1
$\begingroup$

let $X \sim N(0,\sigma^2)$

now let $Y=X^2$

what is $E(Y)$ and $V(Y)$

so $E(Y)=E(X^2)=\sigma^2$

but I can't find $V(Y)$

| cite | improve this question | | | | |
$\endgroup$
  • 2
    $\begingroup$ Surely you can find $V(Y)$ from the definition, because the moments of $Y$ bear a simple relationship to those of $X:$ $$E[Y^k] = E[(X^2)^k] = E[X^{2k}]$$ shows the raw moments of $Y$ are the corresponding raw even moments of $X.$ $\endgroup$ – whuber Mar 13 at 13:47
2
$\begingroup$

Using the definition of variance, you have: $$\operatorname{var}(Y)=E[X^4]-E[X^2]^2$$ You need the 4-th non-central moment of normal RV. Since, the mean is $0$, you can also use central moment, which is $3\sigma^4$. Subtracting $E[X^2]^2=\sigma^4$ makes variance of $Y$ $2\sigma^4$.

Another way is to use the relation between normal, chi-squared and gamma distributions. The squared standard normal is chi-squared distribution with $k=1$, which is at the same time Gamma RV. So, let $X=\sigma Z$, then $Z^2\sim\chi^2(1)\rightarrow \sigma^2 Z^2=X^2 \sim \Gamma(k=1/2,\theta=2\sigma^2)$. So, $Y$ is actually Gamma distributed. Gamma variance is $\operatorname{var}(Y)=k\theta^2=2\sigma^4$.

Another way can be simply evaluating the integral (let $c=\frac{1}{\sqrt{2\pi}\sigma}$):

$$\begin{align}E[X^4]&=c\int_{-\infty}^\infty x^4 \exp(-x^2/2\sigma^2)dx\\&=c\underbrace{\int_{-\infty}^\infty x^4 \exp(-x^2/2\sigma^2)dx}_{u=x^3,dv=x\exp(-x^22\sigma^2)}\\&=3\sigma^2\underbrace{\int_{-\infty}^{\infty}cx^2\exp(-x^2/2\sigma^2)dx}_{E[X^2]}\\&=3\sigma^4\end{align}$$

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.