91影视

We have given the condition

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

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{蟺}}^2}{16}$

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

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

Therefore,

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

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

$E(1-U^2)={鈭�}_0^1(1-u^2)du=\frac{2}{3}$

Therefore,

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