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

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Decision Maths

Applied Mathematics

Geometry

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
$X = {\begin{cases} Z & if Z > x \\ 0 & otherwise \end{cases}$
$Show that E [X] = \frac{1}{\sqrt{2 蟺}} e^{鈭� x^{2} / 2} .$