# Square of normal distribution with specific variance

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?

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

To close this one:

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

with

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

RESPONSE TO QUESTION IN THE COMMENT

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.

• if mu =/= 0, how will this turn out? Commented Aug 31, 2020 at 6:11