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Q. 10.14

Expert-verifiedFound in: Page 431

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/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$Var\left({(1-{v}^{2})}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)=var\left({(1-{U}^{2})}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)=\frac{2}{3}-\frac{{\mathrm{\pi}}^{2}}{16}$.

We have given the condition

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

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

Using the basic formula for the variance,

$Var\left({(1-{V}^{2})}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)=E(1-{V}^{2})-E{\left({(1-{V}^{2})}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^{2}$

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

$E{\left({(1-{V}^{2})}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^{2}=\frac{{\mathrm{\pi}}^{2}}{16}$

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

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

Therefore,

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

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

$Var\left({(1-{U}^{2})}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)=E(1-{U}^{2})-E({(1-{U}^{2})}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{)}^{2}$

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

$E({(1-{U}^{2})}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{)}^{2}=\frac{\mathrm{\pi}2}{16}$

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

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

Therefore,

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

So, we have proved

$Var\left({(1-{v}^{2})}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)=var\left({(1-{U}^{2})}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)=\frac{2}{3}-\frac{{\mathrm{\pi}}^{2}}{16}$

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