Mean and variance of Squared Gaussian: $Y=X^2$ where: $X\sim\mathcal{N}(0,\sigma^2)$?

What is the mean and variance of Squared Gaussian: $Y=X^2$ where: $X\sim\mathcal{N}(0,\sigma^2)$?

It is interesting to note that Gaussian R.V here is zero-mean and non-central Chi-square Distribution doesn't work.

Thanks.

• $Y/\sigma$ has Chi-squared distribution with 1 degree of freedom. – Yury Dec 28 '13 at 0:09
• @Yury You mean $Y/\sigma^2$. – iballa Dec 28 '13 at 0:11
• Yes, $Y/\sigma^2$. – Yury Dec 28 '13 at 0:16

Note that $X^2 \sim \sigma^2 \chi^2_1$ where $\chi^2_1$ is the Chi-squared distribution with 1 degree of freedom. Since $E[\chi^2_1] = 1, \text{Var}[\chi^2_1] = 2$ we have $E[X^2] = \sigma^2, \text{Var}[X^2] = 2 \sigma^4$.
• Thanks. Does $X^2 \sim \sigma^2 \chi^2_1$ holds true too if $X\sim\mathcal{N}(\mu,\sigma^2)$ ? – kaka Jan 19 '14 at 2:14
• @kaka No. If $X = \sigma Y + \mu$ where $Y$ is a standard normal, then $X^2 = \sigma^2 Y^2 + 2 \sigma \mu Y + \mu^2$. – iballa Jan 28 '14 at 20:16
We can avoid using the fact that $X^2\sim\sigma^2\chi_1^2$, where $\chi_1^2$ is the chi-squared distribution with $1$ degree of freedom, and calculate the expected value and the variance just using the definition. We have that $$\operatorname E X^2=\operatorname{Var}X=\sigma^2$$ since $\operatorname EX=0$ (see here).
Also, $$\operatorname{Var}X^2=\operatorname EX^4-(\operatorname EX^2)^2.$$ The fourth moment $\operatorname EX^4$ is equal to $3\sigma^4$ (see here). Hence, $$\operatorname{Var}X^2=3\sigma^4-\sigma^4=2\sigma^4.$$