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Q. 10.14

Expert-verified

A First Course in Probability

Found in: Page 431

Book edition 9th

Author(s) Sheldon M. Ross

Pages 432 pages

ISBN 9780321794772

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Short Answer

In Example 4a, we showed that
E[(1 − V2) 1/2] = E[(1 − U2) 1/2] = π/4
when V is uniform (−1, 1) and U is uniform (0, 1). Now show that
Var[(1 − V2) 1/2] = Var[(1 − U2) 1/2]
and find their common value.

The common value is $V a r ({(1 - v^{2})}^{\frac{1}{2}}) = v a r ({(1 - U^{2})}^{\frac{1}{2}}) = \frac{2}{3} - \frac{π^{2}}{16}$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Information

We have given the condition

$E [{(1 - V^{2})}^{1 / 2}] = E [{(1 - U^{2})}^{1 / 2}] = \frac{π}{4}$

when $V$ is unform $(- 1, 1)$ and $U$ is unform $(0, 1)$ .

Step 2: Simplify

Using the basic formula for the variance,

$V a r ({(1 - V^{2})}^{\frac{1}{2}}) = E (1 - V^{2}) - E {({(1 - V^{2})}^{\frac{1}{2}})}^{2}$

Consider that we know the second expression as it is given. we have

$E {({(1 - V^{2})}^{\frac{1}{2}})}^{2} = \frac{π^{2}}{16}$

Calculating $E (1 - V^{2})$ . since V is uniformly distributed over (-1 , 1), we have

$E (1 - V^{2}) = \int_{- 1}^{1} (1 - v^{2}) \frac{1}{2} d v = \frac{2}{3}$

Therefore,

$V a r ({(1 - V^{2})}^{\frac{1}{2}}) = \frac{2}{3} - \frac{π^{2}}{16}$

Similarly, the same calculation for ${(1 - U^{2})}^{\frac{1}{2}}$ . we have

$V a r ({(1 - U^{2})}^{\frac{1}{2}}) = E (1 - U^{2}) - E ({(1 - U^{2})}^{\frac{1}{2}})^{2}$

Observe that we know the second expression as it is given in the task.

$E ({(1 - U^{2})}^{\frac{1}{2}})^{2} = π$ ²16

Calculating $E (1 - U^{2})$ . since U is uniformly distributed over (0,1), we have

$E (1 - U^{2}) = \int_{0}^{1} (1 - u^{2}) d u = \frac{2}{3}$

Therefore,

$V a r ({(1 - U^{2})}^{\frac{1}{2}}) = \frac{2}{3} - π$ ²16

So, we have proved

$V a r ({(1 - v^{2})}^{\frac{1}{2}}) = v a r ({(1 - U^{2})}^{\frac{1}{2}}) = \frac{2}{3} - \frac{π^{2}}{16}$

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