Q.7.4 7.4. If X and Y have joint densi... [FREE SOLUTION]

魅影直播

A First Course in Probability

Sheldon M. Ross

$Math Studyset 魅影直播 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 7: Q.7.4 (page 352)

7.4. If X and Y have joint density function $f_{X, Y} (x, y) = {\begin{cases} 1 / y, & if 0 < y < 1, 0 < x < y \\ 0, & otherwise \end{cases}$ find
(a) E[X Y]
(b) E[X]
(c) E[Y]

Short Answer

Expert verified

a) $E [X Y] = \frac{1}{6}$

b) $E [X] = \frac{1}{4}$

c) $E [Y] = \frac{1}{2}$

Step by step solution

Part(a) - Step 1: To find

Expectation of $X Y$

Part (a) - Step 2: Explanation

Given : $f_{X, Y} (x, y) = {\begin{cases} 1 / y, & if 0 < y < 1, 0 < x < y \\ 0, & otherwise \end{cases}$

Formula to be used:

$E [X Y] = {鈭�}_{y} {鈭�}_{x} x y 脳 f (x, y) d x d y$

Calculation:
$\begin{aligned} E [X Y] = {鈭�}_{0}^{1} 鈥� {鈭�}_{0}^{y} 鈥� x y 脳 \frac{1}{y} d x d y \\ = {鈭�}_{0}^{1} 鈥� {鈭�}_{0}^{y} 鈥� x d x d y \\ = {{鈭�}_{0}^{1} 鈥� [\frac{x^{2}}{2}]|}_{0}^{y} d y \\ = {鈭�}_{0}^{1} 鈥� (\frac{y^{2}}{2} 鈭� 0) d y \end{aligned}$

Now, integrating w.r.t $Y$

$\begin{aligned} = {[\frac{y^{3}}{2 脳 3}]|}_{0}^{1} \\ = \frac{1^{3}}{6} 鈭� 0 \\ = \frac{1}{6} \end{aligned}$

Hence $E [X Y] = \frac{1}{6}$

Part (b) - Step 3: To find

Expectation ofX

Part (b) - Step 4: Explanation

To find : $E [X]$

Formula to be used:

$E [X Y] = {鈭�}_{y} {鈭�}_{x} x y 脳 f (x, y) d x d y$

Calculation:

$\begin{aligned} E [X] = {鈭�}_{0}^{1} 鈥� {鈭�}_{0}^{y} 鈥� x 脳 \frac{1}{y} d x d y \\ = {鈭�}_{0}^{1} 鈥� ({[\frac{1}{y} 脳 \frac{x^{2}}{2}]|}_{0}^{y}) d x d y \\ = {鈭�}_{0}^{1} 鈥� (\frac{y}{2} 鈭� 0) d y \\ = {[\frac{y^{2}}{2 脳 2}]|}_{0}^{1} = \frac{1^{2}}{4} 鈭� 0 \end{aligned}$

Therefore $E [X] = \frac{1}{4}$

Part (c) : To find

Expectation of $X$

Part(c) : Step 6: Explanation

To find: $E [Y]$

Formula to be used: $E [X Y] = {鈭�}_{y} {鈭�}_{x} x y 脳 f (x, y) d x d y$

Calculation:

$\begin{aligned} E [X] = {鈭�}_{0}^{1} 鈥� {鈭�}_{0}^{y} 鈥� y 脳 \frac{1}{y} d x d y \\ = {鈭�}_{0}^{1} 鈥� ({[x]|}_{0}^{y}) d x d y \\ = {鈭�}_{0}^{1} 鈥� (y 鈭� 0) d y \\ = {[\frac{y^{2}}{2}]|}_{0}^{1} = \frac{1^{2}}{2} 鈭� 0 \end{aligned}$

Therefore $E [Y] = \frac{1}{2}$

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魅影直播

7.4. If X and Y have joint density function $f_{X, Y} (x, y) = {\begin{cases} 1 / y, & if 0 < y < 1, 0 < x < y \\ 0, & otherwise \end{cases}$ find
(a) E[X Y]
(b) E[X]
(c) E[Y]

Short Answer

Step by step solution

Part(a) - Step 1: To find

Part (a) - Step 2: Explanation

Part (b) - Step 3: To find

Part (b) - Step 4: Explanation

Part (c) : To find

Part(c) : Step 6: Explanation

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7.4. If X and Y have joint density function fX,Y(x,y)={1/y,if0<y<1,0<x<y0,otherwisefind(a) E[X Y](b) E[X](c) E[Y]

Short Answer

Step by step solution

Part(a) - Step 1: To find

Part (a) - Step 2: Explanation

Part (b) - Step 3: To find

Part (b) - Step 4: Explanation

Part (c) : To find

Part(c) : Step 6: Explanation

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7.4. If X and Y have joint density function $f_{X, Y} (x, y) = {\begin{cases} 1 / y, & if 0 < y < 1, 0 < x < y \\ 0, & otherwise \end{cases}$ find
(a) E[X Y]
(b) E[X]
(c) E[Y]