What is the distribution of the square of a normally distributed random variable $X^2$ with $X\sim N(0,\sigma^2/4)$?
I know $\chi^2(1)=Z^2$ is a valid argument for when squaring a standard normal distribution, but what about the case of non-unit variance?

  • 1
    $\begingroup$ Why not just calculate this directly from the Normal equation, then plot the resulting function? $\endgroup$
    – user32490
    Commented Apr 10, 2014 at 23:40
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    $\begingroup$ I am looking for a theoretical explanation here... $\endgroup$
    – CodeTrek
    Commented Apr 10, 2014 at 23:40
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    $\begingroup$ Write $Z = \frac{X}{\sigma/2}$... or equivalently $X=\frac{\sigma}{2}\cdot Z$. Can you do it now? $\endgroup$
    – Glen_b
    Commented Apr 11, 2014 at 0:29
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    $\begingroup$ $\sigma^2/4∗\chi^2(1)$? So, nothing of fancy uncentered chi square stuff? $\endgroup$
    – CodeTrek
    Commented Apr 11, 2014 at 0:53
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    $\begingroup$ As long as the mean is $0$, no noncentral chi-square stuff; just plain vanilla scaled $\chi^2$ distribution as Glen_b points out. $\endgroup$ Commented Apr 11, 2014 at 4:29

1 Answer 1


To close this one:

$$ X\sim N(0,\sigma^2/4) \Rightarrow \frac {X^2}{\sigma^2/4}\sim \mathcal \chi^2_1 \Rightarrow X^2 = \frac {\sigma^2}{4}\mathcal \chi^2_1 = Q\sim \text{Gamma}(1/2, \sigma^2/2)$$


$$E(Q) = \frac {\sigma^2}{4},\;\; \text{Var}(Q) = \frac {\sigma^4}{8}$$


If $$X\sim N(\mu,\sigma^2/4)$$

then $$\frac {X^2}{\sigma^2/4} \sim \mathcal \chi^2_{1,NC}(\lambda=\mu^2),$$

where $\mathcal \chi^2_{1,NC}(\lambda)$ represents a Non-Central Chi-square with one degree of freedom, and $\lambda$ is the non-centrality parameter. Then

$$X^2 =\frac{\sigma^2}{4} \mathcal \chi^2_{1,NC}(\lambda)$$

can be treated as a version of the Generalized Chi-square.

  • 11
    $\begingroup$ if mu =/= 0, how will this turn out? $\endgroup$
    – Dom Jo
    Commented Aug 31, 2020 at 6:11

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