Mean and Variance of Squared Gaussian

Mean and Variance of Squared Gaussian

Question

Given $Y=X^2$ where $X \sim \phi(0, \sigma^2)$, what is the mean and variance of the $Y$?

Since $\sigma^2 = E(X^2) - E(X)^2$, and $E(X) = 0$, $E(X^2) = \sigma^2$. Also,

$$VarX^2 = EX^4 - (EX^2)^2$$

and the fourth moment $EX^4$ is equal to $3\sigma^4$ since

$$\begin{aligned} E(X^4) &= \int \frac{x^4}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}}dx \\ &=\int -\frac{\sigma^2x^3}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}}d-\frac{x^2}{2\sigma^2} \\&=\int -\frac{\sigma^2x^3}{\sqrt{2\pi}\sigma}de^{-\frac{x^2}{2\sigma^2}} \\&= \frac{1}{\sqrt{2\pi}\sigma}\left ( -\sigma^2x^3e^{-{\frac{x^2}{2\sigma^2}}}|_{-\infty }^{+\infty } + 3\sigma^2\int x^2e^{-{\frac{x^2}{2\sigma^2}}}dx \right ) \\ &= \frac{1}{\sqrt{2\pi}\sigma} \left ( 0 - 3\sigma^4xe^{-{\frac{x^2}{2\sigma^2}}}|_{-\infty }^{+\infty } + 3\sigma^4\int e^{-{\frac{x^2}{2\sigma^2}}}dx \right ) \\ &= 3\sigma^4 \end{aligned}$$

(actually, $E\left [ X^{2n}\right ] = (2n - 1)!!\sigma^{2n}$, $!!$ is the double factorial

thus

$$\begin{aligned} VarX^2 &= EX^4 - (EX^2)^2 \\ &= 3\sigma^4 - \sigma^4 \\ &= 2\sigma^4 \end{aligned}$$

Reference

1. Mean and variance of Squared Gaussian
2. Proving $E(X^4) = 3\sigma^4$