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Mean and Variance of the Uniform Random Variable. Consider the case of a uniform PDF over an interval \( \left\lbrack {a, b}\right\rbrack \), as in Example 3.1. | \[ \mathbf{E}\left\lbrack X\right\rbrack = {\int }_{-\infty }^{\infty }x{f}_{X}\left( x\right) {dx} \] \[ = {\int }_{a}^{b}x \cdot \frac{1}{b - a}{dx} \] \[ = {\left. \frac{1}{b - a} \cdot \frac{1}{2}{x}^{2}\right| }_{a}^{b} \] \[ = \frac{1}{b - a} \cdot \frac{{b}^{2} - {a}^{2}}{2} \] \[ = \frac{a + b}{2} \] as one exp... | Yes |
The time until a small meteorite first lands anywhere in the Sahara desert is modeled as an exponential random variable with a mean of 10 days. The time is currently midnight. What is the probability that a meteorite first lands some time between \( 6\mathrm{{am}} \) and \( 6\mathrm{{pm}} \) of the first day? | Let \( X \) be the time elapsed until the event of interest, measured in days. Then, \( X \) is exponential, with mean \( 1/\lambda = {10} \), which yields \( \lambda = 1/{10} \) . The desired probability is\n\n\[ \mathbf{P}\left( {1/4 \leq X \leq 3/4}\right) = \mathbf{P}\left( {X \geq 1/4}\right) - \mathbf{P}\left( {X... | Yes |
The Geometric and Exponential CDFs. Let \( X \) be a geometric random variable with parameter \( p \) ; that is, \( X \) is the number of trials to obtain the first success in a sequence of independent Bernoulli trials, where the probability of success is \( p \) . Thus, for \( k = 1,2,\ldots \), we have \( \mathbf{P}\... | \[ {F}^{\text{geo }}\left( n\right) = \mathop{\sum }\limits_{{k = 1}}^{n}p{\left( 1 - p\right) }^{k - 1} = p\frac{1 - {\left( 1 - p\right) }^{n}}{1 - \left( {1 - p}\right) } = 1 - {\left( 1 - p\right) }^{n},\;\text{ for }n = 1,2,\ldots \] | Yes |
The Maximum of Several Random Variables. You are allowed to take a certain test three times, and your final score will be the maximum of the test scores. Thus,\n\n\[ \nX = \max \left\{ {{X}_{1},{X}_{2},{X}_{3}}\right\} \n\]\n\nwhere \( {X}_{1},{X}_{2},{X}_{3} \) are the three test scores and \( X \) is the final score.... | We calculate the PMF indirectly. We first compute the CDF \( {F}_{X}\left( k\right) \) and then obtain the PMF as\n\n\[ \n{p}_{X}\left( k\right) = {F}_{X}\left( k\right) - {F}_{X}\left( {k - 1}\right) ,\;k = 1,\ldots ,{10}.\n\]\n\nWe have\n\n\[ \n{F}_{X}\left( k\right) = \mathbf{P}\left( {X \leq k}\right) \n\]\n\n\[ \n... | Yes |
Using the Normal Table. The annual snowfall at a particular geographic location is modeled as a normal random variable with a mean of \( \mu = {60} \) inches, and a standard deviation of \( \sigma = {20} \) . What is the probability that this year’s snowfall will be at least 80 inches? | Let \( X \) be the snow accumulation, viewed as a normal random variable, and let \[ Y = \frac{X - \mu }{\sigma } = \frac{X - {60}}{20} \] be the corresponding standard normal random variable. We want to find \[ \mathbf{P}\left( {X \geq {80}}\right) = \mathbf{P}\left( {\frac{X - {60}}{20} \geq \frac{{80} - {60}}{20}}\r... | Yes |
Example 3.9. Signal Detection. A binary message is transmitted as a signal that is either -1 or +1 . The communication channel corrupts the transmission with additive normal noise with mean \( \mu = 0 \) and variance \( {\sigma }^{2} \) . The receiver concludes that the signal -1 (or +1) was transmitted if the value re... | An error occurs whenever -1 is transmitted and the noise \( N \) is at least 1 so that \( N + S = N - 1 \geq 0 \), or whenever +1 is transmitted and the noise \( N \) is smaller than -1 so that \( N + S = N + 1 < 0 \) . In the former case, the probability of error is\n\n\[ \mathbf{P}\left( {N \geq 1}\right) = 1 - \math... | Yes |
The exponential random variable is memoryless. Alvin goes to a bus stop where the time \( T \) between two successive buses has an exponential PDF with parameter \( \lambda \) . Suppose that Alvin arrives \( t \) secs after the preceding bus arrival and let us express this fact with the event \( A = \{ T > t\} \) . Let... | We have\n\n\[ \mathbf{P}\left( {X > x \mid A}\right) = \mathbf{P}\left( {T > t + x \mid T > t}\right) \]\n\n\[ = \frac{\mathbf{P}\left( {T > t + x\text{ and }T > t}\right) }{\mathbf{P}\left( {T > t}\right) } \]\n\n\[ = \frac{\mathbf{P}\left( {T > t + x}\right) }{\mathbf{P}\left( {T > t}\right) } \]\n\n\[ = \frac{{e}^{-... | Yes |
Mean and Variance of a Piecewise Constant PDF. Suppose that the random variable \( X \) has the piecewise constant PDF\n\n\[ \n{f}_{X}\left( x\right) = \left\{ \begin{array}{ll} 1/3 & \text{ if }0 \leq x \leq 1 \\ 2/3 & \text{ if }1 < x \leq 2 \\ 0 & \text{ otherwise } \end{array}\right.\n\] | We now use the total expectation theorem to obtain\n\n\[ \n\mathbf{E}\left\lbrack X\right\rbrack = \mathbf{P}\left( {A}_{1}\right) \mathbf{E}\left\lbrack {X \mid {A}_{1}}\right\rbrack + \mathbf{P}\left( {A}_{2}\right) \mathbf{E}\left\lbrack {X \mid {A}_{2}}\right\rbrack = \frac{1}{3} \cdot \frac{1}{2} + \frac{2}{3} \cd... | Yes |
The metro train arrives at the station near your home every quarter hour starting at 6:00 AM. You walk into the station every morning between 7:10 and 7:30 AM, with the time in this interval being a uniform random variable. What is the PDF of the time you have to wait for the first train to arrive? | The time of your arrival, denoted by \( X \), is a uniform random variable on the interval from 7:10 to 7:30; see Fig. 3.15(a). Let \( Y \) be the waiting time. We calculate the PDF \( {f}_{Y} \) using a divide-and-conquer strategy. Let \( A \) and \( B \) be the events\n\n\[ A = \{ 7 : {10} \leq X \leq 7 : {15}\} = \{... | Yes |
Two-Dimensional Uniform PDF. Romeo and Juliet have a date at a given time, and each will arrive at the meeting place with a delay between 0 and 1 hour (recall the example given in Section 1.2). Let \( X \) and \( Y \) denote the delays of Romeo and Juliet, respectively. Assuming that no pairs \( \left( {x, y}\right) \)... | This is an example of a uniform PDF on the unit square. More generally, let us fix some subset \( S \) of the two-dimensional plane. The corresponding uniform joint PDF on \( S \) is defined to be\n\n\[ \n{f}_{X, Y}\left( {x, y}\right) = \left\{ \begin{array}{ll} \frac{1}{\operatorname{area}\text{ of }S} & \text{ if }\... | Yes |
We are told that the joint PDF of the random variables \( X \) and \( Y \) is a constant \( c \) on the set \( S \) shown in Fig. 3.16 and is zero outside. Find the value of \( c \) and the marginal PDFs of \( X \) and \( Y \) . | The area of the set \( S \) is equal to 4 and, therefore, \( {f}_{X, Y}\left( {x, y}\right) = c = 1/4 \), for \( \left( {x, y}\right) \in S \) . To find the marginal PDF \( {f}_{X}\left( x\right) \) for some particular \( x \), we integrate (with respect to \( y \) ) the joint PDF over the vertical line corresponding t... | Yes |
A surface is ruled with parallel lines, which are at distance \( d \) from each other. Suppose that we throw a needle of length \( l \) on the surface at random. What is the probability that the needle will intersect one of the lines? | We assume here that \( l < d \) so that the needle cannot intersect two lines simultaneously. Let \( X \) be the distance from the midpoint of the needle to the nearest of the parallel lines, and let \( \Theta \) be the acute angle formed by the axis of the needle and the parallel lines. We model the pair of random var... | Yes |
John throws a dart at a circular target of radius \( r \) (see Fig. 3.19). We assume that he always hits the target, and that all points of impact \( \left( {x, y}\right) \) are equally likely, so that the joint PDF of the random variables \( X \) and \( Y \) is uniform. | Following Example 3.13, and since the area of the circle is \( \pi {r}^{2} \), we have\n\n\[ \n{f}_{X, Y}\left( {x, y}\right) = \left\{ \begin{array}{ll} \frac{1}{\text{ area of the circle }} & \text{ if }\left( {x, y}\right) \text{ is in the circle,} \\ 0 & \text{ otherwise,} \end{array}\right. \]\n\n\[ \n= \left\{ \b... | Yes |
Let \( X \) be exponentially distributed with mean 1. Once we observe the experimental value \( x \) of \( X \), we generate a normal random variable \( Y \) with zero mean and variance \( x + 1 \) . What is the joint PDF of \( X \) and \( Y \) ? | We have \( {f}_{X}\left( x\right) = {e}^{-x} \), for \( x \geq 0 \), and\n\n\[ \n{f}_{Y \mid X}\left( {y \mid x}\right) = \frac{1}{\sqrt{{2\pi }\left( {x + 1}\right) }}{e}^{-{y}^{2}/2\left( {x + 1}\right) }.\n\]\n\nThus,\n\n\[ \n{f}_{X, Y}\left( {x, y}\right) = {f}_{X}\left( x\right) {f}_{Y \mid X}\left( {y \mid x}\rig... | Yes |
A lightbulb produced by the General Illumination Company is known to have an exponentially distributed lifetime \( Y \) . However, the company has been experiencing quality control problems. On any given day, the parameter \( \lambda \) of the PDF of \( Y \) is actually a random variable, uniformly distributed in the i... | We model the parameter \( \lambda \) as a random variable \( X \), with a uniform distribution. All available information about \( X \) is contained in the conditional PDF \( {f}_{X \mid X}\left( {x \mid y}\right) \) . We view \( y \) as a constant (equal to the observed value of \( Y \) ) and concentrate on the depend... | Yes |
Example 3.19. Let us revisit the signal detection problem considered in 3.9. A signal \( S \) is transmitted and we are given that \( \mathbf{P}\left( {S = 1}\right) = p \) and \( \mathbf{P}\left( {S = - 1}\right) = 1 - p \) . The received signal is \( Y = N + S \), where \( N \) is zero mean normal noise, with varianc... | Conditioned on \( S = s \), the random variable \( Y \) has a normal distribution with mean \( s \) and variance \( {\sigma }^{2} \) . Applying the formula developed above, we obtain\n\n\[ \mathbf{P}\left( {S = 1 \mid Y = y}\right) = \frac{{p}_{S}\left( 1\right) {f}_{Y \mid S}\left( {y \mid 1}\right) }{{f}_{Y}\left( y\... | Yes |
Let \( X \) and \( Y \) be described by a uniform PDF on the unit square. The joint CDF is given by\n\n\[ \n{F}_{X, Y}\left( {x, y}\right) = \mathbf{P}\left( {X \leq x, Y \leq y}\right) = {xy},\;\text{ for }0 \leq x, y \leq 1.\n\] | We then verify that\n\n\[ \n\frac{{\partial }^{2}{F}_{X, Y}}{\partial x\partial y}\left( {x, y}\right) = \frac{{\partial }^{2}\left( {xy}\right) }{\partial x\partial y}\left( {x, y}\right) = 1 = {f}_{X, Y}\left( {x, y}\right) ,\n\]\n\nfor all \( \left( {x, y}\right) \) in the unit square. | Yes |
Let \( X \) be uniform on \( \left\lbrack {0,1}\right\rbrack \) . Find the PDF of \( Y = \sqrt{X} \) . | Note that \( Y \) takes values between 0 and 1 . For every \( y \in \left\lbrack {0,1}\right\rbrack \), we have\n\n\[ \n{F}_{Y}\left( y\right) = \mathbf{P}\left( {Y \leq y}\right) = \mathbf{P}\left( {\sqrt{X} \leq y}\right) = \mathbf{P}\left( {X \leq {y}^{2}}\right) = {y}^{2},\;0 \leq y \leq 1.\n\]\n\nWe then different... | Yes |
John Slow is driving from Boston to the New York area, a distance of 180 miles. His average speed is uniformly distributed between 30 and 60 miles per hour. What is the PDF of the duration of the trip? | Let \( X \) be the speed and let \( Y = g\left( X\right) \) be the trip duration:\n\n\[ g\left( X\right) = \frac{180}{X}. \]\n\nTo find the CDF of \( Y \), we must calculate\n\n\[ \mathbf{P}\left( {Y \leq y}\right) = \mathbf{P}\left( {\frac{180}{X} \leq y}\right) = \mathbf{P}\left( {\frac{180}{y} \leq X}\right) . \]\n\... | Yes |
Let \( Y = g\left( X\right) = {X}^{2} \), where \( X \) is a random variable with known PDF. For any \( y \geq 0 \), we have\n\n\[ \n{F}_{Y}\left( y\right) = \mathbf{P}\left( {Y \leq y}\right) \n\]\n\n\[ \n= \mathbf{P}\left( {{X}^{2} \leq y}\right) \n\]\n\n\[ \n= \mathbf{P}\left( {-\sqrt{y} \leq X \leq \sqrt{y}}\right)... | and therefore, by differentiating and using the chain rule,\n\n\[ \n{f}_{Y}\left( y\right) = \frac{1}{2\sqrt{y}}{f}_{X}\left( \sqrt{y}\right) + \frac{1}{2\sqrt{y}}{f}_{X}\left( {-\sqrt{y}}\right) ,\;y \geq 0. \n\] | Yes |
A linear function of an exponential random variable. Suppose that \( X \) is an exponential random variable with PDF\n\n\[ \n{f}_{X}\left( x\right) = \left\{ \begin{array}{ll} \lambda {e}^{-{\lambda x}} & \text{ if }x \geq 0 \\ 0 & \text{ otherwise } \end{array}\right.\n\]\n\nwhere \( \lambda \) is a positive parameter... | \[ \n{f}_{Y}\left( y\right) = \left\{ \begin{array}{ll} \frac{\lambda }{\left| a\right| }{e}^{-\lambda \left( {y - b}\right) /a} & \text{ if }\left( {y - b}\right) /a \geq 0, \\ 0 & \text{ otherwise. } \end{array}\right.\n\] | Yes |
A linear function of a normal random variable is normal. Suppose that \( X \) is a normal random variable with mean \( \mu \) and variance \( {\sigma }^{2} \), and let \( Y = {aX} + b \), where \( a \) and \( b \) are some scalars. | We have\n\n\[ \n{f}_{X}\left( x\right) = \frac{1}{\sqrt{2\pi }\sigma }{e}^{-{\left( x - \mu \right) }^{2}/2{\sigma }^{2}}. \n\]\n\nTherefore,\n\n\[ \n{f}_{Y}\left( y\right) = \frac{1}{\left| a\right| }{f}_{X}\left( \frac{y - b}{a}\right) \n\]\n\n\[ \n= \frac{1}{\left| a\right| }\frac{1}{\sqrt{2\pi }\sigma }{e}^{-{\left... | Yes |
To check the PDF formula, let us apply it to the problem of Example 3.22. In the region of interest, \( x \in \left\lbrack {{30},{60}}\right\rbrack \), we have \( h\left( y\right) = {180}/y \), and | \[ \frac{d{F}_{X}}{dh}\left( {h\left( y\right) }\right) = \frac{1}{30},\;\left| {\frac{dh}{dy}\left( y\right) }\right| = \frac{180}{{y}^{2}}. \] Thus, in the region of interest \( y \in \left\lbrack {3,6}\right\rbrack \), the PDF formula yields \[ {f}_{Y}\left( y\right) = {f}_{X}\left( {h\left( y\right) }\right) \left|... | Yes |
Let \( Y = g\left( X\right) = {X}^{2} \), where \( X \) is a continuous uniform random variable in the interval \( (0,1\rbrack \) . Within this interval, \( g \) is monotonic, and its inverse is \( h\left( y\right) = \sqrt{y} \) . Thus, for any \( y \in (0,1\rbrack \), we have | \[
\left| {\frac{dh}{dy}\left( y\right) }\right| = \frac{1}{2\sqrt{y}},\;{f}_{X}\left( \sqrt{y}\right) = 1
\]
and
\[
{f}_{Y}\left( y\right) = \left\{ \begin{array}{ll} \frac{1}{2\sqrt{y}} & \text{ if }y \in (0,1\rbrack \\ 0 & \text{ otherwise. } \end{array}\right.
\] | Yes |
Two archers shoot at a target. The distance of each shot from the center of the target is uniformly distributed from 0 to 1 , independently of the other shot. What is the PDF of the distance of the losing shot from the center? | Let \( X \) and \( Y \) be the distances from the center of the first and second shots, respectively. Let also \( Z \) be the distance of the losing shot:\n\n\[ Z = \max \{ X, Y\} \]\n\nWe know that \( X \) and \( Y \) are uniformly distributed over \( \left\lbrack {0,1}\right\rbrack \), so that for all \( z \in \left\... | Yes |
Let \( X \) and \( Y \) be independent random variables that are uniformly distributed on the interval \( \left\lbrack {0,1}\right\rbrack \) . What is the PDF of the random variable \( Z = Y/X \) ? | We will find the PDF of \( Z \) by first finding its CDF and then differentiating. We consider separately the cases \( 0 \leq z \leq 1 \) and \( z > 1 \) . As shown in Fig. 3.25, we have\n\n\[ {F}_{Z}\left( z\right) = \mathbf{P}\left( {\frac{Y}{X} \leq z}\right) = \left\{ \begin{array}{ll} z/2 & \text{ if }0 \leq z \le... | Yes |
Romeo and Juliet have a date at a given time, and each, independently, will be late by an amount of time that is exponentially distributed with parameter \( \lambda \) . What is the PDF of the difference between their times of arrival? | Let us denote by \( X \) and \( Y \) the amounts by which Romeo and Juliet are late, respectively. We want to find the PDF of \( Z = X - Y \), assuming that \( X \) and \( Y \) are independent and exponentially distributed with parameter \( \lambda \) . We will first calculate the CDF \( {F}_{Z}\left( z\right) \) by co... | Yes |
Example 4.1. Let\n\n\\[ \n{p}_{X}\left( x\right) = \left\{ \begin{array}{ll} 1/2, & \\text{ if }x = 2 \\ 1/6, & \\text{ if }x = 3 \\ 1/3, & \\text{ if }x = 5 \end{array}\\right. \n\\]\n\nThen, the corresponding transform is | \\[ \nM\\left( s\\right) = \\frac{1}{2}{e}^{2s} + \\frac{1}{6}{e}^{3s} + \\frac{1}{3}{e}^{5s} \n\\] | Yes |
Example 4.2. The Transform of a Poisson Random Variable. Consider a Poisson random variable \( X \) with parameter \( \lambda \) :\n\n\[ \n{p}_{X}\left( x\right) = \frac{{\lambda }^{x}{e}^{-\lambda }}{x!},\;x = 0,1,2,\ldots \n\] | The corresponding transform is given by\n\n\[ \nM\left( s\right) = \mathop{\sum }\limits_{{x = 0}}^{\infty }{e}^{sx}\frac{{\lambda }^{x}{e}^{-\lambda }}{x!}. \n\]\n\nWe let \( a = {e}^{s}\lambda \) and obtain\n\n\[ \nM\left( s\right) = {e}^{-\lambda }\mathop{\sum }\limits_{{x = 0}}^{\infty }\frac{{a}^{x}}{x!} = {e}^{-\... | Yes |
The Transform of an Exponential Random Variable. Let \( X \) be an exponential random variable with parameter \( \lambda \) :\n\n\[ \n{f}_{X}\left( x\right) = \lambda {e}^{-{\lambda x}},\;x \geq 0.\n\]\n\nThen, | \[ \nM\left( s\right) = \lambda {\int }_{0}^{\infty }{e}^{sx}{e}^{-{\lambda x}}{dx}\n\]\n\n\[ \n= \lambda {\int }_{0}^{\infty }{e}^{\left( {s - \lambda }\right) x}{dx}\n\]\n\n\[ \n= {\left. \lambda \frac{{e}^{\left( {s - \lambda }\right) x}}{s - \lambda }\right| }_{0}^{\infty }\;\text{ (if }s < \lambda \text{ ) }\n\]\n... | Yes |
The Transform of a Linear Function of a Random Variable. Let \( {M}_{X}\left( s\right) \) be the transform associated with a random variable \( X \) . Consider a new random variable \( Y = {aX} + b \) . We then have | \[ {M}_{Y}\left( s\right) = \mathbf{E}\left\lbrack {e}^{s\left( {{aX} + b}\right) }\right\rbrack = {e}^{sb}\mathbf{E}\left\lbrack {e}^{saX}\right\rbrack = {e}^{sb}{M}_{X}\left( {sa}\right) . \] | Yes |
The Transform of a Normal Random Variable. Let \( X \) be a normal random variable with mean \( \mu \) and variance \( {\sigma }^{2} \) . To calculate the corresponding transform, we first consider the special case of the standard normal random variable \( Y \), where \( \mu = 0 \) and \( {\sigma }^{2} = 1 \), and then... | The PDF of the standard normal is\n\n\[ \n{f}_{Y}\left( y\right) = \frac{1}{\sqrt{2\pi }}{e}^{-{y}^{2}/2} \n\]\n\nand its transform is\n\n\[ \n{M}_{Y}\left( s\right) = {\int }_{-\infty }^{\infty }\frac{1}{\sqrt{2\pi }}{e}^{-{y}^{2}/2}{e}^{sy}{dy} \n\]\n\n\[ \n= \frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }{e}^{-\l... | Yes |
Example 4.6. We saw earlier (Example 4.1) that the PMF\n\n\\[ \n{p}_{X}\left( x\right) = \\left\\{ \\begin{array}{ll} 1/2, & \\text{ if }x = 2 \\\\ 1/6, & \\text{ if }x = 3 \\\\ 1/3, & \\text{ if }x = 5 \\end{array}\\right.\n\\]\n\nhas the transform\n\\[ \nM\\left( s\\right) = \\frac{1}{2}{e}^{2s} + \\frac{1}{6}{e}^{3s... | Thus,\n\n\\[ \n\\mathbf{E}\\left\\lbrack X\\right\\rbrack = {\\left. \\frac{d}{ds}M\\left( s\\right) \\right| }_{s = 0}\n\\]\n\n\\[ \n= \\frac{1}{2}2{e}^{2s} + {\\left. \\frac{1}{6}3{e}^{3s} + \\frac{1}{3}5{e}^{5s}\\right| }_{s = 0}\n\\]\n\n\\[ \n= \\frac{1}{2} \\cdot 2 + \\frac{1}{6} \\cdot 3 + \\frac{1}{3} \\cdot 5\n... | Yes |
We are told that the transform associated with a random variable \( X \) is \( M\left( s\right) = \frac{1}{4}{e}^{-s} + \frac{1}{2} + \frac{1}{8}{e}^{4s} + \frac{1}{8}{e}^{5s} \). | Since \( M\left( s\right) \) is a sum of terms of the form \( {e}^{sx} \), we can compare with the general formula \( M\left( s\right) = \mathop{\sum }\limits_{x}{e}^{sx}{p}_{X}\left( x\right) \) and infer that \( X \) is a discrete random variable. The different values that \( X \) can take can be read from the corres... | Yes |
The Transform of a Geometric Random Variable. We are told that the transform associated with random variable \( X \) is of the form\n\n\[ M\left( s\right) = \frac{p{e}^{s}}{1 - \left( {1 - p}\right) {e}^{s}}, \]\n\nwhere \( p \) is a constant in the range \( 0 < p < 1 \). We wish to find the distribution of \( X \). | We recall the formula for the geometric series:\n\n\[ \frac{1}{1 - \alpha } = 1 + \alpha + {\alpha }^{2} + \cdots \]\n\nwhich is valid whenever \( \left| \alpha \right| < 1 \). We use this formula with \( \alpha = \left( {1 - p}\right) {e}^{s} \), and for \( s \) sufficiently close to zero so that \( \left( {1 - p}\rig... | Yes |
Example 4.9. The Transform of a Mixture of Two Distributions. The neighborhood bank has three tellers, two of them fast, one slow. The time to assist a customer is exponentially distributed with parameter \( \lambda = 6 \) at the fast tellers, and \( \lambda = 4 \) at the slow teller. Jane enters the bank and chooses a... | We have\n\n\[ \n{f}_{X}\left( x\right) = \frac{2}{3} \cdot 6{e}^{-{6x}} + \frac{1}{3} \cdot 4{e}^{-{4x}},\;x \geq 0. \n\]\n\nThen,\n\n\[ \nM\left( s\right) = {\int }_{0}^{\infty }{e}^{sx}\left( {\frac{2}{3}6{e}^{-{6x}} + \frac{1}{3}4{e}^{-{4x}}}\right) {dx} \n\]\n\n\[ \n= \frac{2}{3}{\int }_{0}^{\infty }{e}^{sx}6{e}^{-... | Yes |
The Transform of the Binomial. Let \( {X}_{1},\ldots ,{X}_{n} \) be independent Bernoulli random variables with a common parameter \( p \) . Then, | \[ {M}_{{X}_{i}}\left( s\right) = \left( {1 - p}\right) {e}^{0s} + p{e}^{1s} = 1 - p + p{e}^{s},\;\text{ for all }i. \] The random variable \( Y = {X}_{1} + \cdots + {X}_{n} \) is binomial with parameters \( n \) and \( p \) . Its transform is given by \[ {M}_{Y}\left( s\right) = {\left( 1 - p + p{e}^{s}\right) }^{n}. ... | Yes |
The Sum of Independent Poisson Random Variables is Poisson. Let \( X \) and \( Y \) be independent Poisson random variables with means \( \lambda \) and \( \mu \), respectively, and let \( W = X + Y \) . | \[ {M}_{X}\left( s\right) = {e}^{\lambda \left( {{e}^{s} - 1}\right) },\;{M}_{Y}\left( s\right) = {e}^{\mu \left( {{e}^{s} - 1}\right) }, \] and \[ {M}_{W}\left( s\right) = {M}_{X}\left( s\right) {M}_{Y}\left( s\right) = {e}^{\lambda \left( {{e}^{s} - 1}\right) }{e}^{\mu \left( {{e}^{s} - 1}\right) } = {e}^{\left( {\la... | Yes |
The Sum of Independent Normal Random Variables is Normal. Let \( X \) and \( Y \) be independent normal random variables with means \( {\mu }_{x} \) , \( {\mu }_{y} \), and variances \( {\sigma }_{x}^{2},{\sigma }_{y}^{2} \), respectively. Let \( W = X + Y \) . | \[ {M}_{X}\left( s\right) = {e}^{\frac{{\sigma }_{x}^{2}{s}^{2}}{2} + {\mu }_{x}s},\;{M}_{Y}\left( s\right) = {e}^{\frac{{\sigma }_{y}^{2}{s}^{2}}{2} + {\mu }_{y}s}, \] and \[ {M}_{W}\left( s\right) = {e}^{\frac{\left( {{\sigma }_{x}^{2} + {\sigma }_{y}^{2}}\right) {s}^{2}}{2} + \left( {{\mu }_{x} + {\mu }_{y}}\right) ... | Yes |
Let \( X \) and \( Y \) be independent and have PMFs given by\n\n\[ \n{p}_{X}\left( x\right) = \left\{ {\begin{array}{ll} \frac{1}{3} & \text{ if }x = 1,2,3, \\ 0 & \text{ otherwise }, \end{array}\;{p}_{Y}\left( y\right) = \left\{ \begin{array}{ll} \frac{1}{2} & \text{ if }x = 0, \\ \frac{1}{3} & \text{ if }x = 1, \\ \... | We calculate \( {p}_{W}\left( w\right) \) for each of the values \( w = 1,2,3,4,5 \) using the convolution formula. We have\n\n\[ \n{p}_{W}\left( 1\right) = \mathop{\sum }\limits_{x}{p}_{X}\left( x\right) {p}_{Y}\left( {1 - x}\right) = {p}_{X}\left( 1\right) \cdot {p}_{Y}\left( 0\right) = \frac{1}{3} \cdot \frac{1}{2} ... | Yes |
The random variables \( X \) and \( Y \) are independent and uniformly distributed in the interval \( \left\lbrack {0,1}\right\rbrack \) . The PDF of \( W = X + Y \) is | \[ {f}_{W}\left( w\right) = {\int }_{-\infty }^{\infty }{f}_{X}\left( x\right) {f}_{Y}\left( {w - x}\right) {dx}. \] The integrand \( {f}_{X}\left( x\right) {f}_{Y}\left( {w - x}\right) \) is nonzero (and equal to 1) for \( 0 \leq x \leq 1 \) and \( 0 \leq w - x \leq 1 \) . Combining these two inequalities, the integra... | Yes |
Let the random variables \( X \) and \( Y \) have a joint PDF which is equal to 2 for \( \left( {x, y}\right) \) belonging to the triangle indicated in Fig. 4.6(a), and zero everywhere else. In order to compute \( \mathbf{E}\left\lbrack {X \mid Y = y}\right\rbrack \), we first need to obtain the conditional density of ... | We have\n\n\[ \n{f}_{Y}\left( y\right) = {\int }_{-\infty }^{\infty }{f}_{X, Y}\left( {x, y}\right) {dx} = {\int }_{0}^{1 - y}{2dx} = 2\left( {1 - y}\right) ,\;0 \leq y \leq 1, \n\] \n\nand \n\n\[ \n{f}_{X \mid Y}\left( {x \mid y}\right) = \frac{{f}_{X, Y}\left( {x, y}\right) }{{f}_{Y}\left( y\right) } = \frac{1}{1 - y... | Yes |
We saw that \( \mathbf{E}\left\lbrack {X \mid Y = y}\right\rbrack = \left( {1 - y}\right) /2 \) . Hence, \( \mathbf{E}\left\lbrack {X \mid Y}\right\rbrack \) is the random variable \( \left( {1 - Y}\right) /2 \) : | \[ \mathbf{E}\left\lbrack {\mathbf{E}\left\lbrack {X \mid Y}\right\rbrack }\right\rbrack = \left\{ \begin{array}{ll} \mathop{\sum }\limits_{y}\mathbf{E}\left\lbrack {X \mid Y = y}\right\rbrack {p}_{Y}\left( y\right) , & Y\text{ discrete,} \\ {\int }_{-\infty }^{\infty }\mathbf{E}\left\lbrack {X \mid Y = y}\right\rbrack... | Yes |
In Example 4.15, we found \( \mathbf{E}\left\lbrack {X \mid Y}\right\rbrack = \left( {1 - Y}\right) /2 \) [see Fig. 4.6(b)]. Taking expectations of both sides, and using the law of iterated expectations to evaluate the left-hand side, we obtain \( \mathbf{E}\left\lbrack X\right\rbrack = \left( {1 - \mathbf{E}\left\lbra... | Because of symmetry, we must have \( \mathbf{E}\left\lbrack X\right\rbrack = \mathbf{E}\left\lbrack Y\right\rbrack \) . Therefore, \( \mathbf{E}\left\lbrack X\right\rbrack = \left( {1 - \mathbf{E}\left\lbrack X\right\rbrack }\right) /2 \), which yields \( \mathbf{E}\left\lbrack X\right\rbrack = 1/3 \) . | Yes |
Example 4.16. We start with a stick of length \( \ell \) . We break it at a point which is chosen randomly and uniformly over its length, and keep the piece that contains the left end of the stick. We then repeat the same process on the stick that we were left with. What is the expected length of the stick that we are ... | Let \( Y \) be the length of the stick after we break for the first time. Let \( X \) be the length after the second time. We have \( \mathbf{E}\left\lbrack {X \mid Y}\right\rbrack = Y/2 \), since the breakpoint is chosen uniformly over the length \( Y \) of the remaining stick. For a similar reason, we also have \( \m... | Yes |
Averaging Quiz Scores by Section. A class has \( n \) students and the quiz score of student \( i \) is \( {x}_{i} \) . The average quiz score is\n\n\[ m = \frac{1}{n}\mathop{\sum }\limits_{{i = 1}}^{n}{x}_{i} \]\n\nThe class consists of \( S \) sections, with \( {n}_{s} \) students in section \( s \) . The average sco... | \[ \mathop{\sum }\limits_{{s = 1}}^{S}\frac{{n}_{s}}{n}{m}_{s} = \mathop{\sum }\limits_{{s = 1}}^{S}\frac{{n}_{s}}{n} \cdot \frac{1}{{n}_{s}}\mathop{\sum }\limits_{{\text{stdnts. }i\text{ in sec. }s}}{x}_{i} \]\n\n\[ = \frac{1}{n}\mathop{\sum }\limits_{{s = 1}}^{S}\mathop{\sum }\limits_{{\text{stdnts. }i\text{ in sec. ... | Yes |
Consider again the problem where we break twice a stick of length \( \ell \), at randomly chosen points, with \( Y \) being the length of the stick after the first break and \( X \) being the length after the second break. We calculated the mean of \( X \) as \( \ell /4 \), and now let us use the law of conditional var... | We have \( \mathbf{E}\left\lbrack {X \mid Y}\right\rbrack = Y/2 \), so since \( Y \) is uniformly distributed between 0 and \( \ell \) ,\n\n\[ \operatorname{var}\left( {\mathbf{E}\left\lbrack {X \mid Y}\right\rbrack }\right) = \operatorname{var}\left( {Y/2}\right) = \frac{1}{4}\operatorname{var}\left( Y\right) = \frac{... | Yes |
Computing Variances by Conditioning. Consider a continuous random variable \( X \) with the PDF given in Fig. 4.7. We define an auxiliary random variable \( Y \) as follows:\n\n\[ Y = \left\{ \begin{array}{ll} 1, & \text{ if }x < 1 \\ 2, & \text{ of }x \geq 1 \end{array}\right. \] | Here, \( \mathbf{E}\left\lbrack {X \mid Y}\right\rbrack \) takes the values \( 1/2 \) and \( 3/2 \), with probabilities \( 1/3 \) and \( 2/3 \), respectively. Thus, the mean of \( \mathbf{E}\left\lbrack {X \mid Y}\right\rbrack \) is \( 7/6 \) . Therefore,\n\n\[ \operatorname{var}\left( {\mathbf{E}\left\lbrack {X \mid Y... | Yes |
A remote village has three gas stations, and each one of them is open on any given day with probability \( 1/2 \), independently of the others. The amount of gas available in each gas station is unknown and is uniformly distributed between 0 and 1000 gallons. We wish to characterize the distribution of the total amount... | The number \( N \) of open gas stations is a binomial random variable with \( p = \) \( 1/2 \) and the corresponding transform is\n\n\[ \n{M}_{N}\left( s\right) = {\left( 1 - p + p{e}^{s}\right) }^{3} = \frac{1}{8}{\left( 1 + {e}^{s}\right) }^{3}. \n\]\n\nThe transform \( {M}_{X}\left( s\right) \) associated with the a... | Yes |
Sum of a Geometric Number of Independent Exponential Random Variables. Jane visits a number of bookstores, looking for Great Expectations. Any given bookstore carries the book with probability \( p \), independently of the others. In a typical bookstore visited, Jane spends a random amount of time, exponentially distri... | The total number \( N \) of bookstores visited is geometrically distributed with parameter \( p \) . Hence, the total time \( Y \) spent in bookstores is the sum of a geometrically distributed number \( N \) of independent exponential random variables \( {X}_{1},{X}_{2},\ldots \) We have\n\[ \mathbf{E}\left\lbrack Y\ri... | Yes |
Example 4.23. Sum of a Geometric Number of Independent Geometric Random Variables. This example is a discrete counterpart of the preceding one. We let \( N \) be geometrically distributed with parameter \( p \) . We also let each random variable \( {X}_{i} \) be geometrically distributed with parameter \( q \) . We ass... | We have\n\n\[ \n{M}_{N}\left( s\right) = \frac{p{e}^{s}}{1 - \left( {1 - p}\right) {e}^{s}},\;{M}_{X}\left( s\right) = \frac{q{e}^{s}}{1 - \left( {1 - q}\right) {e}^{s}}. \n\] \n\nTo determine \( {M}_{Y}\left( s\right) \), we start with the formula for \( {M}_{N}\left( s\right) \) and replace each occurrence of \( {e}^... | Yes |
The pair of random variables \( \left( {X, Y}\right) \) takes the values \( \left( {1,0}\right) ,\left( {0,1}\right) \) , \( \left( {-1,0}\right) \), and \( \left( {0, - 1}\right) \), each with probability \( 1/4 \) (see Fig. 4.9). Thus, the marginal PMFs of \( X \) and \( Y \) are symmetric around 0, and \( \mathbf{E}... | \[ \operatorname{cov}\left( {X, Y}\right) = \mathbf{E}\left\lbrack {\left( {X - \mathbf{E}\left\lbrack X\right\rbrack }\right) \left( {Y - \mathbf{E}\left\lbrack Y\right\rbrack }\right) }\right\rbrack = \mathbf{E}\left\lbrack {XY}\right\rbrack = 0, \] | Yes |
Consider \( n \) independent tosses of a biased coin with probability of a head equal to \( p \) . Let \( X \) and \( Y \) be the numbers of heads and of tails, respectively, and let us look at the correlation of \( X \) and \( Y \) . Here, for all possible pairs of values \( \left( {x, y}\right) \), we have \( x + y =... | We have\n\n\[ \operatorname{cov}\left( {X, Y}\right) = \mathbf{E}\left\lbrack {\left( {X - \mathbf{E}\left\lbrack X\right\rbrack }\right) \left( {Y - \mathbf{E}\left\lbrack Y\right\rbrack }\right) }\right\rbrack \]\n\n\[ = - \mathbf{E}\left\lbrack {\left( X - \mathbf{E}\left\lbrack X\right\rbrack \right) }^{2}\right\rb... | Yes |
Consider the hat problem discussed in Section 2.5, where \( n \) people throw their hats in a box and then pick a hat at random. Let us find the variance of \( X \), the number of people that pick their own hat. | We have\n\n\[ \nX = {X}_{1} + \cdots + {X}_{n} \]\n\nwhere \( {X}_{i} \) is the random variable that takes the value 1 if the \( i \) th person selects his/her own hat, and takes the value 0 otherwise. Noting that \( {X}_{i} \) is Bernoulli with parameter \( p = \mathbf{P}\left( {{X}_{i} = 1}\right) = 1/n \), we obtain... | Yes |
Let \( X \) be uniformly distributed in the interval \( \left\lbrack {4,{10}}\right\rbrack \) and suppose that we observe \( X \) with some random error \( W \), that is, we observe the experimental value of the random variable\n\n\[ Y = X + W \]\n\nWe assume that \( W \) is uniformly distributed in the interval \( \le... | We have \( {f}_{X}\left( x\right) = 1/6 \) for \( 4 \leq x \leq {10} \), and \( {f}_{X}\left( x\right) = 0 \), elsewhere. Conditioned on \( X \) being equal to some \( x, Y \) is the same as \( x + W \), and is uniform over the interval \( \left\lbrack {x - 1, x + 1}\right\rbrack \) . Thus, the joint PDF is given by\n\... | Yes |
Let us say that the observed random variable \( Y \) is uninformative if the mean squared estimation error \( \mathbf{E}\left\lbrack {\widetilde{X}}^{2}\right\rbrack = \operatorname{var}\left( \widetilde{X}\right) \) is the same as the unconditional variance \( \operatorname{var}\left( X\right) \) of \( X \) . When is ... | Using the formula\n\n\[\n\operatorname{var}\left( X\right) = \operatorname{var}\left( \widehat{X}\right) + \operatorname{var}\left( \widetilde{X}\right)\n\]\n\nwe see that \( Y \) is uninformative if and only if \( \operatorname{var}\left( \widehat{X}\right) = 0 \) . The variance of a random variable is zero if and onl... | Yes |
Let \( X \) have a normal distribution with zero mean and unit variance. Let \( z \) be independent of \( X \), with \( \mathbf{P}\left( {Z = 1}\right) = \mathbf{P}\left( {Z = - 1}\right) = 1/2 \) . Let \( Y = {ZX} \), which is also normal with zero mean (why?). Furthermore, | \[ \mathbf{E}\left\lbrack {XY}\right\rbrack = \mathbf{E}\left\lbrack {Z{X}^{2}}\right\rbrack = \mathbf{E}\left\lbrack Z\right\rbrack \mathbf{E}\left\lbrack {X}^{2}\right\rbrack = 0 \times 1 = 0, \] so \( X \) and \( Y \) are uncorrelated. On the other hand \( X \) and \( Y \) are clearly dependent. (For example, if \( ... | Yes |
Let \( N \) be the first time in which we have a success immediately following a previous success. (That is, \( N \) is the first \( i \) for which \( {X}_{i - 1} = {X}_{i} = 1 \) .) What is the probability \( \mathbf{P}\left( {{X}_{N + 1} = {X}_{N + 2} = 0}\right) \) that there are no successes in the two trials that ... | Intuitively, once the condition \( {X}_{N - 1} = {X}_{N} = 1 \) is satisfied, from then on, the future of the process still consists of independent Bernoulli trials. Therefore the probability of an event that refers to the future of the process is the same as in a fresh-starting Bernoulli process, so that \( \mathbf{P}... | Yes |
(a) \( T = \) the time index of the first idle slot; | We recognize \( T \) as a geometrically distributed random variable with parameter \( 1 - p \) . Its PMF is\n\n\[ \n{p}_{T}\left( k\right) = {p}^{k - 1}\left( {1 - p}\right) ,\;k = 1,2,\ldots \n\]\n\nIts mean and variance are\n\n\[ \n\mathbf{E}\left\lbrack T\right\rbrack = \frac{1}{1 - p},\;\operatorname{var}\left( T\r... | Yes |
In each minute of basketball play, Alice commits a single foul with probability \( p \) and no foul with probability \( 1 - p \) . The number of fouls in different minutes are assumed to be independent. Alice will foul out of the game once she commits her sixth foul, and will play 30 minutes if she does not foul out. W... | We model fouls as a Bernoulli process with parameter \( p \) . Alice’s playing time \( Z \) is equal to \( {Y}_{6} \), the time until the sixth foul, except if \( {Y}_{6} \) is larger than 30, in which case, her playing time is 30, the duration of the game; that is, \( Z = \min \left\{ {{Y}_{6},{30}}\right\} \) . The r... | Yes |
As a rule of thumb, the Poisson/binomial approximation\n\n\[ \n{e}^{-\lambda }\frac{{\lambda }^{k}}{k!} \approx \frac{n!}{\left( {n - k}\right) !k!}{p}^{k}{\left( 1 - p\right) }^{n - k},\;k = 0,1,\ldots, n, \n\]\n\nis valid to several decimal places if \( n \geq {100}, p \leq {0.01} \), and \( \lambda = {np} \). To che... | Gary Kasparov, the world chess champion (as of 1999) plays against 100 amateurs in a large simultaneous exhibition. It has been estimated from past experience that Kasparov wins in such exhibitions \( {99}\% \) of his games on the average (in precise probabilistic terms, we assume that he wins each game with probabilit... | Yes |
A packet consisting of a string of \( n \) symbols is transmitted over a noisy channel. Each symbol has probability \( p = {0.0001} \) of being transmitted in error, independently of errors in the other symbols. How small should \( n \) be in order for the probability of incorrect transmission (at least one symbol in e... | Each symbol transmission is viewed as an independent Bernoulli trial. Thus, the probability of a positive number \( S \) of errors in the packet is\n\n\[ 1 - \mathbf{P}\left( {S = 0}\right) = 1 - {\left( 1 - p\right) }^{n}. \]\n\nFor this probability to be less than 0.001, we must have \( 1 - {\left( 1 - {0.0001}\right... | Yes |
You get email according to a Poisson process at a rate of \( \lambda = {0.2} \) messages per hour. You check your email every hour. What is the probability of finding 0 and 1 new messages? | These probabilities can be found using the Poisson PMF \( {\left( \lambda \tau \right) }^{k}{e}^{-{\lambda \tau }}/k \), with \( \tau = 1 \), and \( k = 0 \) or \( k = 1 \) :\n\n\[ \mathbf{P}\left( {0,1}\right) = {e}^{-{0.2}} = {0.819},\;\mathbf{P}\left( {1,1}\right) = {0.2} \cdot {e}^{-{0.2}} = {0.164} \] | No |
Sum of Independent Poisson Random Variables. Arrivals of customers at the local supermarket are modeled by a Poisson process with a rate of \( \lambda = {10} \) customers per minute. Let \( M \) be the number of customers arriving between 9:00 and 9:10. Also, let \( N \) be the number of customers arriving between 9:30... | We notice that \( M \) is Poisson with parameter \( \mu = {10} \cdot {10} = {100} \) and \( N \) is Poisson with parameter \( \nu = {10} \cdot 5 = {50} \) . Furthermore, \( M \) and \( N \) are independent. As shown in Section 4.1, using transforms, \( M + N \) is Poisson with parameter \( \mu + \nu = {150} \) . We wil... | Yes |
During rush hour, from 8 am to 9 am, traffic accidents occur according to a Poisson process with a rate \( \mu \) of 5 accidents per hour. Between 9 am and 11 am, they occur as an independent Poisson process with a rate \( \nu \) of 3 accidents per hour. What is the PMF of the total number of accidents between 8 am and... | This is the sum of two independent Poisson random variables with parameters 5 and \( 3 \cdot 2 = 6 \), respectively. Since the sum of independent Poisson random variables is also Poisson, the total number of accidents has a Poisson PMF with parameter \( 5 + 6 = {11} \) . | Yes |
When you enter the bank, you find that all three tellers are busy serving other customers, and there are no other customers in queue. Assume that the service times for you and for each of the customers being served are independent identically distributed exponential random variables. What is the probability that you wi... | The answer is \( 1/3 \) . To see this, focus at the moment when you start service with one of the tellers. Then, the remaining time of each of the other two customers being served, as well as your own remaining time, have the same PDF. Therefore, you and the other two customers have equal probability \( 1/3 \) of being... | Yes |
You call the IRS hotline and you are told that you are the 56th person in line, excluding the person currently being served. Callers depart according to a Poisson process with a rate of \( \lambda = 2 \) per minute. How long will you have to wait on the average until your service starts, and what is the probability you... | By the memoryless property, the remaining service time of the person currently being served is exponentially distributed with parameter 2 . The service times of the 55 persons ahead of you are also exponential with the same parameter, and all of these random variables are independent. Thus, your waiting time \( Y \) is... | Yes |
A packet that arrives at a node of a data network is either a local packet which is destined for that node (this happens with probability \( p \) ), or else it is a transit packet that must be relayed to another node (this happens with probability \( 1 - p \) ). Packets arrive according to a Poisson process with rate \... | We verify that the process of local packet arrivals satisfies the defining properties of a Poisson process. Since \( \lambda \) and \( p \) are constant (do not change with time), the first property (time homogeneity) clearly holds. Furthermore, there is no dependence between what happens in disjoint time intervals, ve... | Yes |
Merging of Poisson Processes. People with letters to mail arrive at the post office according to a Poisson process with rate \( {\lambda }_{1} \), while people with packages to mail arrive according to an independent Poisson process with rate \( {\lambda }_{2} \). As stated earlier the merged process, which includes ar... | First, it should be clear that the merged process satisfies the time-homogeneity property. Furthermore, since different intervals in each of the two arrival processes are independent, the same property holds for the merged process. Let us now focus on a small interval of length \( \delta \) . Ignoring terms that are ne... | Yes |
Two light bulbs have independent and exponentially distributed lifetimes \( {T}^{\left( 1\right) } \) and \( {T}^{\left( 2\right) } \), with parameters \( {\lambda }_{1} \) and \( {\lambda }_{2} \) , respectively. What is the distribution of the first time \( Z = \min \left\{ {{T}^{\left( 1\right) },{T}^{\left( 2\right... | We can treat this as an exercise in derived distributions. For all \( z \geq 0 \), we have,\n\n\[ \n{F}_{Z}\left( z\right) = \mathbf{P}\left( {\min \left\{ {{T}^{\left( 1\right) },{T}^{\left( 2\right) }}\right\} \leq z}\right) \n\]\n\n\[ \n= 1 - \mathbf{P}\left( {\min \left\{ {{T}^{\left( 1\right) },{T}^{\left( 2\right... | Yes |
Three light bulbs have independent exponentially distributed lifetimes with a common parameter \( \lambda \) . What is the expectation of the time until the last bulb burns out? | We think of the times when each bulb burns out as the first arrival times in independent Poisson processes. In the beginning, we have three bulbs, and the merged process has rate \( {3\lambda } \) . Thus, the time \( {T}_{1} \) of the first burnout is exponential with parameter \( {3\lambda } \), and mean \( 1/{3\lambd... | Yes |
Alice is taking a probability class and in each week she can be either up-to-date or she may have fallen behind. If she is up-to-date in a given week, the probability that she will be up-to-date (or behind) in the next week is 0.8 (or 0.2 , respectively). If she is behind in the given week, the probability that she wil... | Let us introduce states 1 and 2, and identify them with being up-to-date and behind, respectively. Then, the transition probabilities are\n\n\[ \n{p}_{11} = {0.8},\;{p}_{12} = {0.2},\;{p}_{21} = {0.6},\;{p}_{22} = {0.4}, \n\]\n\nand the transition probability matrix is\n\n\[ \n\left\lbrack \begin{array}{ll} {0.8} & {0.... | Yes |
A fly moves along a straight line in unit increments. At each time period, it moves one unit to the left with probability 0.3, one unit to the right with probability 0.3, and stays in place with probability 0.4, independently of the past history of movements. A spider is lurking at positions 1 and \( m \): if the fly l... | Let us introduce states \( 1,2,\ldots, m \), and identify them with the corresponding positions of the fly. The nonzero transition probabilities are\n\n\[ \n{p}_{11} = 1,\;{p}_{mm} = 1, \n\]\n\n\[ \n{p}_{ij} = \left\{ {\begin{array}{ll} {0.3} & \text{ if }j = i - 1\text{ or }j = i + 1, \\ {0.4} & \text{ if }j = i, \end... | Yes |
Example 6.3. For the spider and fly example (Example 6.2), we have\n\n\\[ \n\\mathbf{P}\\left( {{X}_{1} = 2,{X}_{2} = 2,{X}_{3} = 3,{X}_{4} = 4 \\mid {X}_{0} = 2}\\right) = {p}_{22}{p}_{22}{p}_{23}{p}_{34} = {\\left( {0.4}\\right) }^{2}{\\left( {0.3}\\right) }^{2}.\n\\]\n\nWe also have\n\n\\[ \n\\mathbf{P}\\left( {{X}_... | Note that in order to calculate a probability of this form, in which there is no conditioning on a fixed initial state, we need to specify a probability law for the initial state \\( {X}_{0} \\) . | Yes |
Consider a two-state Markov chain with transition probabilities\n\n\[ \n{p}_{11} = {0.8},\;{p}_{12} = {0.2}, \n\]\n\n\[ \n{p}_{21} = {0.6},\;{p}_{22} = {0.4}. \n\]\n\n[This is the same as the chain of Example 6.1 (cf. Fig. 6.1).] The balance equations take the form\n\n\[ \n{\pi }_{1} = {\pi }_{1}{p}_{11} + {\pi }_{2}{p... | Note that the above two equations are dependent, since they are both equivalent to\n\n\[ \n{\pi }_{1} = 3{\pi }_{2} \n\]\n\nThis is a generic property, and in fact it can be shown that one of the balance equations depends on the remaining equations (see the theoretical problems). However, we know that the \( {\pi }_{j}... | Yes |
An absent-minded professor has two umbrellas that she uses when commuting from home to office and back. If it rains and an umbrella is available in her location, she takes it. If it is not raining, she always forgets to take an umbrella. Suppose that it rains with probability \( p \) each time she commutes, independent... | From the second equation, we obtain \( {\pi }_{1} = {\pi }_{2} \), which together with the first equation \( {\pi }_{0} = \left( {1 - p}\right) {\pi }_{2} \) and the normalization equation \( {\pi }_{0} + {\pi }_{1} + {\pi }_{2} = 1 \), yields\n\n\[ {\pi }_{0} = \frac{1 - p}{3 - p},\;{\pi }_{1} = \frac{1}{3 - p},\;{\pi... | Yes |
A superstitious professor works in a circular building with \( m \) doors, where \( m \) is odd, and never uses the same door twice in a row. Instead he uses with probability \( p \) (or probability \( 1 - p \) ) the door that is adjacent in the clockwise direction (or the counterclockwise direction, respectively) to t... | We introduce a Markov chain with the following \( m \) states:\n\nState \( i \) : Last door used is door \( i,\;i = 1,\ldots, m \) .\n\nThe transition probability graph of the chain is given in Fig. 6.10, for the case \( m = 5 \) . The transition probability matrix is\n\n\[ \left\lbrack \begin{matrix} 0 & p & 0 & 0 & \... | Yes |
A machine can be either working or broken down on a given day. If it is working, it will break down in the next day with probability \( b \), and will continue working with probability \( 1 - b \). If it breaks down on a given day, it will be repaired and be working in the next day with probability \( r \), and will co... | We introduce a Markov chain with the following two states:\n\nState 1: Machine is working, State 2: Machine is broken down.\n\nThe transition probability graph of the chain is given in Fig. 6.11. The transition probability matrix is\n\n\[ \left\lbrack \begin{matrix} 1 - b & b \\ r & 1 - r \end{matrix}\right\rbrack \]\n... | Yes |
Consider a variation of Example 6.7. If the machine remains broken for a given number of \( \ell \) days, despite the repair efforts, it is replaced by a new working machine. To model this as a Markov chain, we replace the single state 2, corresponding to a broken down machine, with several states that indicate the num... | \[{\pi }_{1} = \left( {1 - b}\right) {\pi }_{1} + r\left( {{\pi }_{\left( 2,1\right) } + \cdots + {\pi }_{\left( 2,\ell - 1\right) }}\right) + {\pi }_{\left( 2,\ell \right) },\]\n\n\[{\pi }_{\left( 2,1\right) } = b{\pi }_{1}\]\n\n\[{\pi }_{\left( 2, i\right) } = \left( {1 - r}\right) {\pi }_{\left( 2, i - 1\right) },\;... | Yes |
Example 6.9. (Random Walk with Reflecting Barriers) A person walks along a straight line and, at each time period, takes a step to the right with probability \( b \), and a step to the left with probability \( 1 - b \). The person starts in one of the positions \( 1,2,\ldots, m \), but if he reaches position 0 (or posi... | The local balance equations are\n\n\[ \n{\pi }_{i}b = {\pi }_{i + 1}\left( {1 - b}\right) ,\;i = 1,\ldots, m - 1.\n\]\n\nThus, \( {\pi }_{i + 1} = \rho {\pi }_{i} \), where\n\n\[ \n\rho = \frac{b}{1 - b}\n\]\n\nand we can express all the \( {\pi }_{j} \) in terms of \( {\pi }_{1} \), as\n\n\[ \n{\pi }_{i} = {\rho }^{i ... | Yes |
Example 6.10. (Birth-Death Markov Chains - Queueing) Packets arrive at a node of a communication network, where they are stored in a buffer and then transmitted. The storage capacity of the buffer is \( m \) : if \( m \) packets are already present, any newly arriving packets are discarded. We discretize time in very s... | The local balance equations are\n\n\[ \n{\pi }_{i}b = {\pi }_{i + 1}d,\;i = 0,1,\ldots, m - 1.\n\]\n\nWe define\n\n\[ \n\rho = \frac{b}{d}\n\]\n\nand obtain \( {\pi }_{i + 1} = \rho {\pi }_{i} \), which leads to \( {\pi }_{i} = {\rho }^{i}{\pi }_{0} \) for all \( i \) . By using the normalization equation \( 1 = {\pi }... | Yes |
Consider the Markov chain shown in Fig. 6.17(a). We would like to calculate the probability that the state eventually enters the recurrent class \( \{ 4,5\} \) starting from one of the transient states. | The absorption probabilities \( {a}_{i} \) of eventually reaching state \( s = 6 \) starting from state \( i \), satisfy the following equations:\n\n\[ \n{a}_{2} = {0.2}{a}_{1} + {0.3}{a}_{2} + {0.4}{a}_{3} + {0.1}{a}_{6}, \n\]\n\n\[ \n{a}_{3} = {0.2}{a}_{2} + {0.8}{a}_{6} \n\]\n\nUsing the facts \( {a}_{1} = 0 \) and ... | Yes |
Consider the spiders-and-fly model of Example 6.2. This corresponds to the Markov chain shown in Fig. 6.19. The states correspond to possible fly positions, and the absorbing states 1 and \( m \) correspond to capture by a spider. Let us calculate the expected number of steps until the fly is captured. | We have\n\n\[ \mu_1 = \mu_m = 0 \]\n\nand\n\n\[ \mu_i = 1 + 0.3 \cdot \mu_{i-1} + 0.4 \cdot \mu_i + 0.3 \cdot \mu_{i+1}, \text{ for } i = 2, \ldots, m-1. \]\n\nWe can solve these equations in a variety of ways, such as for example by successive substitution. As an illustration, let \( m = 4 \), in which case, the equat... | Yes |
Example 6.17. (The M/M/1 Queue) Packets arrive at a node of a communication network according to a Poissson process with rate \( \lambda \) . The packets are stored at an infinite capacity buffer and are then transmitted one at a time. The transmission time of a packet is exponentially distributed with parameter \( \mu... | We will model this system using a continuous-time process with state \( X\left( t\right) \) equal to the number of packets in the system at time \( t \) [if \( X\left( t\right) > 0 \), then \( X\left( t\right) - 1 \) packets are waiting in the queue and one packet is under transmission]. The state increases by one when... | No |
Example 6.19. (The \( M/M/m \) and \( M/M/\infty \) Queues) The \( M/M/m \) queueing system is identical to the \( M/M/1 \) system except that \( m \) packets can be simultaneously transmitted (i.e., the transmission line of the node has \( m \) transmission channels). A packet at the head of the queue is routed to any... | By writing down the local balance equations for the steady-state probabilities \( {\pi }_{n} \), we obtain\n\n\[ \lambda {\pi }_{n - 1} = \left\{ \begin{array}{ll} {n\mu }{\pi }_{n} & \text{ if }n \leq m \\ {m\mu }{\pi }_{n} & \text{ if }n > m \end{array}\right. \]\n\nFrom these equations, we obtain\n\n\[ {\pi }_{n} = ... | Yes |
Let \( X \) be uniformly distributed on the interval \( \left\lbrack {0,4}\right\rbrack \) and note that \( \mathbf{E}\left\lbrack X\right\rbrack = 2 \) . Then, the Markov inequality asserts that\n\n\[ \mathbf{P}\left( {X \geq 2}\right) \leq \frac{2}{2} = 1,\;\mathbf{P}\left( {X \geq 3}\right) \leq \frac{2}{3} = {0.67}... | By comparing with the exact probabilities\n\n\[ \mathbf{P}\left( {X \geq 2}\right) = {0.5},\;\mathbf{P}\left( {X \geq 3}\right) = {0.25},\;\mathbf{P}\left( {X \geq 4}\right) = 0, \] \n\nwe see that the bounds provided by the Markov inequality can be quite loose. | Yes |
As in Example 7.1, let \( X \) be uniformly distributed on \( \left\lbrack {0,4}\right\rbrack \) . Let us use the Chebyshev inequality to bound the probability that \( \left| {X - 2}\right| \geq 1 \) . | We have \( {\sigma }^{2} = {16}/{12} = 4/3 \), and\n\n\[ \mathbf{P}\left( {\left| {X - 2}\right| \geq 1}\right) \leq \frac{4}{3} \]\n\nwhich is not particularly informative. | Yes |
Consider an event \( A \) defined in the context of some probabilistic experiment. Let \( p = \mathbf{P}\left( A\right) \) be the probability of that event. We consider \( n \) independent repetitions of the experiment, and let \( {M}_{n} \) be the fraction of time that event \( A \) occurred; in this context, \( {M}_{... | The weak law applies and shows that when \( n \) is large, the empirical frequency is most likely to be within \( \epsilon \) of \( p \) . Loosely speaking, this allows us to say that empirical frequencies are faithful estimates of \( p \) . Alternatively, this is a step towards interpreting the probability \( p \) as ... | No |
Consider a sequence of independent random variables \( {X}_{n} \) that are uniformly distributed over the interval \( \left\lbrack {0,1}\right\rbrack \), and let\n\n\[ \n{Y}_{n} = \min \left\{ {{X}_{1},\ldots ,{X}_{n}}\right\} \n\]\n\nThe sequence of values of \( {Y}_{n} \) cannot increase as \( n \) increases, and it ... | Indeed, for \( \epsilon > 0 \), we have using the independence of the \( {X}_{n} \) ,\n\n\[ \n\mathbf{P}\left( {\left| {{Y}_{n} - 0}\right| \geq \epsilon }\right) = \mathbf{P}\left( {{X}_{1} \geq \epsilon ,\ldots ,{X}_{n} \geq \epsilon }\right) \n\]\n\n\[ \n= \mathbf{P}\left( {{X}_{1} \geq \epsilon }\right) \cdots \mat... | Yes |
Let \( Y \) be an exponentially distributed random variable with parameter \( \lambda = 1 \) . For any positive integer \( n \), let \( {Y}_{n} = Y/n \) . (Note that these random variables are dependent.) We wish to investigate whether the sequence \( {Y}_{n} \) converges to zero. | For \( \epsilon > 0 \), we have\n\n\[ \mathbf{P}\left( {\left| {{Y}_{n} - 0}\right| \geq \epsilon }\right) = \mathbf{P}\left( {{Y}_{n} \geq \epsilon }\right) = \mathbf{P}\left( {Y \geq {n\epsilon }}\right) = {e}^{-{n\epsilon }}.\n\]\n\nIn particular,\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathbf{P}\left(... | Yes |
Consider a sequence of discrete random variables \( {Y}_{n} \) with the following distribution:\n\n\[ \mathbf{P}\left( {{Y}_{n} = y}\right) = \left\{ \begin{array}{ll} 1 - \frac{1}{n}, & \text{ for }y = 0, \\ \frac{1}{n}, & \text{ for }y = {n}^{2}, \\ 0, & \text{ elsewhere. } \end{array}\right. \]\n\nFor every \( \epsi... | and \( {Y}_{n} \) converges to zero in probability. On the other hand, \( \mathbf{E}\left\lbrack {Y}_{n}\right\rbrack = {n}^{2}/n = n \) , which goes to infinity as \( n \) increases. | Yes |
We load on a plane 100 packages whose weights are independent random variables that are uniformly distributed between 5 and 50 pounds. What is the probability that the total weight will exceed 3000 pounds? | It is not easy to calculate the CDF of the total weight and the desired probability, but an approximate answer can be quickly obtained using the central limit theorem.\n\nWe want to calculate \( \mathbf{P}\left( {{S}_{100} > {3000}}\right) \), where \( {S}_{100} \) is the sum of the 100 packages. The mean and the varia... | Yes |
A machine processes parts, one at a time. The processing times of different parts are independent random variables, uniformly distributed on \( \left\lbrack {1,5}\right\rbrack \) . We wish to approximate the probability that the number of parts processed within 320 time units is at least 100 . | Let us call \( {N}_{320} \) this number. We want to calculate \( \mathbf{P}\left( {{N}_{320} \geq {100}}\right) \) . There is no obvious way of expressing the random variable \( {N}_{320} \) as the sum of independent random variables, but we can proceed differently. Let \( {X}_{i} \) be the processing time of the \( i ... | Yes |
Let us revisit the polling problem in Example 7.4. We poll \( n \) voters and record the fraction \( {M}_{n} \) of those polled who are in favor of a particular candidate. If \( p \) is the fraction of the entire voter population that supports this candidate, then \[ {M}_{n} = \frac{{X}_{1} + \cdots + {X}_{n}}{n} \] wh... | Because of the symmetry of the normal PDF around the mean, we have \[ \mathbf{P}\left( {\left| {{M}_{n} - p}\right| \geq \epsilon }\right) \approx 2\mathbf{P}\left( {{M}_{n} - p \geq \epsilon }\right) \] The variance \( p\left( {1 - p}\right) /n \) of \( {M}_{n} - p \) depends on \( p \) and is therefore unknown. We no... | Yes |
Let \( {S}_{n} \) be a binomial random variable with parameters \( n = {36} \) and \( p = {0.5} \) . An exact calculation yields\n\n\[ \mathbf{P}\left( {{S}_{n} \leq {21}}\right) = \mathop{\sum }\limits_{{k = 0}}^{{21}}\left( \begin{matrix} {36} \\ k \end{matrix}\right) {\left( {0.5}\right) }^{36} = {0.8785}. \] | The central limit approximation, without the above discussed refinement, yields\n\n\[ \mathbf{P}\left( {{S}_{n} \leq {21}}\right) \approx \Phi \left( \frac{{21} - {np}}{\sqrt{{np}\left( {1 - p}\right) }}\right) = \Phi \left( \frac{{21} - {18}}{3}\right) = \Phi \left( 1\right) = {0.8413}. \]\n\nUsing the proposed refine... | Yes |
Consider a discrete-time arrival process. The set of times is partitioned into consecutive intervals of the form \( {I}_{k} = \left\{ {{2}^{k},{2}^{k} + 1,\ldots ,{2}^{k + 1} - 1}\right\} \) . Note that the length of \( {I}_{k} \) is \( {2}^{k} \), which increases with \( k \) . During each interval \( {I}_{k} \) , the... | We have \( \mathbf{P}\left( {{Y}_{n} \neq 0}\right) = 1/{2}^{k} \), if \( n \in {I}_{k} \) . Note that as \( n \) increases, it belongs to intervals \( {I}_{k} \) with increasingly large indices \( k \) . Consequently, \[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathbf{P}\left( {{Y}_{n} \neq 0}\right) = \mathop{... | Yes |
Let \( {X}_{1},{X}_{2},\ldots \) be a sequence of independent random variables that are uniformly distributed on \( \left\lbrack {0,1}\right\rbrack \), and let \( {Y}_{n} = \min \left\{ {{X}_{1},\ldots ,{X}_{n}}\right\} \) . We wish to show that \( {Y}_{n} \) converges to 0, with probability 1 . | In any execution of the experiment, the sequence \( {Y}_{n} \) is nonincreasing, i.e., \( {Y}_{n + 1} \leq {Y}_{n} \) for all \( n \) . Since this sequence is bounded below by zero, it must have a limit, which we denote by \( Y \) . Let us fix some \( \epsilon > 0 \) . If \( Y \geq \epsilon \), then \( {X}_{i} \geq \ep... | Yes |
Lemma 2.3 Recall the notation of Definition 2.1. If \( Y \) and \( {Y}^{\prime } \) are conditional expectations of \( X \) given \( \mathcal{B} \), then \( Y = {Y}^{\prime } \) almost surely. | Proof. Indeed, \( Y \) and \( {Y}^{\prime } \) are both measurable with respect to \( \mathcal{B} \) and they have the same integral over every event belonging to \( \mathcal{B} \) . | Yes |
Theorem 2.4 Let \( X : \\left( {\\Omega ,\\mathcal{A},\\mathbb{P}}\\right) \\rightarrow \\left( {\\mathbb{R},\\mathcal{B}_{\\mathbb{R}}}\\right) \) be a square-integrable random variable. Let \( \\mathcal{B} \) be a sub- \( \\sigma \) -field of \( \\mathcal{A} \). There exists, among all square-integrable random variab... | Proof. The space of square-integrable random variables which are measurable with respect to \( \\mathcal{B} \) is the subspace \( {L}^{2}\\left( {\\Omega ,\\mathcal{B},\\mathbb{P}}\\right) \) of \( {L}^{2}\\left( {\\Omega ,\\mathcal{A},\\mathbb{P}}\\right) \). Since the \( {L}^{2} \) distance on \( {L}^{2}\\left( {\\Om... | Yes |
For all reals \( a, b \), the random variable \( a{X}_{1} + b{X}_{2} \) admits a conditional expectation given \( \mathcal{B} \) and \( \mathbb{E}\left\lbrack {a{X}_{1} + b{X}_{2} \mid \mathcal{B}}\right\rbrack = a\mathbb{E}\left\lbrack {{X}_{1} \mid \mathcal{B}}\right\rbrack + b\mathbb{E}\left\lbrack {{X}_{2} \mid \ma... | The random variable \( a\mathbb{E}\left\lbrack {{X}_{1} \mid \mathcal{B}}\right\rbrack + b\mathbb{E}\left\lbrack {{X}_{2} \mid \mathcal{B}}\right\rbrack \) is integrable and \( \mathcal{B} \) -measurable. The linearity of the integral implies immediately that it is a conditional expectation of \( a{X}_{1} + b{X}_{2} \)... | Yes |
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