31
$\begingroup$

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?

$\endgroup$
6
  • 1
    $\begingroup$ Why not just calculate this directly from the Normal equation, then plot the resulting function? $\endgroup$
    – user32490
    Apr 10, 2014 at 23:40
  • 2
    $\begingroup$ I am looking for a theoretical explanation here... $\endgroup$
    – CodeTrek
    Apr 10, 2014 at 23:40
  • 1
    $\begingroup$ Write $Z = \frac{X}{\sigma/2}$... or equivalently $X=\frac{\sigma}{2}\cdot Z$. Can you do it now? $\endgroup$
    – Glen_b
    Apr 11, 2014 at 0:29
  • 1
    $\begingroup$ $\sigma^2/4∗\chi^2(1)$? So, nothing of fancy uncentered chi square stuff? $\endgroup$
    – CodeTrek
    Apr 11, 2014 at 0:53
  • 1
    $\begingroup$ As long as the mean is $0$, no noncentral chi-square stuff; just plain vanilla scaled $\chi^2$ distribution as Glen_b points out. $\endgroup$ Apr 11, 2014 at 4:29

1 Answer 1

42
$\begingroup$

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.

$\endgroup$
1
  • 11
    $\begingroup$ if mu =/= 0, how will this turn out? $\endgroup$
    – Dom Jo
    Aug 31, 2020 at 6:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.