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?
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Richard Hardy
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CodeTrek
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1Why not just calculate this directly from the Normal equation, then plot the resulting function? – Apr 10 '14 at 23:40
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2I am looking for a theoretical explanation here... – CodeTrek Apr 10 '14 at 23:40
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1Write $Z = \frac{X}{\sigma/2}$... or equivalently $X=\frac{\sigma}{2}\cdot Z$. Can you do it now? – Glen_b Apr 11 '14 at 00:29
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1$\sigma^2/4∗\chi^2(1)$? So, nothing of fancy uncentered chi square stuff? – CodeTrek Apr 11 '14 at 00:53
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1As long as the mean is $0$, no noncentral chi-square stuff; just plain vanilla scaled $\chi^2$ distribution as Glen_b points out. – Dilip Sarwate Apr 11 '14 at 04:29
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I think one of the above comments is wrong. It is not true that $$ X = \sigma/2 *Z$$. – Jan 31 '16 at 06:35
1 Answers
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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.
Alecos Papadopoulos
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