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A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 7: Q.50 (page 356)

The joint density of $X$ and $Y$ is given by $f (x, y) = \frac{e^{- x / y} e^{- y}}{y}$ , $0 < x < 鈭�, 0 < y < 鈭�$ , Compute $E [X^{2} 鈭� Y = y]$ .

Short Answer

Expert verified

$E [X^{2} 鈭� Y = y] = 2 y^{2}$

Step by step solution

01

Given information

Given in the question that, The joint density of $X$ and $Y$ is given by

$f (x, y) = \frac{e^{- x / y} e^{- y}}{y}, 0 < x < 鈭�, 0 < y < 鈭�$ .

02

Explanation

The joint density of $X$ and $Y$ is given by $f (x, y) = \frac{e^{- \frac{x}{y}} e^{- y}}{y}$ where $0 < x < 鈭�, 0 < y < 鈭�$ . We have to compute $E [X^{2} 鈭� Y = y]$ .

From the definition we have:

$E [X 鈭� Y = y] = {鈭�}_{- 鈭�}^{鈭�} x f_{x 鈭� y} (x 鈭� y) d x$

Therefore it simply follows:

$E [X^{2} 鈭� Y = y] = {鈭�}_{- 鈭�}^{鈭�} x^{2} 路 \frac{f (x, y)}{f_{Y} (y)} d x$

$= {鈭�}_{0}^{鈭�} x^{2} 路 \frac{\frac{e^{- \frac{x}{y}} e^{- y}}{y}}{{鈭�}_{- 鈭�}^{鈭�} f (x, y) d x} d x$

$= {鈭�}_{0}^{鈭�} x^{2} 路 \frac{e^{- \frac{x}{y}} e^{- y}}{y e^{- y}} d x$

$= \frac{1}{y} {鈭�}_{0}^{鈭�} x^{2} 路 e^{- \frac{x}{y}} d x$

$= \frac{1}{y} 路 2 y^{3} = 2 y^{2}$

03

Final answer

$E [X^{2} 鈭� Y = y] = 2 y^{2}$

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Most popular questions from this chapter

A population is made up of $r$ disjoint subgroups. Let $p_{i}$ denote the proportion of the population that is in subgroup $i, i = 1, 鈥�, r$ . If the average weight of the members of subgroup $i$ is $w_{i}, i = 1, 鈥�, r$ , what is the average weight of the members of the population?

A deck of n cards numbered 1 through n is thoroughly shuf铿俥d so that all possible n! orderings can be assumed to be equally likely. Suppose you are to make n guesses sequentially, where the ith one is a guess of the card in position i. Let N denote the number of correct guesses.

(a) If you are not given any information about your earlier guesses, show that for any strategy, E[N]=1.

(b) Suppose that after each guess you are shown the card that was in the position in question. What do you think is the best strategy? Show that under this strategy

$\begin{aligned} E [N] & = \frac{1}{n} + \frac{1}{n 鈭� 1} + 鈰� + 1 \\ 鈮� {鈭�}_{1}^{n} \frac{1}{x} d x = \log n \end{aligned}$

(c) Supposethatyouaretoldaftereachguesswhetheryou are right or wrong. In this case, it can be shown that the strategy that maximizes E[N] is one that keeps on guessing the same card until you are told you are correct and then changes to a new card. For this strategy, show that

$\begin{aligned} E [N] & = 1 + \frac{1}{2!} + \frac{1}{3!} + 鈰� + \frac{1}{n!} \\ 鈮� e 鈭� 1 \end{aligned}$

Hint: For all parts, express N as the sum of indicator (that is, Bernoulli) random variables.

The joint density function of $X$ and $Y$ is given by

$f (x, y) = \frac{1}{y} e^{- (y + x / y)}, x > 0, y > 0$

Find $E [X], E [Y]$ and show that $C o v (X, Y) = 1$

Suppose that the expected number of accidents per week at an industrial plant is $5$ . Suppose also that the numbers of workers injured in each accident are independent random variables with a common mean of $2.5$ . If the number of workers injured in each accident is independent of the number of accidents that occur, compute the expected number of workers injured in a week .

Suppose that each of the elements of $S = {1, 2, 鈥�, n}$ is to be colored either red or blue. Show that if $A_{1}, 鈥�, A_{r}$ are subsets of $S$ , there is a way of doing the coloring so that at most ${鈭�}_{i = 1}^{r} (1 / 2)^{|A_{i}| - 1}$ of these subsets have all their elements the same color (where $| A |$ denotes the number of elements in the set $A$ ).

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