Q.5.12 聽Use the identity of Theoretica... [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 5: Q.5.12 (page 215)

Use the identity of Theoretical Exercise 5.5 to derive E[X2] when X is an exponential random variable with parameter 位.

Short Answer

Expert verified

Thus,

${鈭�}_{0}^{鈭�} 2 x P (X > x) d x = {鈭�}_{0}^{鈭�} 2 x e^{鈭� 位 x} d x = 2 {鈭�}_{0}^{鈭�} x e^{鈭� 位 x} d x$

$Use partial integration . Let u = x and d v = e^{鈭� 位 x} . Hence, we have that d u = d x and v = 鈭� \frac{e^{鈭� 位 x}}{位} . Thus$

$2 {鈭�}_{0}^{鈭�} x e^{鈭� 位 x} d x = 鈭� {2 x \frac{e^{鈭� 位 x}}{位} |}_{0}^{鈭�} + \frac{2}{位} {鈭�}_{0}^{鈭�} e^{鈭� 位 x} d x = {\frac{2}{位} 鈰� (鈭� \frac{e^{鈭� 位 x}}{位}) |}_{0}^{鈭�} = \frac{2}{位^{2}}$

Step by step solution

Given Information

Derive E[X2] when X is an exponential random variable with parameter 位.

Explanation

From exercise 5, we have that.

$E (X^{2}) = {鈭�}_{0}^{鈭�} 2 x P (X > x) d x$

$Now, we are given that X ~ Expo 鈦� (位) . Hence, we have that$

$P (X > x) = 1 鈭� P (X 鈮� x) = 1 鈭� (1 鈭� e^{鈭� 位 x}) = e^{鈭� 位 x}$

Thus,

${鈭�}_{0}^{鈭�} 2 x P (X > x) d x = {鈭�}_{0}^{鈭�} 2 x e^{鈭� 位 x} d x = 2 {鈭�}_{0}^{鈭�} x e^{鈭� 位 x} d x$

$Use partial integration . Let u = x and d v = e^{鈭� 位 x} . Hence, we have that d u = d x and v = 鈭� \frac{e^{鈭� 位 x}}{位} . Thus$

Hence, we have proved that

$E (X^{2}) = \frac{2}{位^{2}}$

Final Answer

Thus,

${鈭�}_{0}^{鈭�} 2 x P (X > x) d x = {鈭�}_{0}^{鈭�} 2 x e^{鈭� 位 x} d x = 2 {鈭�}_{0}^{鈭�} x e^{鈭� 位 x} d x$

$Use partial integration . Let u = x and d v = e^{鈭� 位 x} . Hence, we have that d u = d x and v = 鈭� \frac{e^{鈭� 位 x}}{位} . Thus$

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91影视

Use the identity of Theoretical Exercise 5.5 to derive E[X2] when X is an exponential random variable with parameter 位.

Short Answer

Step by step solution

Given Information

Explanation

Final Answer

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