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I have a normal distribution $X$~$N(\mu,\sigma^2)$. Is there an exact value for the mean and standard deviation of $X^2$?

Thanks

kjetil b halvorsen
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Minas
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1 Answers1

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This sounds like a self-study question, so I’ll show you how to derive this.

The mean is the expected value, so $\mathbb{E}X=\int_{\mathbb{R}}xf_X(x)dx$. Variance is related to the expected value of both $X$ and $X^2$: $$var(X)=\mathbb{E}[X^2]-(\mathbb{E}X)^2$$

The expected value of a transformation $g$ of a variable is given by $\int_{\mathbb{R}}g(x)f_X(x)dx$.

You want to evaluate $\mathbb{E}X$ and $\mathbb{E}[X^2]$, where $X=Y^2$ and $Y\sim N(\mu, \sigma^2)$.

Stitch these together and either integrate by parts by hand or stick the integrals in WolframAlpha. You’ll take the square root of variance to get standard deviation.

If $f(x)$ is the PDF of the normal distribution, you’ll be interested in $\int x^2f(x)dx$ and $\int x^4 f(x)dx$, both integrated over all real numbers $\mathbb{R}$.

Dave
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    Dave, I appreciate your reply. Does this imply that that the two 'Y''s in Y^2 that I am analysing are independent or not? – Minas May 17 '20 at 22:44
  • The two $Y$s aren’t just dependent; they’re the same random variable! $Y^2$ for $Y\sim N(\mu,\sigma^2)$ is different from $XY$ for $X,Y\sim N(\mu,\sigma^2)$. – Dave May 17 '20 at 23:28
  • Behave like a $\chi^2$ distribution with 1 DOF... I see. Is there a version of $\chi^2$ distribution for more powers? Like $X^1.5$ (or $X^0.5$)? Because I can't seem to find any. – Minas May 18 '20 at 10:32
  • Well if $\chi^2=\sum_i X_i^2$ for $iid$ $X_i\sim N(0,1)$ and $\chi =\sqrt{\sum_i X_i^2}$, then I think it’s reasonable to define $\chi^{\alpha} = \big( \sum_i X_i^2\big)^{\alpha/2}$. There’s (probably) a way to define these as special cases of gamma distributions. – Dave May 18 '20 at 11:03
  • Thank you Dave, that's very helpful. – Minas May 18 '20 at 14:45