# 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
Apr 10 '14 at 23:40
• I am looking for a theoretical explanation here... Apr 10 '14 at 23:40
• Write $Z = \frac{X}{\sigma/2}$... or equivalently $X=\frac{\sigma}{2}\cdot Z$. Can you do it now? Apr 11 '14 at 0:29
• $\sigma^2/4∗\chi^2(1)$? So, nothing of fancy uncentered chi square stuff? Apr 11 '14 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. Apr 11 '14 at 4:29

$$X\sim N(0,\sigma^2/4) \Rightarrow \frac {X^2}{\sigma^2/4}\sim \mathcal \chi^2_1 \Rightarrow X^2 = \frac {\sigma^2}{4}\mathcal \chi^2_1 = Q\sim \text{Gamma}(1/2, \sigma^2/2)$$
$$E(Q) = \frac {\sigma^2}{4},\;\; \text{Var}(Q) = \frac {\sigma^4}{8}$$