Content check — Probability (Grinstead and Snell), Chapter 7: Sums of Independent Random Variables

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Chapter 7: Sums of Independent Random Variables

13 problems · 0 figures

Exercise 7.1.1 §7.1 source ↗

A die is rolled three times. Find the probability that the sum of the outcomes is

(a) greater than 9.

(b) an odd number.

Answer

(a) $.625$

(b) $.5$

Exercise 7.1.3 §7.1 source ↗

Let $X_{1}$ and $X_{2}$ be independent random variables with common distribution $p_{X} = (\begin{matrix} 0 & 1 & 2 \\ 1 / 8 & 3 / 8 & 1 / 2 \end{matrix}) .$ Find the distribution of the sum $X_{1} + X_{2}$ ⁠.

Answer

$(\begin{matrix} 0 & 1 & 2 & 3 & 4 \\ \frac{1}{64} & \frac{3}{32} & \frac{17}{64} & \frac{3}{8} & \frac{1}{4} \end{matrix})$

Exercise 7.1.5 §7.1 source ↗

Consider the following two experiments: the first has outcome $X$ taking on the values 0, 1, and 2 with equal probabilities; the second results in an (independent) outcome $Y$ taking on the value 3 with probability 1/4 and 4 with probability 3/4. Find the distribution of

(a) $Y + X$ ⁠.

(b) $Y - X$ ⁠.

Answer

(a) $(\begin{matrix} 3 & 4 & 5 & 6 \\ \frac{1}{12} & \frac{4}{12} & \frac{4}{12} & \frac{3}{12} \end{matrix})$ (b) $(\begin{matrix} 1 & 2 & 3 & 4 \\ \frac{1}{12} & \frac{4}{12} & \frac{4}{12} & \frac{3}{12} \end{matrix})$

Exercise 7.1.7 §7.1 source ↗

(a) A die is rolled three times with outcomes $X_{1}$ ⁠, $X_{2}$ ⁠, and $X_{3}$ ⁠. Let $Y_{3}$ be the maximum of the values obtained. Show that $P (Y_{3} \leq j) = P {(X_{1} \leq j)}^{3} .$ Use this to find the distribution of $Y_{3}$ ⁠. Does $Y_{3}$ have a bell-shaped distribution?

(b) Now let $Y_{n}$ be the maximum value when $n$ dice are rolled. Find the distribution of $Y_{n}$ ⁠. Is this distribution bell-shaped for large values of $n$ ⁠?

Answer

(a) $P (Y_{3} \leq j) =$ $P (X_{1} \leq j, X_{2} \leq j, X_{3} \leq j) =$ $P {(X_{1} \leq j)}^{3}$ ⁠. Thus $p_{Y_{3}} = (\begin{matrix} 1 & 2 & 3 & 4 & 5 & 6 \\ \frac{1}{216} & \frac{7}{216} & \frac{19}{216} & \frac{37}{216} & \frac{61}{216} & \frac{91}{216} \end{matrix}) .$ This distribution is not bell-shaped.

(b) In general, $P (Y_{n} \leq j) = P {(X_{1} \leq j)}^{3} = {(\frac{j}{n})}^{n} .$ Therefore, $P (Y_{n} = j) = {(\frac{j}{n})}^{n} - {(\frac{j - 1}{n})}^{n} .$ This distribution is not bell-shaped for large $n$ ⁠.

Exercise 7.1.9 §7.1 source ↗

Prove that you cannot load two dice in such a way that the probabilities for any sum from 2 to 12 are the same. (Be sure to consider the case where one or more sides turn up with probability zero.)

Answer

Let $p_{1}, \dots, p_{6}$ be the probabilities for one die and $q_{1}, \dots, q_{6}$ be the probabilities for the other die. Assume first that all probabilities are positive. Then $p_{1} q_{1} > p_{1} q_{6}$ ⁠, since there is only one way to get a 2 and several ways to get a 7. Thus $q_{1} > q_{6}$ ⁠. In the same way $q_{6} q_{6} > q_{1} p_{6}$ and so $q_{6} > q_{1}$ ⁠. This is a contradiction. If any of the sides has probability 0, then we can renumber them so that it is side 1. But then the probability of a 2 is 0 and so all sums would have to have probability 0, which is impossible.

Here’s a fancy way to prove it. Define the polynomials $p (x) = \sum_{k = 0}^{5} p_{(k + 1)} x^{k}$ and $q (x) = \sum_{k = 0}^{5} q_{(k + 1)} x^{k} .$ Then we must have $p (x) q (x) = \sum_{k = 0}^{10} \frac{x^{k}}{11} = \frac{(1 - x^{11})}{(1 - x)} .$ The left side is the product of two fifth degree polynomials. A fifth degree polynomial must have a real root which will not be 0 if $p_{1} > 0$ ⁠. Consider the right side as a polynomial. For $x$ to be a non-zero root of this polynomial it would have to be a real eleventh root of unity other than 1, and there are no such roots. Hence again we have a contradiction.

Exercise 7.2.3 §7.2 source ↗

Suppose again that $Z = X + Y$ ⁠. Find $f_{Z}$ if

(a) $f_{X} (x) = f_{Y} (x) = {\begin{matrix} x / 2, & if 0 < x < 2, \\ 0, & otherwise . \end{matrix}$

(b) $f_{X} (x) = f_{Y} (x) = {\begin{matrix} (1 / 2) (x - 3), & if 3 < x < 5, \\ 0, & otherwise . \end{matrix}$

(c) $f_{X} (x) = {\begin{matrix} 1 / 2, & if 0 < x < 2, \\ 0, & otherwise, \end{matrix}$ $f_{Y} (x) = {\begin{matrix} x / 2, & if 0 < x < 2, \\ 0, & otherwise . \end{matrix}$

(d) What can you say about the set $E = {z : f_{Z} (z) > 0}$ in each case?

Answer

(a) $f_{Z} (x) = {\begin{matrix} x^{3} / 24, & if 0 \leq x \leq 2; \\ x - x^{3} / 24 - 4 / 3, & if 2 \leq x \leq 4. \end{matrix}$ (b) $f_{Z} (x) = {\begin{matrix} (x^{3} - 18 x^{2} + 108 x - 216) / 24, & if 6 \leq x \leq 8; \\ (- x^{3} + 18 x^{2} - 84 x + 40) / 24, & if 8 \leq x \leq 10. \end{matrix}$ (c) $f_{Z} (x) = {\begin{matrix} x^{2} / 8, & if 0 \leq x \leq 2; \\ 1 / 2 - {(x - 2)}^{2} / 8, & if 2 \leq x \leq 4. \end{matrix}$

Exercise 7.2.5 §7.2 source ↗

Suppose that $X$ and $Y$ are independent and $Z = X + Y$ ⁠. Find $f_{Z}$ if

(a) $f_{X} (x) = {\begin{matrix} λ e^{- λ x}, & if x > 0, \\ 0, & otherwise. \end{matrix}$ $f_{Y} (x) = {\begin{matrix} μ e^{- μ x}, & if x > 0, \\ 0, & otherwise. \end{matrix}$

(b) $f_{X} (x) = {\begin{matrix} λ e^{- λ x}, & if x > 0, \\ 0, & otherwise. \end{matrix}$
$f_{Y} (x) = {\begin{matrix} 1, & if 0 < x < 1, \\ 0, & otherwise. \end{matrix}$

Answer

(a) $f_{Z} (x) = {\begin{matrix} \frac{λ μ}{μ + λ} e^{λ x}, & x < 0; \\ \frac{λ μ}{μ + λ} e^{- μ x}, & x \geq 0. \end{matrix}$ (b) $f_{Z} (x) = {\begin{matrix} 1 - e^{- λ x}, & 0 < x < 1; \\ (e^{λ} - 1) e^{- λ x}, & x \geq 1. \end{matrix}$

Exercise *7.2.7 §7.2 source ↗

Suppose that $R^{2} = X^{2} + Y^{2}$ ⁠. Find $f_{R^{2}}$ and $f_{R}$ if

$\begin{matrix} f_{X} (x) & = \frac{1}{\sqrt{2 π} σ_{1}} e^{- {(x - μ_{1})}^{2} / 2 σ_{1}^{2}} \\ f_{Y} (x) & = \frac{1}{\sqrt{2 π} σ_{2}} e^{- {(x - μ_{2})}^{2} / 2 σ_{2}^{2}} . \end{matrix}$

Answer

We first find the density for $X^{2}$ when $X$ has a general normal density $f_{X} (x) = \frac{1}{σ \sqrt{2 π}} e^{- {(x - μ)}^{2} / 2 σ^{2}} d x .$ Then (see Theorem 1 of Chapter 5, Section 5.2 and the discussion following) we have $f_{X}^{2} (x) = \frac{1}{σ \sqrt{2 π}} \frac{1}{2 \sqrt{x}} \exp (- x / 2 σ^{2} - μ^{2} / 2 σ^{2}) (\exp (\sqrt{x} μ / σ^{2}) + \exp (- \sqrt{x} μ / σ^{2})) .$ Replacing the last two exponentials by their series representation, we have $f_{X}^{2} (x) = e^{- μ / 2 σ^{2}} \sum_{r = 0}^{\infty} {(\frac{μ}{2 σ^{2}})}^{r} \frac{1}{r!} G (1 / 2 σ^{2}, r + 1 / 2, x),$ where $G (a, p, x) = \frac{a^{p}}{Γ (p)} e^{- a x} x^{p - 1}$ is the gamma density. We now consider the original problem with $X_{1}$ and $X_{2}$ two random variables with normal density with parameters $μ_{1}, σ_{1}$ and $μ_{2}, σ_{2}$ ⁠. This is too much generality for us, and we shall assume that the variances are equal, and then for simplicity we shall assume they are 1. Let $c = \sqrt{μ_{1}^{2} + μ_{2}^{2}} .$ We introduce the new random variables $Z_{1} = \frac{1}{c} (μ_{1} X_{1} + μ_{2} X_{2}),$ $Z_{2} = \frac{1}{c} (μ_{2} X_{1} - μ_{1} X_{2}) .$ Then $Z_{1}$ is normal with mean $c$ and variance 1 and $Z_{2}$ is normal with mean 0 and variance 1. Thus, $f_{Z_{1}^{2}} = e^{- c^{2} / 2} \sum_{r = 0}^{\infty} {(\frac{c^{2}}{2})}^{r} \frac{1}{r!} G (1 / 2, r + 1 / 2, x),$ and $f_{Z_{2}^{2}} = G (1 / 2,1 / 2, x) .$ Convoluting these two densities and using the fact that the convolution of a gamma density $G (a, p, x)$ and $G (a, q, x)$ is a gamma density $G (a, p + q, x)$ we finally obtain $f_{Z_{1}^{2} + Z_{2}^{2}} = f_{X_{1}^{2} + X_{2}^{2}} = e^{- c^{2} / 2} \sum_{r = 0}^{\infty} {(\frac{c^{2}}{2})}^{r} \frac{1}{r!} G (1 / 2, r + 1, x) .$ (This derivation is adapted from that of C.R. Rao in his book Advanced Statistical Methods in Biometric Research, Wiley, l952.)

Exercise 7.2.9 §7.2 source ↗

Assume that the service time for a customer at a bank is exponentially distributed with mean service time 2 minutes. Let $X$ be the total service time for 10 customers. Estimate the probability that $X > 22$ minutes.

Answer

$P (X_{10} > 22) = .341$ by numerical integration. This also could be estimated by simulation.

Exercise 7.2.11 §7.2 source ↗

A company buys 100 lightbulbs, each of which has an exponential lifetime of 1000 hours. What is the expected time for the first of these bulbs to burn out? (See Exercise 10.)

Exercise 7.2.10 in the book ↗

Let $X_{1}$ ⁠, $X_{2}$ ⁠, …, $X_{n}$ be $n$ independent random variables each of which has an exponential density with mean $μ$ ⁠. Let $M$ be the minimum value of the $X_{j}$ ⁠. Show that the density for $M$ is exponential with mean $μ / n$ ⁠. Hint: Use cumulative distribution functions.

Answer

10 hours

Exercise 7.2.13 §7.2 source ↗

Particles are subject to collisions that cause them to split into two parts with each part a fraction of the parent. Suppose that this fraction is uniformly distributed between 0 and 1. Following a single particle through several splittings we obtain a fraction of the original particle $Z_{n} = X_{1} \cdot X_{2} \cdot \dots \cdot X_{n}$ where each $X_{j}$ is uniformly distributed between 0 and 1. Show that the density for the random variable $Z_{n}$ is $f_{n} (z) = \frac{1}{(n - 1)!} {(- \log z)}^{n - 1} .$ Hint: Show that $Y_{k} = - \log X_{k}$ is exponentially distributed. Use this to find the density function for $S_{n} = Y_{1} + Y_{2} + \dots + Y_{n}$ ⁠, and from this the cumulative distribution and density of $Z_{n} = e^{- S_{n}}$ ⁠.

Answer

$Y_{1} = - \log (X_{1})$ has an exponential density $f_{Y_{1}} (x) = e^{- x}$ ⁠. Thus $S_{n}$ has the gamma density $f_{S_{n}} (x) = \frac{x^{n - 1} e^{- x}}{(n - 1)!} .$ Therefore $f_{Z_{n}} (x) = \frac{1}{(n - 1)!} {(\log \frac{1}{x})}^{n - 1} .$

Exercise 7.2.19 §7.2 source ↗

The support of a function $f (x)$ is defined to be the set ${x : f (x) > 0} .$ Suppose that $X$ and $Y$ are two continuous random variables with density functions $f_{X} (x)$ and $f_{Y} (y)$ ⁠, respectively, and suppose that the supports of these density functions are the intervals $[a, b]$ and $[c, d]$ ⁠, respectively. Find the support of the density function of the random variable $X + Y$ ⁠.

Answer

The support of $X + Y$ is $[a + c, b + d]$ ⁠.

Exercise 7.2.21 §7.2 source ↗

Let $X_{1}$ ⁠, $X_{2}$ ⁠, …, $X_{n}$ be a sequence of independent random variables, all having a common density function $f_{X}$ ⁠. Let $A = S_{n} / n$ be their average. Find $f_{A}$ if

(a) $f_{X} (x) = (1 / \sqrt{2 π}) e^{- x^{2} / 2}$ (normal density).

(b) $f_{X} (x) = e^{- x}$ (exponential density).
Hint: Write $f_{A} (x)$ in terms of $f_{S_{n}} (x)$ ⁠.

Answer

(a) $f_{A} (x) = \frac{1}{\sqrt{2 π n}} e^{- x^{2} / (2 n)} .$ (b) $f_{A} (x) = n^{n} x^{n} e^{- n x} / (n - 1)! .$

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Exercises and answers are adapted from Grinstead and Snell's Introduction to Probability by The CHANCE Project, which is free to read and download, and are used under the GNU FDL licence. They are taken from the GNU version’s LaTeX source (4 July 2006), with answers from the authors’ published answers to the odd-numbered exercises, and set for the web, so they may differ slightly from the printed page. This adaptation is released under the same licence and is neither affiliated with nor endorsed by Grinstead and Snell or American Mathematical Society. Copyright (C) 2006 Peter G. Doyle. This work is a version of Grinstead and Snell's ‘Introduction to Probability, 2nd edition’, published by the American Mathematical Society, Copyright (C) 2003 Charles M. Grinstead and J. Laurie Snell.