Q.7.16 聽Let Z be a standard normal ran... [FREE SOLUTION]

91影视

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.7.16 (page 353)

Let Z be a standard normal random variable,and, for a 铿亁ed x, set
$X = {\begin{cases} Z & if Z > x \\ 0 & otherwise \end{cases}$
$Show that E [X] = \frac{1}{\sqrt{2 蟺}} e^{鈭� x^{2} / 2} .$

Short Answer

Expert verified

$E (X) = {鈭�}_{z > x} z 鈰� d P_{Z} = {鈭�}_{z > x} z f_{Z} (z) d z = {鈭�}_{x}^{鈭�} z \frac{1}{\sqrt{2 蟺}} e^{鈭� \frac{z^{2}}{2}} d z$

Let's solve this integral using the substitution $u = Z^{2} / 2$ which implies $d u = z d z .$

\frac{1}{\sqrt{2 蟺}} {鈭�}_{\frac{x^{2}}{2}}^{鈭�} e^{鈭� u} d u = \frac{1}{\sqrt{2 蟺}} e^{鈭� \frac{x^{2}}{2}}

Step by step solution

Given Information

Given in the question that

Let Z be a standard normal random variable

$X = {\begin{cases} Z & if Z > x \\ 0 & otherwise \end{cases}$

$E [X] = \frac{1}{\sqrt{2 蟺}} e^{鈭� x^{2} / 2} .$

Explanation

Observe that $X$ can be written as $X = Z 路 I_{(Z > x)}$ . So, the mean of $X$ is

$E (X) = {鈭�}_{z > x} z 鈰� d P_{Z} = {鈭�}_{z > x} z f_{Z} (z) d z = {鈭�}_{x}^{鈭�} z \frac{1}{\sqrt{2 蟺}} e^{鈭� \frac{z^{2}}{2}} d z$

Let's solve this integral using the substitution $u = z^{2} / 2$ which implies $d u = z d z$ . So, the integral above is equal to

$\frac{1}{\sqrt{2 蟺}} {鈭�}_{\frac{u^{2}}{2}}^{鈭�} e^{鈭� u} d u = \frac{1}{\sqrt{2 蟺}} e^{鈭� \frac{x^{2}}{2}}$

So, we have proved that

$E [X] = \frac{1}{\sqrt{2 蟺}} e^{鈭� x^{2} / 2} .$

Final Answer

$E (X) = {鈭�}_{z > x} z 鈰� d P_{Z} = {鈭�}_{z > x} z f_{Z} (z) d z = {鈭�}_{x}^{鈭�} z \frac{1}{\sqrt{2 蟺}} e^{鈭� \frac{z^{2}}{2}} d z$

Let's solve this integral using the substitution $u = Z^{2} / 2$ which implies $d u = z d z$

$\frac{1}{\sqrt{2 蟺}} {鈭�}_{\frac{x^{2}}{2}}^{鈭�} e^{鈭� u} d u = \frac{1}{\sqrt{2 蟺}} e^{鈭� \frac{x^{2}}{2}}$

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91影视

Let Z be a standard normal random variable,and, for a 铿亁ed x, set
$X = {\begin{cases} Z & if Z > x \\ 0 & otherwise \end{cases}$
$Show that E [X] = \frac{1}{\sqrt{2 蟺}} e^{鈭� x^{2} / 2} .$

Short Answer

Step by step solution

Given Information

Explanation

Final Answer

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91影视

Let Z be a standard normal random variable,and, for a 铿亁ed x, setX={ZifZ>x0otherwiseShow thatE[X]=12蟺e鈭�x2/2.

Short Answer

Step by step solution

Given Information

Explanation

Final Answer

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Probability and Statistics

Theoretical and Mathematical Physics

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Discrete Mathematics

Pure Maths

Study anywhere. Anytime. Across all devices.

Let Z be a standard normal random variable,and, for a 铿亁ed x, set
$X = {\begin{cases} Z & if Z > x \\ 0 & otherwise \end{cases}$
$Show that E [X] = \frac{1}{\sqrt{2 蟺}} e^{鈭� x^{2} / 2} .$