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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.

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  • $\begingroup$ $Y/\sigma$ has Chi-squared distribution with 1 degree of freedom. $\endgroup$
    – Yury
    Dec 28, 2013 at 0:09
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    $\begingroup$ @Yury You mean $Y/\sigma^2$. $\endgroup$
    – iballa
    Dec 28, 2013 at 0:11
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    $\begingroup$ Yes, $Y/\sigma^2$. $\endgroup$
    – Yury
    Dec 28, 2013 at 0:16

2 Answers 2

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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, $$ \operatorname{Var}X^2=3\sigma^4-\sigma^4=2\sigma^4. $$

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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 $E[X^2] = \sigma^2, \text{Var}[X^2] = 2 \sigma^4$.

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  • $\begingroup$ Thanks. Does $X^2 \sim \sigma^2 \chi^2_1$ holds true too if $X\sim\mathcal{N}(\mu,\sigma^2)$ ? $\endgroup$
    – kaka
    Jan 19, 2014 at 2:14
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    $\begingroup$ @kaka No. If $X = \sigma Y + \mu$ where $Y$ is a standard normal, then $X^2 = \sigma^2 Y^2 + 2 \sigma \mu Y + \mu^2$. $\endgroup$
    – iballa
    Jan 28, 2014 at 20:16

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