors
# return linear model fit for given sample size and k predictors
x <- data.frame(matrix( rnorm(n*k), nrow=n))
names(x) <- paste("x", seq(k), sep="")
x$y <- rnorm(n)
lm(y~., data=x)
}
```
The `fitmodel` function takes two arguments `n` for the sample size and `k` for the number of predictors. I am not counting the constant as a predictor, but it is estimated.
I then generates random data and fits a regression model predicting a y variable from `k` predictor variables and returns the fit.
Given that you mentioned in your question that you were interested in whether 10 predictors is too much, the following function calls show what happens when the sample size is 9, 10, 11, and 12 respectively. I.e., sample size is one less than the number of predictors to two more than the number of predictors
```
summary(fitmodel(n=9, k=10))
summary(fitmodel(n=10, k=10))
summary(fitmodel(n=11, k=10))
summary(fitmodel(n=12, k=10))
```
### > summary(fitmodel(n=9, k=10))
```
Call:
lm(formula = y ~ ., data = x)
Residuals:
ALL 9 residuals are 0: no residual degrees of freedom!
Coefficients: (2 not defined because of singularities)
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.31455 NA NA NA
x1 0.34139 NA NA NA
x2 -0.45924 NA NA NA
x3 0.42474 NA NA NA
x4 -0.87727 NA NA NA
x5 -0.07884 NA NA NA
x6 -0.03900 NA NA NA
x7 1.08482 NA NA NA
x8 0.62890 NA NA NA
x9 NA NA NA NA
x10 NA NA NA NA
Residual standard error: NaN on 0 degrees of freedom
Multiple R-squared: 1, Adjusted R-squared: NaN
F-statistic: NaN on 8 and 0 DF, p-value: NA
```
Sample size is one less than the number of predictors.
It is only possible to estimate 9 parameters, one of which is the constant.
### > summary(fitmodel(n=10, k=10))
```
Call:
lm(formula = y ~ ., data = x)
Residuals:
ALL 10 residuals are 0: no residual degrees of freedom!
Coefficients: (1 not defined because of singularities)
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.1724 NA NA NA
x1 -0.3615 NA NA NA
x2 -0.4670 NA NA NA
x3 -0.6883 NA NA NA
x4 -0.1744 NA NA NA
x5 -1.0331 NA NA NA
x6 0.3886 NA NA NA
x7 -0.9886 NA NA NA
x8 0.2778 NA NA NA
x9 0.4616 NA NA NA
x10 NA NA NA NA
Residual standard error: NaN on 0 degrees of freedom
Multiple R-squared: 1, Adjusted R-squared: NaN
F-statistic: NaN on 9 and 0 DF, p-value: NA
```
Sample size is the same as the number of predictors.
It is only possible to estimate 10 parameters, one of which is the constant.
### > summary(fitmodel(n=11, k=10))
```
Call:
lm(formula = y ~ ., data = x)
Residuals:
ALL 11 residuals are 0: no residual degrees of freedom!
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.9638 NA NA NA
x1 -0.8393 NA NA NA
x2 -1.5061 NA NA NA
x3 -0.4917 NA NA NA
x4 0.3251 NA NA NA
x5 4.4212 NA NA NA
x6 0.7614 NA NA NA
x7 -0.4195 NA NA NA
x8 0.2142 NA NA NA
x9 -0.9264 NA NA NA
x10 -1.2286 NA NA NA
Residual standard error: NaN on 0 degrees of freedom
Multiple R-squared: 1, Adjusted R-squared: NaN
F-statistic: NaN on 10 and 0 DF, p-value: NA
```
Sample size is one more than the number of predictors.
All parameters are estimated including the constant.
### > summary(fitmodel(n=12, k=10))
```
Call:
lm(formula = y ~ ., data = x)
Residuals:
1 2 3 4 5 6 7 8 9 10 11
0.036530 -0.042154 -0.009044 -0.117590 0.171923 -0.007976 0.050542 -0.011462 0.010270 0.000914 -0.083533
12
0.001581
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.14680 0.11180 1.313 0.4144
x1 0.02498 0.09832 0.254 0.8416
x2 1.01950 0.13602 7.495 0.0844 .
x3 -1.76290 0.26094 -6.756 0.0936 .
x4 0.44832 0.16283 2.753 0.2218
x5 -0.76818 0.15651 -4.908 0.1280
x6 -0.33209 0.18554 -1.790 0.3244
x7 1.62276 0.21562 7.526 0.0841 .
x8 -0.47561 0.18468 -2.575 0.2358
x9 1.70578 0.31547 5.407 0.1164
x10 3.25415 0.46447 7.006 0.0903 .
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.2375 on 1 degrees of freedom
Multiple R-squared: 0.995, Adjusted R-squared: 0.9452
F-statistic: 19.96 on 10 and 1 DF, p-value: 0.1726
```
Sample size is two more than the number of predictors, and it is finally possible to estimate the fit of the overall model.
| null | CC BY-SA 3.0 | null | 2011-07-10T03:59:17.047 | 2011-07-10T04:14:29.670 | 2017-04-13T12:44:33.237 | -1 | 183 | null |
12856 | 2 | null | 12853 | 5 | null | In reply to your title "When do we combine dimensionality reduction with clustering?" rather than the full question. One possible reason is obvious: when we want to secure agaist outliers. K-means algo, if without initial centers hint, takes k most apart points in the cloud as initial centers, and right these are likely to be outliers. Preacting by PCA neutralizes outliers which lie along junior components - by projecting them onto the few senior components which are retained in PCA.
| null | CC BY-SA 4.0 | null | 2011-07-10T06:29:29.610 | 2018-06-19T06:28:26.880 | 2018-06-19T06:28:26.880 | 3277 | 3277 | null |
12857 | 1 | 12870 | null | 24 | 13727 | I want to generate two variables. One is binary outcome variable (say success / failure) and the other is age in years. I want age to be positively correlated with success. For example there should be more successes in the higher age segments than in lower. Ideally I should be in position to control degree of correlation. How do I do that?
Thanks
| Generate random correlated data between a binary and a continuous variable | CC BY-SA 3.0 | null | 2011-07-10T08:25:40.963 | 2016-06-07T09:53:43.733 | 2011-07-11T01:33:29.657 | 183 | 333 | [
"correlation",
"random-variable",
"random-generation",
"binary-data"
] |
12858 | 2 | null | 12857 | 30 | null | You can simulate [the logistic regression model](http://en.wikipedia.org/wiki/Logistic_regression).
More precisely, you can first generate values for the age variable (for example using a uniform distribution) and then compute probabilities of success using
$$\pi ( x ) = \frac{\exp(\beta_0 + \beta_1 x)}{1 + \exp(\beta_0 + \beta_1 x)}$$
where $\beta_0$ and $\beta_1$ are constant regression coefficients to be specified. In particular, $\beta_1$ controls the magnitude of association between success and age.
Having values of $\pi$, you can now generate values for the binary outcome variable using the Bernoulli distribution.
Illustrative example in R:
```
n <- 10
beta0 <- -1.6
beta1 <- 0.03
x <- runif(n=n, min=18, max=60)
pi_x <- exp(beta0 + beta1 * x) / (1 + exp(beta0 + beta1 * x))
y <- rbinom(n=length(x), size=1, prob=pi_x)
data <- data.frame(x, pi_x, y)
names(data) <- c("age", "pi", "y")
print(data)
age pi y
1 44.99389 0.4377784 1
2 38.06071 0.3874180 0
3 48.84682 0.4664019 1
4 24.60762 0.2969694 0
5 39.21008 0.3956323 1
6 24.89943 0.2988003 0
7 51.21295 0.4841025 1
8 43.63633 0.4277811 0
9 33.05582 0.3524413 0
10 30.20088 0.3331497 1
```
| null | CC BY-SA 3.0 | null | 2011-07-10T09:10:40.877 | 2014-12-24T01:16:45.310 | 2014-12-24T01:16:45.310 | 919 | 3019 | null |
12859 | 1 | null | null | 1 | 56 | I have few training on Hidden Markov Model. But, I intend to solve my problem by HMM. I would like to have your helps/directions to me.
Here, I have two variables to define the 8 one-dimension space (`coordinate.1`, `coordinate.2`). In this one-dimension space, there are two sequences of values (shared and specific). This means I would like to detect/guess/predict the regions (defined by two coordinates variables) which are significantly dominant (with higher values) by shared/specific in some consecutive cells (units of the `coordinate.2` in the dummy). I would like to get the coordinates (the 1st two variables) of such dominant regions for shared or specific.
I use R to construct a dummy (the real data are more complex than this
Here is my data set:
```
mydata <- data.frame(coordinate.1=rep(1:8, each=25),
coordinate.2=rep(seq(100, 2500, 100), 8),
shared=rep(c(100,30,100), c(5,15,5)),
specific=rep(c(25,90,20,30), c(5,7,8,5)))
```
Here is how I happen to make a plot:
```
library(ggplot2)
pdf("shared_specific.pdf", width = 14, height = 8)
p.test<-ggplot(mydata, aes(coordinate.2)) +
geom_line(aes(y = shared, colour = "shared")) +
geom_line(aes(y = specific, colour = "specific")) +
facet_grid(coordinate.1 ~., scales = "free_x") +
scale_x_continuous("coordinate.2") +
scale_y_continuous("shared and specific")
p.test
dev.off()
```
| How to predict significant dominant regions of two sequence of numeric values by Hidden Markov Model? | CC BY-SA 3.0 | null | 2011-07-10T09:40:40.977 | 2022-06-22T20:52:25.993 | 2022-06-22T20:52:25.993 | 11887 | 4650 | [
"r",
"regression",
"biostatistics",
"hidden-markov-model"
] |
12860 | 1 | 12863 | null | 4 | 1141 | I was not sure where to put this question, so I put it here. Feel free to move it to another stack exchange site moderators.
Lets say I have a 10 gigs of pictures (or for that matter any type of data, please don't answer the question specifically relating to pictures). Let's pretend that I have the fastest computer in the world and all the time to use it on my hands.
What lossless compression algorithm should I use to compress these files as much as theoretically possible?
Also, if there is currently a program that will do this, please provide a link.
| Ultimate compression algorithm | CC BY-SA 3.0 | null | 2011-07-10T09:48:45.180 | 2017-05-02T12:06:10.690 | 2011-07-11T00:28:07.280 | 5343 | 5343 | [
"compression"
] |
12861 | 1 | 13246 | null | 9 | 2623 | Given a data scatterplot I can plot the data's principal components on it, as axes tiled with points which are principal components scores. You can see an example plot with the cloud (consisting of 2 clusters) and its first principle component. It is drawn easily: raw component scores are computed as data-matrix x eigenvector(s); coordinate of each score point on the original axis (V1 or V2) is score x cos-between-the-axis-and-the-component (which is the element of the eigenvector).
My question: Is it possible somehow to draw a discriminant in a similar fashion? Look at my pic please. I'd like to plot now the discriminant between two clusters, as a line tiled with discriminant scores (after discriminant analysis) as points. If yes, what could be the algo?
| Plotting a discriminant as line on scatterplot | CC BY-SA 3.0 | null | 2011-07-10T10:21:48.367 | 2011-07-19T12:30:01.880 | null | null | 3277 | [
"pca",
"discriminant-analysis",
"scatterplot"
] |
12862 | 2 | null | 12854 | 6 | null | In principle, there is no limit per se to how many predictors you can have. You can estimate 2 billion "betas" in principle. But what happens in practice is that without sufficient data, or sufficient prior information, it will not prove a very fruitful exercise. No particular parameters will be determined very well, and you will not learn much from the analysis.
Now if you don't have a lot of prior information about your model (model structure, parameter values, noise, etc.) then you will need the data to provide this information. This is usually the most common situation, which makes sense, because you usually need a pretty good reason to collect data (and spend $$$) about something you already know pretty well. If this is your situation, then a reasonable limit is to have a large number of observations per parameter. You have 12 parameters (10 slope betas, 1 intercept, and a noise parameter), so anything over 100 observations should be able to determine your parameters well enough to be able to make some conclusions.
But there is no "hard and fast" rules. With only 10 predictors you should have no trouble with computation time (get a better computer if you do). It mainly means just doing more work, because you have 11 dimensions of data to absorb - making it difficult to visualise the data. The basic principles from regression with only 1 dependent variable aren't really that different.
The problem with bonferroni correction is that for it to be a reasonable way to adjust your significance level without sacrificing too much power, you require the hypothesis that you are correcting for to be independent (i.e. learning that one hypothesis is true tells you nothing about whether another hypothesis is true). This is not true for the standard "t-test" in multiple regression for a co-efficient being zero, for example. The test statistic depends on what else in the model - which is a roundabout way of saying the hypothesis are dependent. Or, a more frequentist way of saying this is that the sampling distribution of the t-value conditional on the ith predictor being zero depends on what other parameters are zero. So using the bonferroni correction here may be actually be giving you a lower "overall" significance level than what you think.
| null | CC BY-SA 3.0 | null | 2011-07-10T10:50:56.927 | 2011-07-10T10:50:56.927 | null | null | 2392 | null |
12863 | 2 | null | 12860 | 6 | null | The answer depends on the content of your images. As there is no free lunch in lossless compression you cannot create a lossless compression algorithm which generally performs good on all input images. I.e. if you tune your compression algorithm so that it performs good on certain kind of images then there are always images where it must perform badly, meaning that it increases the filesize compared to the uncompressed representation. So you should have an idea of the image content that you are going to process.
The next question would be if you can afford lossy compression or if you require lossless compression.
In case of typical digital photos [JPEG 2000](http://en.wikipedia.org/wiki/JPEG_2000) is a good candidate, as it supports both, lossy and lossless compression and is tuned for photo content.
For lossy compression there is also the very real possibility of advances in encoder technology, e.g. the recent alternative [JPEG encoder Guetzli](https://research.googleblog.com/2017/03/announcing-guetzli-new-open-source-jpeg.html) by Google, which makes better use of specifics in human visual perception to allocate more bits to features that make a difference in perception.
For images with big areas of the same color and sharp edges, as diagrams and graphs or stylized maps, [PNG](http://en.wikipedia.org/wiki/Portable_Network_Graphics) is a good match. PNG is a lossless file format, supports transparency and achieves good compression for b/w images.
Also wikipedia has a [comparison of image file formats](http://en.wikipedia.org/wiki/Comparison_of_graphics_file_formats).
In the spirit of [Kolmogorov complexity](http://en.wikipedia.org/wiki/Kolmogorov_complexity) there might be images that can be compressed much further by finding an algorithm which generates the image but usually this applies only in special cases like fractals or simple raytraced CG images, not for typical digital photos.
Arbitrary (non-image) data
For general data [Arithmetic coding](http://en.wikipedia.org/wiki/Arithmetic_coding) is a good choice, as it can achieve nearly optimal compression (with respect to the occurance proportion of symbols in the data), when the alphabet for data representation suits the data. (E.g. a spectral representation of small chunks of typical music recordings is usually better suited for compression than a time series representation).
| null | CC BY-SA 3.0 | null | 2011-07-10T11:17:48.973 | 2017-05-02T12:06:10.690 | 2017-05-02T12:06:10.690 | 4360 | 4360 | null |
12864 | 1 | null | null | 0 | 857 | I'm interested in plotting the estimator of the standard deviation in a Poisson regression. The variance is $Var(y)=\phi⋅V(\mu)$ where $\phi=1$ and $V(\mu)=\mu$. So the variance should be $Var(y)=V(\mu)=\mu$. (I'm just interested in how the variance should be, so if overdispersion occurs $(\hat{\phi} \ne 1)$, I don't care about it). Thus an estimator of the variance should be $Var(\widehat{y})=V(\widehat{μ})=\widehat{\mu}$ and an estimator of the standard deviation should be $\sqrt{Var(\widehat{y})}=\sqrt{V(\widehat{\mu})}=\sqrt{\widehat{\mu}}=\sqrt{exp(x\widehat{\beta})}=exp(x\widehat{\beta}/2)$ when using the canonical link. Is this correct? I haven't found a discussion about standard deviation in the context with poisson regression yet, that's why I'm asking.
So here is an easy example (which makes no sense btw) of what I'm talking about.
```
>data1<-function(x){x^(2)}
>numberofdrugs<-data(1:84)
>data2<-function(x){x}
>healthvalue<-data2(1:84)
>
>plot(healthvalue,numberofdrugs)
>
>test<-glm(numberofdrugs~healthvalue, family=poisson)
>summary(test) #beta0=5.5 beta1=0.042
>
>mu<-function(x){exp(5.5+0.042*x)}
>
>plot(healthvalue,numberofdrugs)
>curve(mu, add=TRUE, col="purple", lwd=2)
>
> #the purple curve is the estimator for mu and it's also the estimator of the
> #variance,but if I'd like to plot the (not constant) standard deviation I just
> #take the squaroot of the variance. So it is var(y)=mu=exp(Xb) and thus the
> #standard deviation is sqrt(exp(Xb))
>
>sd<-function(x){sqrt(exp(5.5+0.042*x))}
>curve(sd, col="green", lwd=2)
>
```
Is the the green curve the correct estimator of the standard deviation in a Poisson regression? It should be, no?
| Standard deviation when estimating a Poisson regression using R | CC BY-SA 3.0 | 0 | 2011-07-10T12:01:37.650 | 2011-07-11T16:14:31.580 | 2011-07-11T16:14:31.580 | 919 | 4496 | [
"r",
"regression",
"poisson-distribution",
"generalized-linear-model",
"count-data"
] |
12865 | 2 | null | 12854 | 12 | null | I often look at this from the standpoint of whether a model fitted with a certain number of parameters is likely to yield predictions out-of-sample that are as accurate as predictions made on the original model development sample. Calibration curves, mean squared errors of X*Beta, and indexes of predictive discrimination are some of the measures typically used. This is where some of the rules of thumb come from, such as the 15:1 rule (an effective sample size of 15 per parameter examined or estimated).
Regarding multiplicity, a perfect adjustment for multiplicity, assuming the model holds and distributional assumptions are met, is the global test that all betas (other than the intercept) are zero. This is typically tested using a likelihood ratio or an F test.
There are two overall approaches to model development that tend to work well. (1) Have an adequate sample size and fit the entire pre-specified model, and (2) used penalized maximum likelihood estimation to allow only as many effective degrees of freedom in the the regression as the current sample size will support. [Stepwise variable selection without penalization should play no role, as this is known not to work.]
| null | CC BY-SA 3.0 | null | 2011-07-10T12:51:10.197 | 2011-07-10T12:51:10.197 | null | null | 4253 | null |
12866 | 1 | null | null | 2 | 302 | I'm getting confused with the analysis of one of my experiments. Here is how the results look like.
There were P participants in the experiments and they were split into two groups P1 and P2. Every student had to generate some events E for a set of steps S. Based on this I could count the quality of the answer for every student p for every step s.
The quality of participant p for step s was defined as: total number (generated by all the participants P) of unique events for the given step s - number of unique events generated by participant p for the step s.
So now I have a matrix with the qualities for every participant p and step s.
What measure should I use to compare the overall quality of all the groups? I would like to know which group generated results of better overall quality.
I started to count root mean squared error, but I'm not sure if this is a good way to do this.
| How to compare results from two groups? | CC BY-SA 3.0 | null | 2011-07-10T13:16:49.760 | 2017-08-10T10:33:08.913 | 2017-08-10T10:33:08.913 | 95892 | 5344 | [
"statistical-significance",
"multiple-comparisons",
"experiment-design",
"mean"
] |
12867 | 1 | null | null | 5 | 5594 | Recently I am conducting a research on the relationship between motivation/attitude variables (Gardner's model) and English language proficiency in the Philippines. I encountered a problem: missing values. I used a 160-item scale in my study, consisting of around 10 subscales, where each item has a 7-point Likert-type response set, with values from 1 to 7. Some respondents failed to answer some items.
I'd like to try "Multiple Imputation" using SPSS 18. But I have some questions, hope you can help out:
- For example, the variable "Interest in foreign languages" is measured by a 10-item (Q1-Q10) scale, but some respondents left a few items unanswered. And again, "Attitudes toward English-speaking people" is measured by 8-item (e.g., Q11-Q18) scale. I wonder if I can impute missing values on a dataset with variable names such as, "ID, sex, age, Q1, Q2, Q3, Q4,...Q18, Final grade"? Or do I really have to add up the items first to get a subscale score before "Multiple Imputation"?
- Do I have to recode those negatively worded items before "Multiple Imputation"? For example, if Q1, Q3, Q5, Q7, Q9 are negatively worded, do I have to recode them first?
- It seems AMOS 18 cannot do "Calculate Estimates" on those imputed data. Do you think I should just average the five imputed values for each missing data to get a new value, from which I can build a new dataset so that AMOS 18 will have to handle only one complete dataset, rather than the five imputed datasets plus the original? Is averaging the five imputed values the right way of "POOLING"?
| Multiple imputation on single subscale item or subscale scores? | CC BY-SA 3.0 | null | 2011-07-10T14:44:06.160 | 2017-10-21T17:23:44.257 | 2011-08-11T07:34:38.003 | 5739 | 5345 | [
"spss",
"scales",
"missing-data",
"data-imputation",
"structural-equation-modeling"
] |
12868 | 2 | null | 12860 | 1 | null | I you dont care about the time it take to compress it`s really hard to do better then DLI for image compression [https://sites.google.com/site/dlimagecomp/](https://sites.google.com/site/dlimagecomp/)
| null | CC BY-SA 3.0 | null | 2011-07-10T17:55:14.450 | 2011-07-10T17:55:14.450 | null | null | 5244 | null |
12869 | 1 | 13077 | null | 6 | 5803 | Suppose I run Multidimensional Scaling and I got the resulting plot. Can anybody suggest me how to interpret the plot. Please find one of my result below. Here I've 5 concepts which I run the MDS based on 10 variables. I'll be grateful if somebody can help me out.
Regards,
Ari
| Interpretation of MDS factor plot | CC BY-SA 3.0 | null | 2011-07-10T19:14:14.727 | 2013-03-25T01:28:08.177 | 2011-07-11T01:38:19.370 | 183 | 4278 | [
"r",
"multivariate-analysis",
"multidimensional-scaling"
] |
12870 | 2 | null | 12857 | 21 | null | @ocram's approach will certainly work. In terms of the dependence properties it's somewhat restrictive though.
Another method is to use a copula to derive a joint distribution. You can specify marginal distributions for success and age (if you have existing data this is especially simple) and a copula family. Varying the parameters of the copula will yield different degrees of dependence, and different copula families will give you various dependence relationships (e.g. strong upper tail dependence).
A recent overview of doing this in R via the copula package is available [here](http://www.jstatsoft.org/v21/i04/paper). See also the discussion in that paper for additional packages.
You don't necessarily need an entire package though; here's a simple example using a Gaussian copula, marginal success probability 0.6, and gamma distributed ages. Vary r to control the dependence.
```
r = 0.8 # correlation coefficient
sigma = matrix(c(1,r,r,1), ncol=2)
s = chol(sigma)
n = 10000
z = s%*%matrix(rnorm(n*2), nrow=2)
u = pnorm(z)
age = qgamma(u[1,], 15, 0.5)
age_bracket = cut(age, breaks = seq(0,max(age), by=5))
success = u[2,]>0.4
round(prop.table(table(age_bracket, success)),2)
plot(density(age[!success]), main="Age by Success", xlab="age")
lines(density(age[success]), lty=2)
legend('topright', c("Failure", "Success"), lty=c(1,2))
```
Output:
Table:
```
success
age_bracket FALSE TRUE
(0,5] 0.00 0.00
(5,10] 0.00 0.00
(10,15] 0.03 0.00
(15,20] 0.07 0.03
(20,25] 0.10 0.09
(25,30] 0.07 0.13
(30,35] 0.04 0.14
(35,40] 0.02 0.11
(40,45] 0.01 0.07
(45,50] 0.00 0.04
(50,55] 0.00 0.02
(55,60] 0.00 0.01
(60,65] 0.00 0.00
(65,70] 0.00 0.00
(70,75] 0.00 0.00
(75,80] 0.00 0.00
```
| null | CC BY-SA 3.0 | null | 2011-07-10T20:18:37.080 | 2011-07-10T20:18:37.080 | null | null | 26 | null |
12871 | 2 | null | 12860 | 0 | null | The more you know about what you want to compress, the more assumptions you can make and, consequently, the better you can compress. Also you have to decide between lossy and lossless. That is the most important part.
| null | CC BY-SA 3.0 | null | 2011-07-10T22:16:09.570 | 2011-07-10T22:16:09.570 | null | null | 4479 | null |
12873 | 1 | 13290 | null | 11 | 7341 | I am trying to tackle a problem which deals with the imputation of missing data from a panel data study(Not sure if I am using 'panel data study' correctly - as I learned it today.) I have total death count data for years 2003 to 2009, all the months, male & female, for 8 different districts and for 4 age groups.
The dataframe looks something like this:
```
District Gender Year Month AgeGroup TotalDeaths
Northern Male 2006 11 01-4 0
Northern Male 2006 11 05-14 1
Northern Male 2006 11 15+ 83
Northern Male 2006 12 0 3
Northern Male 2006 12 01-4 0
Northern Male 2006 12 05-14 0
Northern Male 2006 12 15+ 106
Southern Female 2003 1 0 6
Southern Female 2003 1 01-4 0
Southern Female 2003 1 05-14 3
Southern Female 2003 1 15+ 136
Southern Female 2003 2 0 6
Southern Female 2003 2 01-4 0
Southern Female 2003 2 05-14 1
Southern Female 2003 2 15+ 111
Southern Female 2003 3 0 2
Southern Female 2003 3 01-4 0
Southern Female 2003 3 05-14 1
Southern Female 2003 3 15+ 141
Southern Female 2003 4 0 4
```
For the 10 months spread over 2007 and 2008 some of the total deaths from all districts were not recorded. I am trying to estimate these missing value through a multiple imputation method. Either using Generalized Linear Models or SARIMA models.
My biggest issue is the use of software and the coding. I asked a question on Stackoverflow, where I want to extract the data into smaller groups such as this:
```
District Gender Year Month AgeGroup TotalDeaths
Northern Male 2003 1 01-4 0
Northern Male 2003 2 01-4 1
Northern Male 2003 3 01-4 0
Northern Male 2003 4 01-4 3
Northern Male 2003 5 01-4 4
Northern Male 2003 6 01-4 6
Northern Male 2003 7 01-4 5
Northern Male 2003 8 01-4 0
Northern Male 2003 9 01-4 1
Northern Male 2003 10 01-4 2
Northern Male 2003 11 01-4 0
Northern Male 2003 12 01-4 1
Northern Male 2004 1 01-4 1
Northern Male 2004 2 01-4 0
```
Going to
```
Northern Male 2006 11 01-4 0
Northern Male 2006 12 01-4 0
```
But someone suggested I should rather bring my question here - perhaps ask for a direction? Currently I am unable to enter this data as a proper time-series/panel study into R. My eventual aim is to use this data and the `amelia2` package with its functions to impute for missing `TotalDeaths` for certain months in 2007 and 2008, where the data is missing.
Any help, how to do this and perhaps suggestions on how to tackle this problem would be gratefully appreciated.
If this helps, I am trying to follow a similar approach to what Clint Roberts did in his PhD [Thesis](http://etd.ohiolink.edu/send-pdf.cgi/Roberts%20Clint.pdf?osu1211910310).
EDIT:
After creating the 'time' and 'group' variable as suggested by @Matt:
```
> head(dat)
District Gender Year Month AgeGroup Unnatural Natural Total time group
1 Khayelitsha Female 2001 1 0 0 6 6 1 Khayelitsha.Female.0
2 Khayelitsha Female 2001 1 01-4 1 3 4 1 Khayelitsha.Female.01-4
3 Khayelitsha Female 2001 1 05-14 0 0 0 1 Khayelitsha.Female.05-14
4 Khayelitsha Female 2001 1 15up 8 73 81 1 Khayelitsha.Female.15up
5 Khayelitsha Female 2001 2 0 2 9 11 2 Khayelitsha.Female.0
6 Khayelitsha Female 2001 2 01-4 0 2 2 2 Khayelitsha.Female.01-4
```
As you notice, there's actually further detail 'Natural' and 'Unnatural'.
| Multiple imputation for missing count data in a time series from a panel study | CC BY-SA 3.0 | null | 2011-07-10T22:37:27.063 | 2012-09-25T21:05:31.673 | 2011-07-21T21:48:24.990 | 5346 | 5346 | [
"r",
"time-series",
"panel-data",
"data-imputation"
] |
12874 | 2 | null | 8358 | 1 | null | Alternative 1. Estimate and compare the stochastic properties of the series (test of equal level, trend etc). Alternative 2. If independence between years (independent samples), you can rank between treatments, i.e. treatment with lowest death rate = 1, otherwise 0 which gives you a digitom dependent variable. Then use a constant and dummy (for one of the selected treatments) as the RHS regressors. Method: probit, logistic estimator.
| null | CC BY-SA 3.0 | null | 2011-07-10T22:44:25.930 | 2011-07-11T18:52:19.143 | 2011-07-11T18:52:19.143 | 3762 | 3762 | null |
12875 | 1 | null | null | 5 | 1183 | Consider the following model
$$
X \sim |\mathcal{N}(X;0,1)|
\qquad
Y|X \sim Q(Y;X)
$$
where I define $Q(Y=-x|X=x)$ with probability mass $\int_{-\infty}^{-x}\mathcal{N}(x;0,1)dx$, $Q(Y=+x|X=x)$ with probability mass $1-\int_{x}^{\infty}\mathcal{N}(x;0,1)dx$, and the density of $Q(Y|X=x)$ to be one of a truncated Normal distribution in $(-x,x)$.
Assume now that an unknown $x_{unk}$ is sampled from $|\mathcal{N}(X;0,1)|$ and that $y$ is sampled from $Q(Y|X=x_{unk})$. I am given $y$ and would like to approximate the posterior distribution, $P(X|Y=y)$.
Using importance sampling, I would like to take $N$ sample values for $X$ then weight them according to the "probability" of $Y|X$ such that for sufficiently many samples, $$\mathbb{E}[X|Y=y] \approx \frac{1}{\sum_{i}w_{i}} \sum_{i} x_{i} w_{i}$$
>
Were $Q(Y|X)$ entirely continuous, I would use $w_{i} =
> Q(Y=y|X=x_{i})$. Were it entirely discrete, I would use the
probability mass function instead. As $Y|X$ is distributed according
to a sort of hybrid, what should my weights be?
| Importance sampling in mixed, discrete/continuous variables | CC BY-SA 3.0 | null | 2011-07-10T23:29:12.343 | 2015-11-01T21:03:32.760 | 2015-11-01T21:03:32.760 | 7224 | 4742 | [
"mixed-model",
"sampling",
"importance",
"conditioning"
] |
12876 | 2 | null | 12853 | 7 | null | This is by no means a complete answer, the question you should be asking is "what kind of distances are preserved when doing dimensionality reduction?". Since clustering algorithms such as K-means operate only on distances, the right distance metric to use (theoretically) is the distance metric which is preserved by the dimensionality reduction. This way, the dimensionality reduction step can be seen as a computational shortcut to cluster the data in a lower dimensional space. (also to avoid local minima, etc)
There are many subtleties here which I will not pretend to understand, (local distances vs global distances, how relative distances are distorted, etc) but I think this is the right direction to to think about these things theoretically.
| null | CC BY-SA 3.0 | null | 2011-07-11T00:06:02.007 | 2011-07-11T00:06:02.007 | null | null | 1760 | null |
12877 | 2 | null | 12875 | 0 | null | [edited]
Restating: The problem you'd like to solve is to infer the posterior over $X$, $P(X|Y)$, using particle filtering. You draw samples $X_i$ from $P(X)$ and then convert them to particles $Y_i \sim Q(Y|X_i)$, i.e., by sampling a $Y$ particle for each $X$ particle from the conditional distribution $Q(Y|X)$. You then need to apply Bayes' rule now to get the posterior of interest:
$P(X|Y) \propto Q(Y|X)P(X)$.
You sampled your initial particles from $P(X)$, so all you need to do is weight them by weights $w_i = Q(Y_i|X_i)$ and you're done! Is that it?
If so, then I think the problem is this: Q(Y|X) has delta masses as a function of Y. But NOT as a function of $X$! That is, there is a point mass at $\pm x$ in the conditional distribution $Q(Y|X_i)$ as a function of $Y$ for a fixed value $X_i$. However, that's not the weight you need here. You want the likelihood, which is $Q(Y_i|X)$ considered as a function of $X$. (Namely, you want to evaluate that function at $X=X_i$).
This latter thing (the likelihood) has no point masses. I assume the sampling procedure for $Y|X$ is something like "add some Gaussian noise and then theshold". You actually get a continuous likelihood function over $X$ for each value of $Y_i$. How you compute it will depend on $Y_i$:
- If $Y_i=x$ (the upper boundary), then the likelihood involves normcdf: it's the probability that $X$ plus some noise ended up $>x$. So it's a sigmoidal function that goes from zero to 1 as it crosses $x$, with slope that depends on the variance of the noise added to $Y$, and it stays at $+1$ for all larger values of $X$ out to infinity. (Note that this doesn't integrate to 1; it doesn't have to since it's a likelihood!)
- If $Y_i \in (-x,x)$, the likelihood involves the (standard) normpdf$(Y_i,X_i,\sigma^2)$
But in both cases, you have a continuous function over $X$. There's no value of $X$ for which a particular observation of $Y$ has infinitely higher probability than the other values of $X$ (except for trivial examples like $Y_i=-x$ and $X_i\rightarrow +\infty$).
| null | CC BY-SA 3.0 | null | 2011-07-11T01:15:47.597 | 2011-07-11T06:27:40.653 | 2011-07-11T06:27:40.653 | 5289 | 5289 | null |
12878 | 1 | null | null | -1 | 141 | I would like to generate datasets which has 3 treatment groups: A, B, and C based on 3 covariates: $x_{1}$, $x_{2}$ and $x_{3}$. The $x_{1}$ and $x_{2}$ are standard normally distributed and $x_{3}$ is Bernoulli (0.5) distributed. Besides these, assume risk difference between A & B = -0.36, A & C = -0.26 and B & C = 0.10. Can someone give me a hint how to generate these datasets in R?
many thanks in advance
| How to generated datasets for estimating risk differences in multiple groups using R? | CC BY-SA 3.0 | null | 2011-07-11T02:13:26.953 | 2011-07-12T06:49:07.067 | null | null | 4559 | [
"r",
"cross-validation",
"biostatistics"
] |
12879 | 2 | null | 10562 | 1 | null | Some interesting things you could try out:
- Take a look at SigClust - it's an R function that allows you to establish the significance of clustering using bootstrapping/monte carlo simulation. SigClust will provide a p-value for the clustering operation between two sets of points. Theoretically, you can run it at every node of your hierarchical clustering tree, but it tends to be time consuming so maybe at nodes of more than 10 points. In either case, if you SigClust consistently provide high p-values for a clustering of points, then those might highlight the natural clusters you are looking for.
- Try to see whether you can use OREO or optimal re-ordering instead of hierarchical clustering. There isn't an R implementation available as far as I know, but the algorithm does generate very impressive results (at least in the papers that I have read). If you have a background in mathematical programming I'm sure you could get something like this working using CPLEX.
| null | CC BY-SA 3.0 | null | 2011-07-11T02:26:50.783 | 2011-07-11T02:26:50.783 | null | null | 3572 | null |
12880 | 2 | null | 12854 | 10 | null | I would rephrase the question as follows: I have $n$ observations, and $p$ candidate predictors. Assume that the true model is a linear combination of $m$ variables among the $p$ candidate predictors. Is there an upper bound to $m$ (your limit), such that I can still identify this model? Intuitively, if $m$ is too large compared to $n$, or it is large compared to $p$, it may be hard to identify the correct model. In other terms: is there a limit to model selection?
To this question, Candes and Plan give an affirmative answer in their paper ["Near-ideal model selection by $\ell_1$ minimization"](http://www-stat.stanford.edu/~candes/papers/LassoPredict.pdf): $m \le K p \sigma_1/\log(p)$, where $\sigma_1$ is the largest [singular value](http://en.wikipedia.org/wiki/Singular_value) of the matrix of predictors $X$. This is a deep result, and although it relies on several technical conditions, it links the number of observations (through $\sigma_1$) and of $p$ to the number of predictors we can hope to estimate.
| null | CC BY-SA 3.0 | null | 2011-07-11T03:47:12.133 | 2011-07-11T12:39:08.500 | 2011-07-11T12:39:08.500 | 2116 | 30 | null |
12881 | 1 | 12882 | null | 3 | 369 | I am facing a problem of learning clusters from a pairwise dissimilarity matrix. The measures i try are not metrics(not symmetric,no triangular equality), like KL-divergence for example.
Are there Clustering methods i can use to find clusters from these kinds of matrices?
If i have labels available, what kind of cluster validation techniques can i use?
| Are there methods to cluster based on pair wise dissimilarity measures? | CC BY-SA 3.0 | null | 2011-07-11T04:24:38.103 | 2011-07-11T06:39:44.133 | null | null | 4534 | [
"clustering"
] |
12882 | 2 | null | 12881 | 3 | null | Classic hierarchical cluster analysis requires the proximity matrix to be symmetric. So, you have to symmetricize your matrix, for example, averaging elements above and below the diagonal.
If it lacks triangular inequality, you can't use methods such as Ward, centroid, median, which are called "geometric" ones because they need distances be euclidean, - or metric at worst. Other methods, such as nearest neighbour, farthest neighbour, average linkage - you still can use them.
Also, you might choose to rectify your dissimilarities into metric/euclidean by adding constant to all the elements. Iteratively add small value and check when the "double centrat" of the matrix looses negative eigenvalues. When it's lost, your dissimilarities got euclidean. Look [here](https://stats.stackexchange.com/questions/12495/k-means-implementation-with-custom-distance-matrix-in-input/12503#12503) how to do double centering.
| null | CC BY-SA 3.0 | null | 2011-07-11T06:39:44.133 | 2011-07-11T06:39:44.133 | 2017-04-13T12:44:29.923 | -1 | 3277 | null |
12884 | 1 | null | null | 0 | 105 | >
Possible Duplicate:
Working with Likert scales in SPSS
I have a data sample of which recorded individual's perception on whether a specific option is viable or not. The data was recorded on a lickert scale of 1 - 5. So basically, I'm looking to say that the "general perception of the sample is ____"
The mean = 2.83 and std. dev. 1.2007.
Firstly, what is the analysis that I should run for this?
I ran a one-sample t-test in SPSS against 3 as >3 means the perception is favourable. The reported significance was > 0.5 (I divided the reported significance by 2 as SPSS runs a 2-tailed test rather than one tailed which is I believe is what I want).
I'm unsure now whether I ran the right test or how to interpret this finding, if I have done the correct analysis.
| Reporting the general perception of tested sample | CC BY-SA 3.0 | null | 2011-07-11T09:30:03.777 | 2011-07-11T09:30:03.777 | 2017-04-13T12:44:39.283 | -1 | 5325 | [
"spss",
"t-test"
] |
12885 | 1 | 12886 | null | 11 | 2606 | I always believed that time should not be used as a predictor in regressions (incl. gam's) because, then, one would simply "describe" the trend itself. If the aim of a study is to find environmental parameters like temperature etc. that explain the variance in, let´s say, activity of an animal, then I wonder, how can time be of any use? as a proxy for unmeasured parameters?
Some trends in time on activity data of harbor porpoises can be seen here:
-> [How to handle gaps in a time series when doing GAMM?](https://stats.stackexchange.com/questions/12712/how-to-handle-gaps-in-a-time-series-when-doing-gamm)
my problem is: when I include time in my model (measured in julian days), then 90% of all other parameters become insignificant (ts-shrinkage smoother from mgcv kick them out). If I leave time out, then some of them are significant...
The question is: is time allowed as a predictor (maybe even needed?) or is it messing up my analysis?
many thanks in advance
| Is it allowed to include time as a predictor in mixed models? | CC BY-SA 3.0 | null | 2011-07-11T11:12:39.240 | 2011-07-11T12:03:23.497 | 2017-04-13T12:44:44.530 | -1 | 5280 | [
"r",
"time-series",
"mixed-model",
"nonlinear-regression"
] |
12886 | 2 | null | 12885 | 14 | null | Time is allowed; whether it is needed will depend on what you are trying to model? The problem you have is that you have covariates that together appear to fit the trend in the data, which Time can do just as well but using fewer degrees of freedom - hence they get dropped out instead of Time.
If the interest is to model the system, the relationship between the response and the covariates over time, rather than model how the response varies over time, then do not include Time as a covariate. If the aim is to model the change in the mean level of the response, include Time but do not include the covariate. From what you say, it would appear that you want the former, not the latter, and should not include Time in your model. (But do consider the extra info below.)
There are a couple of caveats though. For theory to hold, the residuals should be i.i.d. (or i.d. if you relax the independence assumption using a correlation structure). If you are modelling the response as a function of covariates and they do not adequately model any trend in the data, then the residuals will have a trend, which violates the assumptions of theory, unless the correlation structure fitted can cope with this trend.
Conversely, if you are modelling the trend in the response alone (just including Time), there may be systematic variation in the residuals (about the fitted trend) that is not explained by the trend (Time), and this might also violate the assumptions for the residuals. In such cases you might need to include other covariates to render the residuals i.i.d.
Why is this an issue? Well when you are testing if the trend component, for example, is significant, or whether the effects of covariates are significant, the theory used will assume the residuals are i.i.d. If they aren't i.i.d. then the assumptions won't be met and the p-values will be biased.
The point of all this is that you need to model all the various components of the data such that the residuals are i.i.d. for the theory you use, to test if the fitted components are significant, to be valid.
As an example, consider seasonal data and we want to fit a model that describes the long-term variation in the data, the trend. If we only model the trend and not the seasonal cyclic variation, we are unable to test whether the fitted trend is significant because the residuals will not be i.i.d. For such data, we would need to fit a model with both a seasonal component and a trend component, and a null model that contained just the seasonal component. We would then compare the two models using a generalized likelihood ratio test to assess the significance of the fitted trend. This is done using `anova()` on the `$lme` components of the two models fitted using `gamm()`.
| null | CC BY-SA 3.0 | null | 2011-07-11T12:03:23.497 | 2011-07-11T12:03:23.497 | null | null | 1390 | null |
12887 | 1 | 12894 | null | 14 | 940 | I'm a novice who is going to start reading about data mining. I have basic knowledge of AI and statistics. Since many say that machine learning also plays an important role in data mining, is it necessary to read about machine learning before I could go on with data mining?
| How to begin reading about data mining? | CC BY-SA 3.0 | null | 2011-07-11T13:16:03.470 | 2012-10-08T19:45:17.123 | 2011-07-11T16:17:34.290 | 919 | 5351 | [
"machine-learning",
"references",
"data-mining"
] |
12888 | 1 | null | null | 1 | 342 | Consider the following two problems:
- We want to do linear regression, where we know that the true model is actually linear. In the general case, linear regression includes an offset. However, we know, using prior knowledge, that the true model actually has offset 0. So clearly, it is better to fit the model while constraining that offset is 0.
- We want to estimate a distribution $p(z)$ and find its mode. We estimate $p$ using, let's say, density estimator, to be $\tilde{p}$. Now, again, we know some properties about the mode, such that, for some function $F(z_{mode}) = 0$. Then, (perhaps?) it is better to find the mode of $\tilde{p}(z)$ while constraining $F(z)$ to be $0$, instead of just finding the mode of $\tilde{p}$ without this constraint.
My question is: are there any known results about incorporating this kind of prior knowledge (if it is true)? A result that shows that our solution is better when using the prior knowledge, or so on? I hope it is not too vague. I am interested in reading more about this.
Thanks.
| What kind of results are there about prior knowledge? | CC BY-SA 3.0 | null | 2011-07-11T14:08:51.303 | 2011-08-10T18:12:53.810 | 2011-07-11T15:19:21.820 | null | 5352 | [
"probability",
"knowledge-discovery"
] |
12889 | 2 | null | 12887 | 4 | null | Data mining can be descriptive or predictive.
On the one hand, if you are interested in descriptive data mining, then machine learning won't help.
On the other hand, if you are interested in predictive data mining, then machine learning will help you understand that you try to minimize the unknown risk (expectation of the loss function) when minimizing the empirical risk: you will keep in mind overfitting, generalization error and cross-validation. For instance, for a matter of consistency, the $k$-NN for a training sample of size $n$ should be such that:
- $k$ goes to infinity when $n$ goes to infinity,
- $\frac{k}{n}$ goes to 0 when $n$ goes to infinity.
| null | CC BY-SA 3.0 | null | 2011-07-11T14:14:48.557 | 2011-07-11T14:14:48.557 | null | null | 1351 | null |
12890 | 1 | 12891 | null | 2 | 419 | I have a variable which decays very quickly (exponentially). I want to transform it in a way it maintains "exponentially decaying" property but which will allow me to control the "speed" of a decay.
```
original <- c(100, 80, 70, 65, 64, 63, 63, 63, 63, 62, 62, 61);
plot(original);
target <- c(100, 90, 70, 60, 65, 64, 64, 63, 63, 63, 63);
```
target is decaying much slower. I want to be able to controll for that.
How do I do that?
| Exponentially decaying variable | CC BY-SA 3.0 | null | 2011-07-11T14:30:17.307 | 2011-07-11T15:51:14.443 | 2011-07-11T15:51:14.443 | 333 | 333 | [
"data-transformation"
] |
12891 | 2 | null | 12890 | 3 | null | Exponentiate it.
If your variable decays exponentially then it decays like $y = e^{-\delta t}$, so the rate of decay is $\delta$. Well then $y^{\alpha} = e^{-\delta \alpha t}$ so the new rate of decay is $\delta \alpha$. Choose $\alpha$ to make that product as small or as large as you want.
| null | CC BY-SA 3.0 | null | 2011-07-11T14:35:21.747 | 2011-07-11T14:35:21.747 | null | null | 4856 | null |
12892 | 2 | null | 12890 | 2 | null | Rescale it.
Specifically, I am supposing that by "exponential decay" of $X$ you mean that when $x$ is sufficiently large, $\Pr(X \gt x) \sim \exp(-\lambda x)$ for a positive constant $\lambda$ which measures the "speed" of the decay. Let $\alpha \gt 0$ and write $Y = \alpha X$ for the scaled version of $X$. Then
$$\Pr(Y \gt y) = \Pr(\alpha X \gt y) = \Pr(X \gt y/\alpha) \sim \exp(-\lambda (y/\alpha)) = \exp(-(\lambda/\alpha) y),$$
showing that the speed of decay of $Y$ equals $\lambda/\alpha$.
Thus, if you want to transform $X$ to have a decay rate of $\mu$, solving for $\lambda/\alpha = \mu$ yields
$$Y = \lambda X /\mu$$
as a solution. (There are other transformations of $X$ having $\mu$ as a decay rate, but this is arguably the simplest.)
| null | CC BY-SA 3.0 | null | 2011-07-11T14:42:47.257 | 2011-07-11T14:57:22.817 | 2011-07-11T14:57:22.817 | 919 | 919 | null |
12893 | 2 | null | 12888 | 1 | null | With respect to 1.: as chemometrician, my practical question is: do you really have no offset?
One standard situation that corresponds to your scenario from my field of work is linear (univariate) calibration using the [Beer-Lambert-law](http://en.wikipedia.org/wiki/Beer%E2%80%93Lambert_law):
$$
A = \varepsilon l c
$$
(here $\varepsilon$ is the absorption coefficient, no error term)
In practice, even if you obey the prerequisites the fit will include an offset (blank value) because reality is not as easy as theory. E.g. because you use a double beam instrument to compensate variability in the source intensity, but then you cannot use the same single cuvette for sample and blank, which can (will) cause an offset.
Of course, you may have noise $\gg$ offset, so you can neglect the offset...
Or you say that the true model does have an offset as physical reasons can be given. But taking into account the influence of the measurement my bet is that it will be pretty hard to find an example system that is really supposed to be linear. (Anyone knows an example?)
With respect to the general question:
Depending on the situation, however, a constrained model may be better even if it is only an approximation (keyword: bias variance tradeoff, see e.g. the [Elements of statistical learning](http://www-stat.stanford.edu/~tibs/ElemStatLearn/))
| null | CC BY-SA 3.0 | null | 2011-07-11T15:35:03.393 | 2011-07-11T15:35:03.393 | null | null | 4598 | null |
12894 | 2 | null | 12887 | 12 | null | Being somewhat in this position myself, I'll try to give some insight.
Firstly, download the [Elements of Statistical Learning](http://www-stat.stanford.edu/~tibs/ElemStatLearn/). It presumes calculus and linear algebra, and although it is very technical, it is also extremely well written.
Secondly (or firstly) look at [Andrew Ng's tutorials](http://www.stanford.edu/class/cs229/) on machine learning.
Thirdly, get some data, and start attempting to analyse data. You'll need to split into training and test sets, and then build models on the training set and test them against the test set.
I found the caret package for R very useful for all of this.
After that its practice, practice practice (like almost everything else).
| null | CC BY-SA 3.0 | null | 2011-07-11T16:16:42.503 | 2012-10-08T19:45:17.123 | 2012-10-08T19:45:17.123 | 656 | 656 | null |
12895 | 2 | null | 12888 | 1 | null | The general comment is that in the abstract it is usually better to make the model space as small as possible. This is sort of vague and obvious at the same time (we would love the model space to be a singleton set!). "As small as possible" really constitutes a decision problem (basically weighing the losses associated with inconsistent versus inefficient inferences).
From a frequentist perspective you (roughly) enhance the degrees of freedom by estimating fewer parameters, which will give you lower variance estimates, more powerful tests etc. From a Bayesian perspective, if you have $\theta = (\alpha, \beta)$ and are interested in $\alpha$ then you're after the marginal posterior $p(\alpha|Y) = \int p(\alpha, \beta|Y)d\beta$. Integrating over $\beta$ inflates the variance in this distribution, so if you're really confident that $\beta$ takes a fixed value you'll get more efficient inferences by fixing it (assuming you're correct!).
In practice you must also be concerned about parameter constraints complicating inference too, although this is outside the scope of your question...
| null | CC BY-SA 3.0 | null | 2011-07-11T16:42:04.463 | 2011-07-11T16:42:04.463 | null | null | 26 | null |
12896 | 1 | 12918 | null | 9 | 226 | Imagine that you want to assess the compressibility of a large document very fast. You could randomly pick a subsequence, try to compress it. This can serve as a prediction for the overall compressibility of the document. But how big should your sample be?
We have come up with the following strategy:
- Pick an arbitrary (small) sample size. Measure the compressibility.
- Next, double the sample size and measure the compressibility again. If there is little change (say less than 10%), then conclude that you have reliably determined the compressibility of the document. If not, double the sample size again, and so on.
We are quite certain that this is not a new strategy, and we are wondering whether it is related to some well-known strategy used by statisticians.
("Compression" here is just an example. Basically, we are interested in a metric that has no known nice mathematical properties, so that it is not possible to determine analytically what could be a good sample size. We have no choice but to fall back on such heuristics.)
| How do you call this dynamic sample-size selection strategy? | CC BY-SA 3.0 | null | 2011-07-11T16:52:06.113 | 2022-12-08T14:26:31.060 | 2011-07-11T22:41:43.003 | 930 | 5355 | [
"sampling",
"model-selection",
"sample-size"
] |
12897 | 1 | null | null | 6 | 203 | I'm currently in the planning stages of an experiment that I hope will detect a 0.155% signal variation (relative magnitude) within a reasonable time frame (less than 6 months ideally). I've calculated the rate of (usable) data will be around 68 events per day, though it should be stressed this is a random variable. Now I'm trying to work out how many days will I need to run the detector for to see the variation with a confidence level of $3\sigma$?
Some other details that may (or may not) be relevant include: the variation in the signal is expected to be sinusoidal with a period of 0.5 days. For this reason I reduced my useful event rate to 34 (Ie half) as clearly there is no variation to see when the sinusoidal signal is at or close to the mean value.
I've been googling for a method to predict the size of a data set necessary to see such a small signal variation but have come up with nothing. I would be extremely grateful for any hints / tips anyone could offer.
EDIT: With the greatest thanks to @shabbychef I feel this question has now been satisfactorally answered on physics.SE [here](https://physics.stackexchange.com/questions/12165/how-to-predict-how-much-data-to-collect).
| How to predict how much data to collect? | CC BY-SA 3.0 | null | 2011-07-11T16:52:50.967 | 2011-07-12T14:14:45.797 | 2017-04-13T12:40:16.003 | -1 | 5354 | [
"confidence-interval",
"experiment-design",
"statistical-power"
] |
12898 | 2 | null | 12842 | 16 | null | Some other examples, consider datapoints that form a circle or ellipse, the covariance is 0, but knowing x you narrow y to 2 values. Or data in a square or rectangle. Also data that forms an X or a V or a ^ or < or > will all give covariance 0, but are not independent. If y = sin(x) (or cos) and x covers an integer multiple of periods then cov will equal 0, but knowing x you know y or at least |y| in the ellipse, x, <, and > cases.
| null | CC BY-SA 3.0 | null | 2011-07-11T16:54:40.347 | 2011-07-14T18:13:57.207 | 2011-07-14T18:13:57.207 | 4505 | 4505 | null |
12899 | 2 | null | 12845 | 1 | null | Statements about being better (or worse) than some percentage of others is refered to as percentiles (or quantiles if not limited to percents). These can be calculated by direct comparison to a large number of measured subjects, or based on a theoretical distribution (if you are 2 standard deviation above the mean and the population is approximately normal then you are at about the 95th percentile).
| null | CC BY-SA 3.0 | null | 2011-07-11T17:00:58.237 | 2011-07-11T17:00:58.237 | null | null | 4505 | null |
12900 | 1 | 12991 | null | 165 | 412225 | My understanding is that $R^2$ cannot be negative as it is the square of R. However I ran a simple linear regression in SPSS with a single independent variable and a dependent variable. My SPSS output give me a negative value for $R^2$. If I was to calculate this by hand from R then $R^2$ would be
positive. What has SPSS done to calculate this as negative?
```
R=-.395
R squared =-.156
B (un-standardized)=-1261.611
```
Code I've used:
```
DATASET ACTIVATE DataSet1.
REGRESSION /MISSING LISTWISE /STATISTICS COEFF OUTS R ANOVA
/CRITERIA=PIN(.05) POUT(.10) /NOORIGIN
/DEPENDENT valueP /METHOD=ENTER ageP
```
I get a negative value. Can anyone explain what this means?
| When is R squared negative? | CC BY-SA 3.0 | null | 2011-07-11T17:07:34.553 | 2022-09-11T03:11:22.973 | 2017-11-19T18:48:11.087 | 128677 | 4498 | [
"regression",
"spss",
"r-squared"
] |
12902 | 1 | 12921 | null | 10 | 10482 | I have three sets of time-series data I am looking to compare. They have been taken on 3 separate periods of about 12 days. They are the average, maximum and minimum of head counts taken in a college library during finals weeks. I have had to do mean, max and min because the hourly head counts were not continuous (see [Regular data gaps in a time series](https://stats.stackexchange.com/questions/12222/regular-data-gaps-in-a-time-series)).
Now the data set looks like this. There is one data point (average, max or min) per evening, for 12 evenings. There are 3 semesters the data was taken for, in only the 12-day periods of concern. So for example, Spring 2010, Fall 2010, and May 2011 each have a set of the 12 points. Here's an example chart:
I have overlaid the semesters because I want to see how the patterns change from semester to semester. However, as I have been told in the [linked thread](https://stats.stackexchange.com/questions/12222/regular-data-gaps-in-a-time-series), it's not a good idea to slap the semesters tail-to-head since there is no data in between.
The question is then: What mathematical technique can I use to compare the pattern of attendance for each semester? Is there anything special to time-series that I must do, or can I simply take the percent differences? My goal is to say that library usage over these days is going up or down; I am just not sure what technique(s) I should use to show it.
| Comparison of time series sets | CC BY-SA 3.0 | null | 2011-07-11T17:36:48.267 | 2011-07-12T09:54:06.470 | 2017-04-13T12:44:36.927 | -1 | 5129 | [
"time-series",
"multiple-comparisons",
"trend"
] |
12903 | 2 | null | 12902 | 0 | null | Sarah, Take your 36 numbers (12 values per cycle ; 3 cycles ) and construct a regression model with 11 indicators reflecting possible week-of-the-semester effect and then identify any necessary Intervention Series ( Pulses, Level Shifts ) necessary to render the mean of the residuals to be 0.0 everywhere or at least not statistically significantly different from 0.0. For example if you identify a level shift at period 13 this might be suggest a statistically significant difference between the mean of the first semester i.e. the first 12 values) versus the mean of the last two semesters (last 24 values ). You might be able to draw inference or test the hypothesis of no week of the semester effect. A good time series package might be useful to you in this regard. Failing that you might need to find somebody to provide help in this analytical arena.
| null | CC BY-SA 3.0 | null | 2011-07-11T17:56:30.727 | 2011-07-11T17:56:30.727 | null | null | 3382 | null |
12905 | 2 | null | 12900 | 8 | null | This can happen if you have a time series that is N.i.i.d. and you construct an inappropriate ARIMA model of the form(0,1,0) which is a first difference random walk model with no drift then the variance (sum of squares - SSE ) of the residuals will be larger than the variance (sum of squares SSO) of the original series. Thus the equation 1-SSE/SSO will yield a negative number as SSE execeedS SSO . We have seen this when users simply fit an assumed model or use inadequate procedures to identify/form an appropriate ARIMA structure. The larger message IS that a model can distort (much like a pair of bad glasses ) your vision. Without having access to your data I would otherwise have a problem in explaining your faulty results. Have you brought this to the attention of IBM ?
The idea of an assumed model being counter-productive has been echoed by Harvey Motulsky. Great post Harvey !
| null | CC BY-SA 3.0 | null | 2011-07-11T18:11:52.600 | 2011-07-13T17:57:33.853 | 2011-07-13T17:57:33.853 | 3382 | 3382 | null |
12906 | 1 | null | null | 1 | 1383 | I am not sure whether this is the right place to ask this question, but any possible answer will help.
I want to apply some form of template matching on these set of pictures.
This is the template picture:
This is the picture in which I want to search for the template picture and the black rectangle, frame the template shapes which should be identified within the picture:
I have done all the coding for this in Matlab and I have managed to apply template matching algorithms, sift(scale invariant feature transform), histogram based matching, and also, I have tried to apply a variation of expectation maximization on the blue and red areas.
The problem, as you can see from the images, is that the areas where you can identify the template image are totally deformed. So I should apply some template matching algorithms which should be invariant to scaling, rotation and partial deformation of the template image.
Can you please recommend me some computer vision algorithms or techniques for this type of transform which are invariant to scaling, rotation and partial deformation of the template image shape?
| Template matching involving scale-, rotation- and shape-invariant transformation | CC BY-SA 3.0 | null | 2011-07-11T18:25:33.383 | 2018-08-17T10:14:54.530 | 2018-08-17T10:14:54.530 | 11887 | 5228 | [
"multivariate-analysis",
"image-processing",
"procrustes-analysis"
] |
12907 | 1 | null | null | 5 | 2879 | I have a data set that has 10k documents, each of which is mapped to one and only one of 4k categories. This forms my training set.
My requirement is that when a new, unseen document comes in, I need to be able to identify all the categories (from among the 4k categories) to which it belongs.
What is the best way to go about doing this? My problem conforms to what's generally known as multi-label classification in the literature, and I see plenty of papers, but no readily available software. I was hoping to know if anyone is aware of any off-the-shelf tools that would do this task.
Thanks in advance.
Update 1: Based on some comments below I am updating my question. Although I have 4k categories and only 10k documents, the distribution of these documents into categories is highly skewed - in other words, there are a small number of categories that have most of the training documents. I am OK with retaining categories that have sufficient training data, and throwing away the other categories if need be. The bigger problem is identifying all the categories of unseen document.
| Off the shelf tool for multi-label classification | CC BY-SA 3.0 | null | 2011-07-11T18:56:08.777 | 2011-11-26T00:15:35.597 | 2011-07-11T20:05:36.737 | 3301 | 3301 | [
"machine-learning",
"classification"
] |
12908 | 1 | null | null | 20 | 944 | I'm preparing a presentation about parallel statistics. I plan to illustrate the formulas for distributed computation of the mean and variance with examples involving center of gravity and moment of inertia. I'm wondering if there is a physical interpretation of the third or higher central moments that I can use to help illustrate the general formula.
| Physical/pictoral interpretation of higher-order moments | CC BY-SA 3.0 | null | 2011-07-11T19:28:06.590 | 2022-05-23T22:02:40.423 | null | null | 5358 | [
"interpretation",
"moments",
"parallel-computing"
] |
12909 | 2 | null | 12907 | 1 | null | Try using [R](http://www.r-project.org/). You can use the [factor](http://www.stat.berkeley.edu/classes/s133/factors.html) data type to store what you are calling each document's "label."
Before you use any predictive functions, you'll have to calculate some quantitative attributes about each document. This is going to depend on the nature of your problem, but check out the [Natural Language Processing](http://cran.r-project.org/web/views/NaturalLanguageProcessing.html) section on CRAN for help with this step.
Once you've converted your documents into a dataframe with one row per document, and variables representing that document's classification, as well as covariates describing the document's content and metadata, you can start building predictive models. You might start with [linear discriminant analysis](http://stat.ethz.ch/R-manual/R-patched/library/MASS/html/lda.html).
With 4k classes and over 10k documents, I think it is going to be difficult to develop a good classifier. See if any of the authors you've been reading have written packages for R.
| null | CC BY-SA 3.0 | null | 2011-07-11T19:30:13.597 | 2011-07-11T19:30:13.597 | null | null | 2817 | null |
12910 | 2 | null | 12907 | 4 | null | About multilabel classification, the baseline (but usually quite good) approach is just to make a battery of binary classifiers, each trained to recognize one class versus all other, use them all on each sample and combine their answers.
This is trivial to implement, so almost any tool will do.
However, you have other problem -- an ultrapoor training set. 10k samples in 4k classes gives 2-3 examples per class -- this is almost nothing; I can at most imagine some embarrassing 1-NN method in this setting.
| null | CC BY-SA 3.0 | null | 2011-07-11T19:48:25.877 | 2011-07-11T19:48:25.877 | null | null | null | null |
12911 | 1 | null | null | 1 | 146 | I've got a series of statistics computed on different populations of cells, where each population contains a different genetic deletion.
The measurements I have for each population involve GFP intensity: median intensity, mean intensity, median absolute deviation of intensity, max intensity and min intensity.
I'd like to test whether the distributions are significantly different between each mutant strain and a wild type (no mutation) strain, but I don't have any access to the raw data from which these statistics were calculated.
Can anyone tell me if there are tests like Wilcoxon / Mann-Whitney (or even the Kolmogorov-Smirnov) that function on statistics rather than raw-data? Or could someone suggest an alternative?
| Test for differences in distribution without raw data | CC BY-SA 3.0 | null | 2011-07-11T19:51:02.267 | 2011-07-11T22:14:53.927 | 2011-07-11T22:14:53.927 | 930 | 5048 | [
"r",
"hypothesis-testing",
"genetics"
] |
12912 | 2 | null | 12907 | 5 | null | As @mbq suggests, a battery of binary classifiers is a good place to start. Ridge regression classifiers work pretty well on text classification problems (choose the ridge parameter via leave-one-out cross-validation. If training time is not an issue, also use the bootstrap as a further protection against over-fitting; the committee of boostrap classifiers can be amalgamated into one, so it doesn't have a computational cost at runtime).
However, as there are 4K classes and only 10K samples, it will probably be necessary to look at the hierarchy of classes and try to predict whether the page fits into broad categories first.
| null | CC BY-SA 3.0 | null | 2011-07-11T19:53:57.637 | 2011-07-11T22:06:37.040 | 2011-07-11T22:06:37.040 | 930 | 887 | null |
12913 | 1 | 12915 | null | 3 | 264 | I'm trying to visualize a matrix of 3 datasets.* For the sake of example, let's say I have a list of hats, coats and shoes, and I want to display a 2D grid/visualization of each possible combination.
Is it even possible without turning it into a 5x25 grid?
I had a look at this thread:
[Visualizing multi-dimensional data (LSI) in 2D](https://stats.stackexchange.com/questions/11650/visualizing-multi-dimensional-data-lsi-in-2d), but from what I understand that one is about plotting as opposed to showing the fields in a grid/table of some sorts.
(*) I think i'm using the wrong terminology here, what I mean isn't about drawing points in a graph but rather a way to illustrate the data set in a meaningful matter that still allows for editing
Edit: This might explain what i'm trying to do:
```
+------+------------+-------------+------------+
| | hat | shirt | shoe |
+------+------------+-------------+------------+
| red | red hat | red shirt | red shoe |
+------+------------+-------------+------------+
| blue | blue hat | blue shirt | blue shoe |
+------+------------+-------------+------------+
|green | green hat | green shirt | green shoe |
+------+------------+-------------+------------+
|black | black hat | black short | black shoe |
+------+------------+-------------+------------+
```
... but with 3 data sets instead of just colors and clothes.
| Visualizing a 5x5x5 data set | CC BY-SA 3.0 | null | 2011-07-11T20:19:34.960 | 2011-07-11T22:25:34.680 | 2017-04-13T12:44:44.530 | -1 | 5359 | [
"data-visualization",
"multivariable"
] |
12914 | 2 | null | 12913 | 1 | null | Use a 2D plot, and use a color scale to represent the 3rd dimension. You could also try varying the symbol used to represent your points, or change the size of the marker.
| null | CC BY-SA 3.0 | null | 2011-07-11T21:07:14.357 | 2011-07-11T21:07:14.357 | null | null | 2817 | null |
12915 | 2 | null | 12913 | 2 | null | You could just have 5 rectangles, the the first being
```
+------+----------------------+------------------------+-----------------------+
|Babies| hat | shirt | shoe |
+------+----------------------+------------------------+-----------------------+
|red | red hat for babies | red shirt for babies | red shoe for babies |
+------+----------------------+------------------------+-----------------------+
|blue | blue hat for babies | blue shirt for babies | blue shoe for babies |
+------+----------------------+------------------------+-----------------------+
|green | green hat for babies | green shirt for babies | green shoe for babies |
+------+----------------------+------------------------+-----------------------+
|black | black hat for babies | black short for babies | black shoe for babies |
+------+----------------------+------------------------+-----------------------+
```
And the next being similar but being boys instead of babies, then girls, then men, then women.
| null | CC BY-SA 3.0 | null | 2011-07-11T22:25:34.680 | 2011-07-11T22:25:34.680 | null | null | 2958 | null |
12916 | 2 | null | 12897 | 1 | null | If you can put your question in the form of a t-test, then Lehr's rule can be used to estimate the required sample size. For a 2-sided, one sample t-test at the 0.05 level, one can achieve power of 0.80 by using $n = 8 / \Delta^2$, where $\Delta$ is the 'signal to noise' (mean divided by standard deviation). I cannot figure out if you mean this to be 0.155%, but if so, you're looking at roughly 100K years!
| null | CC BY-SA 3.0 | null | 2011-07-11T23:16:08.253 | 2011-07-11T23:16:08.253 | null | null | 795 | null |
12917 | 2 | null | 12867 | 3 | null | Q1: I suppose you're not asking whether the variable names have anything to do with imputation. I suppose you're asking whether to do multiple imputation on your raw data or on "scales" which means sum scores of raw data variables which are assumed to measure the same construct.
Given that you are concerned about the right method (which is good), well, then how do you want to compute scale scores before you do anything about the missing data? Answer: you need to work with your raw data.
Q2: Multiple imputation works by estimating missing values from values of the other variables in your data set (this being a non-technical explanation). So it does not matter if you recode your variables or not, if by recoding you mean changing the sign of the relation from one var to the other variables. Still I would recommend recoding before the imputation so that you don't get confused afterwards.
Q3: I don't know why AMOS does not calculate estimates. I don't really know AMOS, but the user guide seems to say it should be able to calculate estimates for multiply imputated data sets. But there is a general answer anyway: The whole point of multiple imputation is that you get point estimates but also incorporate the additional variation introduced by the imputation process.
This means: You should not use a single "averaged" data set. Instead you work with all of your imputed data sets. First, you get point estimates for your model parameters by running your model (I suppose a structural equation model) for each of the data sets and taking the mean of the point estimates, and then you compute standard errors by combining between-imputation variance (variance of the parameter estimates between imputed data sets) with within-imputation variance (mean of the estimated standard errors from the different models). See any text on multiple imputation. A short and freely available tutorial is here: [http://rhowell.ba.ttu.edu/Enders-MissingHancock.pdf](http://rhowell.ba.ttu.edu/Enders-MissingHancock.pdf)
| null | CC BY-SA 3.0 | null | 2011-07-11T23:39:47.590 | 2011-07-11T23:39:47.590 | null | null | 5020 | null |
12918 | 2 | null | 12896 | 3 | null | This has been called 'Progressive Sampling', e.g. see this paper [Efficient Progressive Sampling](https://web.archive.org/web/20121016140124/http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.86.9168&rep=rep1&type=pdf).
| null | CC BY-SA 4.0 | null | 2011-07-11T23:53:42.673 | 2022-12-08T14:26:31.060 | 2022-12-08T14:26:31.060 | 362671 | 4797 | null |
12919 | 1 | 12920 | null | 6 | 7720 | In R, the function `skewness` from package `moments` allows one to calculate the skewness of the distribution from a given sample. Does anybody know if there is a ready-to-use function to calculate the skewness of the distribution from a given histogram?
| How to calculate skewness of data from histogram in R? | CC BY-SA 3.0 | null | 2011-07-12T00:08:21.980 | 2016-07-02T06:06:41.307 | 2013-05-13T05:04:54.657 | 805 | 4337 | [
"r",
"mathematical-statistics",
"histogram",
"skewness"
] |
12920 | 2 | null | 12919 | 5 | null | Sounds like you're looking at a published histogram and don't have the actual data?
If that is the case you could calculate a rough skewness figure by doing something like `pub_his <- c(rep(10,16),rep(15,18),rep(20,27)...)`, picking the mid point of each bar and reading the frequency off the graph, then you'd have the data and could use the skewness function.
| null | CC BY-SA 3.0 | null | 2011-07-12T01:02:07.347 | 2013-05-13T07:25:58.697 | 2013-05-13T07:25:58.697 | 582 | 4226 | null |
12921 | 2 | null | 12902 | 8 | null | [Fixed-effects ANOVA](http://en.wikipedia.org/wiki/Analysis_of_variance) (or its linear regression equivalent) provides a powerful family of methods to analyze these data. To illustrate, here is a dataset consistent with the plots of mean HC per evening (one plot per color):
```
| Color
Day | B G R | Total
-------+---------------------------------+----------
1 | 117 176 91 | 384
2 | 208 193 156 | 557
3 | 287 218 257 | 762
4 | 256 267 271 | 794
5 | 169 143 163 | 475
6 | 166 163 163 | 492
7 | 237 214 279 | 730
8 | 588 455 457 | 1,500
9 | 443 428 397 | 1,268
10 | 464 408 441 | 1,313
11 | 470 473 464 | 1,407
12 | 171 185 196 | 552
-------+---------------------------------+----------
Total | 3,576 3,323 3,335 | 10,234
```
ANOVA of `count` against `day` and `color` produces this table:
```
Number of obs = 36 R-squared = 0.9656
Root MSE = 31.301 Adj R-squared = 0.9454
Source | Partial SS df MS F Prob > F
-----------+----------------------------------------------------
Model | 605936.611 13 46610.5085 47.57 0.0000
|
day | 602541.222 11 54776.4747 55.91 0.0000
colorcode | 3395.38889 2 1697.69444 1.73 0.2001
|
Residual | 21554.6111 22 979.755051
-----------+----------------------------------------------------
Total | 627491.222 35 17928.3206
```
The `model` p-value of 0.0000 shows the fit is highly significant. The `day` p-value of 0.0000 is also highly significant: you can detect day to day changes. However, the `color` (semester) p-value of 0.2001 should not be considered significant: you cannot detect a systematic difference among the three semesters, even after controlling for day to day variation.
[Tukey's HSD](http://en.wikipedia.org/wiki/Tukey%27s_range_test) ("honest significant difference") test identifies the following significant changes (among others) in day-to-day means (regardless of semester) at the 0.05 level:
```
1 increases to 2, 3
3 and 4 decrease to 5
5, 6, and 7 increase to 8,9,10,11
8, 9, 10, and 11 decrease to 12.
```
This confirms what the eye can see in the graphs.
Because the graphs jump around quite a bit, there's no way to detect day-to-day correlations (serial correlation), which is the whole point of time series analysis. In other words, don't bother with time series techniques: there's not enough data here for them to provide any greater insight.
One should always wonder how much to believe the results of any statistical analysis. Various diagnostics for heteroscedasticity (such as the [Breusch-Pagan test](http://en.wikipedia.org/wiki/Breusch%E2%80%93Pagan_test)) don't show anything untoward. The residuals don't look very normal--they clump into some groups--so all the p-values have to be taken with a grain of salt. Nevertheless, they appear to provide reasonable guidance and help quantify the sense of the data we can get from looking at the graphs.
You can carry out a parallel analysis on the daily minima or on the daily maxima. Make sure to start with a similar plot as a guide and to check the statistical output.
| null | CC BY-SA 3.0 | null | 2011-07-12T02:09:43.260 | 2011-07-12T09:54:06.470 | 2011-07-12T09:54:06.470 | 2116 | 919 | null |
12922 | 1 | 12925 | null | 5 | 5129 | I am using Silhouette width to compute the best value for k in k-means. As I am performing document clustering, I am calculating the values of `a` and `b` as follows in Python:
```
a = distance(data[index], centroids[clusters[index]], metric=metric, p=p)
b = min([distance(data[index], c) for i,c in enumerate(centroids) if i != currentindex])
score = float(b - a) / max(a, b) if max(a, b) > 0 else 0.0
```
The original formula from the [Wikipedia page](http://en.wikipedia.org/wiki/Silhouette_%28clustering%29) is the following:
I am using a `cosine` similarity measure to compute the distances and was wondering if this formula needs to be changed or it can be left as is. In the above snippet of code, the function `distance` computes the cosine similarity metric.
| Will the silhouette formula change depending on the distance metric? | CC BY-SA 3.0 | null | 2011-07-12T03:03:09.470 | 2013-02-26T18:18:45.257 | 2011-07-12T08:47:22.743 | null | 2164 | [
"clustering",
"data-mining",
"python",
"distance-functions"
] |
12923 | 2 | null | 10540 | 16 | null | I have been looking into the same thing today and found an interpretation [here](http://www.unesco.org/webworld/idams/advguide/Chapt7_1_1.htm). It makes logical sense but I am not sure if we can blindly apply the interpretation for our datasets. In summary, what that article says is the following:
```
0.71-1.0
A strong structure has been found
0.51-0.70
A reasonable structure has been found
0.26-0.50
The structure is weak and could be artificial. Try additional methods of data analysis.
< 0.25
No substantial structure has been found
```
However, it seems like we can use the silhouette width to catch outliers. In a document clustering task that I am currently handling, the ones with negative silhouette width are definite outliers (when cross checked with their semantic meaning). I am not sure if this width will improve after removing outliers (again, this makes logical sense but I have not done this myself).
| null | CC BY-SA 3.0 | null | 2011-07-12T03:06:56.500 | 2011-07-12T03:06:56.500 | null | null | 2164 | null |
12924 | 1 | null | null | 5 | 174 | I'm pretty lost at how to approach this, and particularly with regards to the terminology; so please feel free to point to resources, fix my question name, etc. Thanks.
I used R back in college, I've got Orange installed. I've heard of MiniTab. I've got Excel installed. I've got access to SQL Server 2008 w/ Analysis Services. That's the breath of my toolkit. [I'm a developer, but that's all I know of for statistical analysis.]
I'm marketing a product, and I've got a few datasets.
Customers:
- Customer_id
- Campaign they clicked on, and the date.
- Campaign they converted on, and the date.
- The amount of biz they did with us in Q1 2010
- The amount of biz they did with us in Q2 2010
- The amount of biz they did with us in Q3 2010 ... etc.
Campaigns:
- Day of the week
- Time of Day
- Specific pitch
Results:
- Number of purchases per customer
- Value of purchases per customer
I could easily go hogwild, guessing at patterns and testing them to see if there's any accuracy; but it's not a best use of time. What I'd like to know are a few things...
- A ranking of the various pitches in their effectiveness, and an indication of how much the pitch affects the result.
- Same for the day of the week and time.
- And, trickier -- How much of a correlation does their previous business have to do with who converts now? Is it the previous big spenders that convert? Are they the ones buying the most product?
- And finally, of all the attributes above, how do they rank in their importance to conversions?
I assume these would all come with something like an R-squared indicating their 'accuracy.'
How would I go about starting? What reading would be essential to understanding this? Is there an approach/tool I can use that'll allow me to do a basic version of these sorts of analyses [and what would they be called] rapidly? I'm just looking for trends and hints of where it might be best to focus our energy; it can be rough.
Finally, does anyone know of a resource for basic data mining? analysis? for advertising in particular?
Much obliged,
| Checking what attributes of a population made objects more likely to do certain action | CC BY-SA 3.0 | null | 2011-07-12T03:09:31.770 | 2013-03-04T04:31:05.827 | 2011-07-12T08:46:28.937 | null | 252 | [
"correlation",
"business-intelligence"
] |
12925 | 2 | null | 12922 | 7 | null | Silhouette statistic works on distances, not similarities. One should revert similarities into distances. The general pass to do it: 1) set diagonal as 0, 2) revert sign of elements, 3) find the smallest element and substract it from each element, 4) set diagonal as 0.
For cosine or correlation there is also a geometrically more [correct way](https://stats.stackexchange.com/a/36158/3277): distance = sqrt[2(1-similarity)]; it comes from trigonometric "cosine theorem".
BTW, if you use SPSS you can find a collection of macros on my web-page that compute a number of clustering criterions, including Silhouette.
| null | CC BY-SA 3.0 | null | 2011-07-12T04:40:28.260 | 2013-02-26T18:18:45.257 | 2017-04-13T12:44:21.613 | -1 | 3277 | null |
12927 | 2 | null | 12878 | 1 | null | I am not so sure to get what exactly the problem is. If my answer is not what you expect, then you should probably bring some precision...
Let $p_A$, $p_B$, and $p_C$ the proportions of patients receiving treatments $A$, $B$, and $C$, respectively. The constraints are:
$\left\{
\begin{array}{l}
p_A - p_B = -0.36 \\
p_A - p_C = -0.26 \\
p_B - p_C = 0.10 \\
p_A + p_B + p_C = 1.
\end{array}
\right.$
Observe that equation 3 is a direct consequence of the first two.
This leads to $p_A = 0.126666$, $p_B = 0.486666$, and $p_C = 0.386666$.
Now, the [multinom](http://stat.ethz.ch/R-manual/R-patched/library/stats/html/Multinom.html) function in R can be used to generate treatment allocation.
Hope this helps...
Illustrative example in R that shows the use of the `multinom` function:
```
> pA <- 0.126667
> pB <- 0.486666
> pC <- 0.386667
>
> trt <- numeric(10000)
> for(i in 1:10000)
+ {
+ x <- rmultinom(n=1, size=1, prob=c(pA, pB, pC))
+ trt[i] <- which(x==1)
+ }
>
> table(trt) / 10000
trt
1 2 3
0.1260 0.4838 0.3902
```
| null | CC BY-SA 3.0 | null | 2011-07-12T06:14:47.400 | 2011-07-12T06:49:07.067 | 2011-07-12T06:49:07.067 | 3019 | 3019 | null |
12928 | 1 | 12932 | null | 4 | 874 | I'm reading Hayashi's "Econometrics", and in Chapter 2, page 101 he discusses the following white noise process:
I understand the calculations for the expected mean and variance, but I can't understand why the covariance would be 0 for $i\neq j$.
Since $E(z_i)=0$, the expression for the covariance would be reduced to $E(\cos(iw)\cdotp \cos(jw))$. Would someone mind showing me why this goes to zero? Assuming for the moment that $cos(iw)$ is positive, then $\cos((i+1)w)$ would be negative and so on. But why would there be 0 covariance for any two $i$ and $j$, $i\neq j$? I know this is a simple question, but can't seem to wrap my head around it.
| Calculating covariance for a non-strictly-stationary white noise process | CC BY-SA 3.0 | null | 2011-07-12T06:52:49.600 | 2011-07-12T09:02:41.497 | 2011-07-12T08:45:15.113 | null | 2251 | [
"econometrics",
"covariance-matrix"
] |
12929 | 2 | null | 12900 | 28 | null | Have you forgotten to include an intercept in your regression? I'm not familiar with SPSS code, but on page 21 of Hayashi's Econometrics:
>
If the regressors do not include a constant but (as some regression
software packages do) you nevertheless calculate $R^2$ by the formula
$R^2=1-\frac{\sum_{i=1}^{n}e_i^2}{\sum_{i=1}^{n}(y_i-\bar{y})^2}$
then the $R^2$ can be negative. This is because, without the benefit
of an intercept, the regression could do worse than the sample mean in
terms of tracking the dependent variable (i.e., the numerator could be greater than the denominator).
I'd check and make sure that SPSS is including an intercept in your regression.
| null | CC BY-SA 3.0 | null | 2011-07-12T07:04:06.423 | 2011-07-12T07:04:06.423 | null | null | 2251 | null |
12930 | 1 | 12931 | null | 6 | 2626 | What are some good, established methods for estimating the probability density function (denoted $f(x)$ from here on) of a continuous distribution, given a sample of points $x_1, \ldots, x_n$ drawn from it? I primarily need the PDF for plotting purposes.
The naive approach would be using a histogram, i.e. counting how many points fall into different $[a,b)$ intervals. But this has several problems:
- It doesn't give us $f(\frac{a+b}{2})$, but $\int_a^b f(x) \, dx$, which is not the same, and might look qualitatively different on a plot (e.g. for a Pareto distribution it gives an estimate of the PDF that is not a straight line on log-log scale, this is what I mean by looking qualitatively different).
- It heavily depends on binning, requiring a careful choice of bin size.
- Depending on the distribution, it may require a manual choice of a non-uniform bin size to get something reasonably-looking (e.g. a Pareto distribution requires increasing bins).
I am mainly interested in established methods (please note that I'm not a statistician, I don't have formal training in this, so I may not know about the obvious!), but any ideas are welcome too. E.g. would estimating the CDF by sorting the points, then somehow taking the derivative work? But then the problem is transformed to estimating the derivative of noisy data which is again a difficult problem.
I need this mainly not for fitting the PDF to some function, but for visualizing it.
EDIT: I am in particular interested in techniques that work well for long-tail distributions.
| Estimating PDF of continuous distribution from (few) data points | CC BY-SA 3.0 | null | 2011-07-12T08:12:51.963 | 2011-07-12T09:18:02.110 | 2011-07-12T09:18:02.110 | 4764 | 4764 | [
"data-visualization",
"density-function"
] |
12931 | 2 | null | 12930 | 7 | null | What you are looking for is kernel density estimation. You should find numerous hits on an internet search for these terms, and it is even on [Wikipedia](http://en.wikipedia.org/wiki/Kernel_density_estimation) so that should get you started. If you have R at your disposition, the function `density` provides what you need:
```
histAndDensity<-function(x, ...)
{
retval<-hist(x, freq=FALSE, ...)
lines(density(x, na.rm=TRUE), col="red")
invisible(retval)
}
```
| null | CC BY-SA 3.0 | null | 2011-07-12T08:24:37.677 | 2011-07-12T08:24:37.677 | null | null | 4257 | null |
12932 | 2 | null | 12928 | 3 | null | Use [trigonometric identities](http://en.wikipedia.org/wiki/List_of_trigonometric_identities)(scroll down to angle sum and difference identities):
$$\cos(iw)\cdot\cos(jw)=\frac{1}{2}\left(\cos((i+j)w)+\cos((i-j)w)\right)$$
Then substitute this formula to the following equation:
\begin{align}
E\cos(iw)\cdot\cos(jw)=\frac{1}{2\pi}\int_0^{2\pi}\cos(iw)\cdot\cos(jw)dw
\end{align}
and use the fact that
\begin{align}
\int_0^{2\pi}\cos(kw)dw=0
\end{align}
for $k\in\mathbb{N}$.
| null | CC BY-SA 3.0 | null | 2011-07-12T09:02:41.497 | 2011-07-12T09:02:41.497 | null | null | 2116 | null |
12933 | 1 | 13495 | null | 4 | 3997 | Im using gradient-descent-based algorithm for my problem where
new_value = old_value - Step_size*Gradient
For exit criteria, im determining the change in fn value between iteration i.e.,
if (old_Objective_fn_value - new_Objective_fn_value) <=0.001 exist otherwise continue.
For different Step_size, the algorithm meets the exit criteria at different point. For example, when my Step_size is x the final objective function value is p and when my Step_size is y the final objective function value is q.
I would like to know any logical reason why the algorithm converges at different objective fun values rather than at the same.
How can we make the algorithm converge to the same objective function value irrespective of the step size with the same exit criterion?
| Effects of step size in gradient descent optimisation | CC BY-SA 3.0 | null | 2011-07-12T09:09:32.477 | 2011-07-26T09:06:46.403 | 2011-07-12T11:50:27.390 | null | 4290 | [
"algorithms",
"optimization"
] |
12934 | 2 | null | 9990 | 13 | null | LGM can be translated to a MEM and vice versa, so these models are actually the same. I discuss the comparison in the chapter on LGM in my multilevel book, the draft of that chapter is on my homepage at [http://www.joophox.net/papers/chap14.pdf](http://www.joophox.net/papers/chap14.pdf)
| null | CC BY-SA 3.0 | null | 2011-07-12T09:17:38.710 | 2011-07-12T13:25:24.283 | 2011-07-12T13:25:24.283 | 919 | 5366 | null |
12935 | 1 | 12942 | null | 7 | 206 | Suppose $(\mathscr{X,B,P})$ is a [statistical model](http://en.wikipedia.org/wiki/Statistical_model) and $(\mathscr{X,B,P})^n$ is the corresponding model for sampling with replacement. My text book states that its [incompleteness](http://en.wikipedia.org/wiki/Completeness_%28statistics%29) allows multiple unbiased estimators, where problems such as which estimator is better arises. But I don't understand why the model is incomplete, can anyone give hints? Thank you~
| Why the statistical model of sampling with replacement is incomplete? | CC BY-SA 3.0 | null | 2011-07-12T10:19:15.910 | 2011-07-12T18:06:15.477 | 2011-07-12T10:37:11.870 | 4864 | 4864 | [
"mathematical-statistics"
] |
12936 | 1 | null | null | 3 | 5835 | As the R very beginner I'd like to ask if anyone could help me with the following problem?
```
library(foreign)
reg1 <- read.spss("URL of the spss file")
library(nlme)
model1 <- gls(v1 ~ x1 + x2 + x3 + x4 + x5 + x6, correlation=corARMA(p=1), data=reg1)
```
Error in eval(expr, envir, enclos) : object 'v1' not found
All the variables v1 to x6 are present in the spss file. Thanks in advance for help.
Thanks Nick. I can't share the file. I tried the procedure, but I got the similar result:
>
library(foreign)
library(nlme)
reg1 <- read.spss("URL", to.data.frame=TRUE)
class(reg1) #check that it is indeed a data.frame
[1] "data.frame"
colnames(reg1) #check that all column names are properly imported
....
[17] "x1" "x2" "x3" "x4"
[21] "x5" "x6"
....
[97] "...." "v1" "...." "...."
....
>
model1 <- gls(v1 ~ x1 + x2 + x3 + x4 + x5 + x6, correlation=corARMA(p=1), data=reg1)
Error in eval(expr, envir, enclos) : object 'img4' not found
| "Object not found" error in gls | CC BY-SA 3.0 | null | 2011-07-12T11:26:31.827 | 2011-07-12T12:08:23.383 | 2011-07-12T12:08:23.383 | 4837 | 4837 | [
"spss",
"r"
] |
12937 | 2 | null | 12936 | 6 | null | From the documentation of `?gls`:
`data: an optional data frame`
In relation, the documentation of `?read.spss` shows a parameter `to.data.frame`, whose default is `FALSE`, so typically `reg1` will be a list instead of a `data.frame`.
The code you provided is not useful for testing on my system (as you do not provide the actual dataset), but this will probably work (disregarding library statements):
```
reg1 <- read.spss("URL of the spss file", to.data.frame=TRUE)
class(reg1) #check that it is indeed a data.frame
colnames(reg1) #check that all column names are properly imported
model1 <- gls(v1 ~ x1 + x2 + x3 + x4 + x5 + x6, correlation=corARMA(p=1), data=reg1)
```
| null | CC BY-SA 3.0 | null | 2011-07-12T11:42:42.993 | 2011-07-12T11:42:42.993 | null | null | 4257 | null |
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