Question #211464

Supposex ~N(0,1) and Y=X2 Write down the p.d.f. of Y. Calculate E(Y) and Var(Y).


1
Expert's answer
2021-06-29T07:59:55-0400

If the random variable VN(μ,σ2),σ2>0,V \sim N(\mu, \sigma^2),\sigma^2>0, then the random variable W=(Xμ)2σ2=Z2χ2(1),W=\dfrac{(X-\mu)^2}{\sigma^2}=Z^2\sim \chi^2(1), chi-squared distribution, with ν=1\nu=1 degrees of freedom.


Suppose XN(0,1),X\sim N(0, 1), then Y=(X0)212=X2χ2(1),Y=\dfrac{(X-0)^2}{1^2}=X^2\sim \chi^2(1), chi-squared distribution, with ν=1\nu=1 degrees of freedom.

Its density function is given by


f(y;ν)={12ν/2Γ(ν/2)yν/21ey/2,y>00,elsewheref(y;\nu)= \begin{cases} \dfrac{1}{2^{\nu/2}\Gamma(\nu/2)}y^{\nu/2-1}e^{-y/2},y>0 \\ 0, \text{elsewhere} \end{cases}

Then


f(y;ν)={12πyey/2,y>00,elsewheref(y;\nu)= \begin{cases} \dfrac{1}{\sqrt{2\pi}\sqrt{y}}e^{-y/2},y>0 \\ 0, \text{elsewhere} \end{cases}

The mean of the distribution is equal to the number of degrees of freedom:


E(Y)=μY=ν=1E(Y)=\mu_Y=\nu=1

The variance is equal to two times the number of degrees of freedom:


Var(Y)=σY2=2ν=2Var(Y)=\sigma_Y^2=2\nu=2


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