Q. 5.21 Show that 螕12=蟺Hint: 螕12=鈭�0... [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. 5.21 (page 216)

Show that $螕 (\frac{1}{2}) = \sqrt{蟺}$
Hint: $螕 (\frac{1}{2}) = {鈭�}_{0}^{鈭�} e^{- x} x^{- 1 / 2} d x$ Make the change of variables $y = \sqrt{2 x}$ and then relate the resulting expression to the normal distribution.

Short Answer

Expert verified

Using the proper substitution, find the density of Standard Normal under the integral by expressing $螕 (\frac{1}{2})$ the condition.

Step by step solution

Substitutions of variables.

We have that

$螕 (\frac{1}{2}) = {鈭�}_{0}^{鈭�} x^{\frac{1}{2} - 1} e^{- mtight > x} d x = {鈭�}_{0}^{鈭�} x^{- \frac{1}{2}} e^{- x} d x$

Now, make the substitution $x = \frac{y^{2}}{2}$ which implies that $d x = y d y$ and $y = \sqrt{2 x}$

We have that

${鈭�}_{0}^{鈭�} x^{- \frac{1}{2}} e^{- x} d x = {鈭�}_{0}^{鈭�} \frac{\sqrt{2}}{y} e^{- \frac{y^{2}}{2}} y d y = \sqrt{2} {鈭�}_{0}^{鈭�} e^{- \frac{y^{2}}{2}} d y$

Evaluate the integral.

In order to evaluate the remaining integral, observe that ${鈭�}_{0}^{鈭�} \frac{1}{\sqrt{2 蟺}} e^{- \frac{y^{2}}{2}} d y = \frac{1}{2}$ since the function under the integral is the density of Standard Normal and we integrate it over the positive half of the real line.

So, we have that

$\sqrt{2} {鈭�}_{0}^{鈭�} e^{- \frac{y^{2}}{2}} d y = \sqrt{2} 路 \sqrt{2 蟺} {鈭�}_{0}^{鈭�} \frac{1}{\sqrt{2 蟺}} e^{- \frac{y^{2}}{2}} d y = \sqrt{2} 路 \sqrt{2 蟺} 路 \frac{1}{2} = \sqrt{蟺}$

So, we have proved that

$螕 (\frac{1}{2}) = \sqrt{蟺}$

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91影视

Show that $螕 (\frac{1}{2}) = \sqrt{蟺}$
Hint: $螕 (\frac{1}{2}) = {鈭�}_{0}^{鈭�} e^{- x} x^{- 1 / 2} d x$ Make the change of variables $y = \sqrt{2 x}$ and then relate the resulting expression to the normal distribution.

Short Answer

Step by step solution

Substitutions of variables.

Evaluate the integral.

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91影视

Show that 螕12=蟺Hint: 螕12=鈭�0鈭�e-xx-1/2dxMake the change of variables y=2xand then relate the resulting expression to the normal distribution.

Short Answer

Step by step solution

Substitutions of variables.

Evaluate the integral.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Logic and Functions

Theoretical and Mathematical Physics

Pure Maths

Mechanics Maths

Applied Mathematics

Study anywhere. Anytime. Across all devices.

Show that $螕 (\frac{1}{2}) = \sqrt{蟺}$
Hint: $螕 (\frac{1}{2}) = {鈭�}_{0}^{鈭�} e^{- x} x^{- 1 / 2} d x$ Make the change of variables $y = \sqrt{2 x}$ and then relate the resulting expression to the normal distribution.