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Answers without the blur. Sign up and see all textbooks for free! Q. 10.14

Expert-verified Found in: Page 431 ### A First Course in Probability

Book edition 9th
Author(s) Sheldon M. Ross
Pages 432 pages
ISBN 9780321794772 # In Example 4a, we showed that E[(1 − V2) 1/2] = E[(1 − U2) 1/2] = π/4when 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$Var\left({\left(1-{v}^{2}\right)}^{1}{2}}\right)=var\left({\left(1-{U}^{2}\right)}^{1}{2}}\right)=\frac{2}{3}-\frac{{\mathrm{\pi }}^{2}}{16}$.

See the step by step solution

## Step 1: Given Information

We have given the condition

$E\left[{\left(1-{V}^{2}\right)}^{1/2}\right]=E\left[{\left(1-{U}^{2}\right)}^{1/2}\right]=\frac{\mathrm{\pi }}{4}$

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

## Step 2: Simplify

Using the basic formula for the variance,

$Var\left({\left(1-{V}^{2}\right)}^{1}{2}}\right)=E\left(1-{V}^{2}\right)-E{\left({\left(1-{V}^{2}\right)}^{1}{2}}\right)}^{2}$

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

$E{\left({\left(1-{V}^{2}\right)}^{1}{2}}\right)}^{2}=\frac{{\mathrm{\pi }}^{2}}{16}$

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

$E\left(1-{V}^{2}\right)={\int }_{-1}^{1}\left(1-{v}^{2}\right)\frac{1}{2}dv=\frac{2}{3}$

Therefore,

$Var\left({\left(1-{V}^{2}\right)}^{1}{2}}\right)=\frac{2}{3}-\frac{{\mathrm{\pi }}^{2}}{16}$

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

$Var\left({\left(1-{U}^{2}\right)}^{1}{2}}\right)=E\left(1-{U}^{2}\right)-E\left({\left(1-{U}^{2}\right)}^{1}{2}}{\right)}^{2}$

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

$E\left({\left(1-{U}^{2}\right)}^{1}{2}}{\right)}^{2}=\frac{\mathrm{\pi }2}{16}$

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

$E\left(1-{U}^{2}\right)={\int }_{0}^{1}\left(1-{u}^{2}\right)du=\frac{2}{3}$

Therefore,

$Var\left({\left(1-{U}^{2}\right)}^{1}{2}}\right)=\frac{2}{3}-\frac{\mathrm{\pi }2}{16}$

So, we have proved

$Var\left({\left(1-{v}^{2}\right)}^{1}{2}}\right)=var\left({\left(1-{U}^{2}\right)}^{1}{2}}\right)=\frac{2}{3}-\frac{{\mathrm{\pi }}^{2}}{16}$ ### Want to see more solutions like these? 