Chapter 7: Q.7.16 (page 353)

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

Short Answer

Expert verified

$E (X) = \int_{z > x} z \cdot d P_{Z} = \int_{z > x} z f_{Z} (z) d z = \int_{x}^{\infty} z \frac{1}{\sqrt{2 π}} e^{- \frac{z^{2}}{2}} d z$

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

\frac{1}{\sqrt{2 π}} \int_{\frac{x^{2}}{2}}^{\infty} e^{- u} d u = \frac{1}{\sqrt{2 π}} e^{- \frac{x^{2}}{2}}

Step by step solution

Given Information

Given in the question that

Let Z be a standard normal random variable

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

$E [X] = \frac{1}{\sqrt{2 π}} e^{- x^{2} / 2} .$

Explanation

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

$E (X) = \int_{z > x} z \cdot d P_{Z} = \int_{z > x} z f_{Z} (z) d z = \int_{x}^{\infty} z \frac{1}{\sqrt{2 π}} e^{- \frac{z^{2}}{2}} d z$

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

$\frac{1}{\sqrt{2 π}} \int_{\frac{u^{2}}{2}}^{\infty} e^{- u} d u = \frac{1}{\sqrt{2 π}} e^{- \frac{x^{2}}{2}}$

So, we have proved that

$E [X] = \frac{1}{\sqrt{2 π}} e^{- x^{2} / 2} .$

Final Answer

$E (X) = \int_{z > x} z \cdot d P_{Z} = \int_{z > x} z f_{Z} (z) d z = \int_{x}^{\infty} z \frac{1}{\sqrt{2 π}} e^{- \frac{z^{2}}{2}} d z$

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

$\frac{1}{\sqrt{2 π}} \int_{\frac{x^{2}}{2}}^{\infty} e^{- u} d u = \frac{1}{\sqrt{2 π}} e^{- \frac{x^{2}}{2}}$

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

Short Answer

Step by step solution

Given Information

Explanation

Final Answer

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Let Z be a standard normal random variable,and, for a ﬁxed x, setX={ZifZ>x0otherwiseShow thatE[X]=12πe−x2/2.

Short Answer

Step by step solution

Given Information

Explanation

Final Answer

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Most popular questions from this chapter

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