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If $X \sim \mathcal{N}(0,\sigma^2)$, then $X^2$ is distributed according to a scaled chi-square distribution.

If $X \sim \mathcal{N}(\mu,1)$, then $X^2$ is distributed according to a noncentral chi-square distribution.

But what about the case when $X \sim \mathcal{N}(\mu,\sigma^2)$, is the distribution of $X^2$ known in this situation?

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    $\begingroup$ Must be a scaled noncentral $\chi^2(1)$. Also, did you mean $\sigma$ or $\sigma^2$ in the last line? $\endgroup$ Nov 17, 2020 at 12:36
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    $\begingroup$ Observe that $X/\sigma\sim\mathcal{N}(\mu/\sigma,1)$ and (immediately) draw your conclusion. $\endgroup$
    – whuber
    Nov 17, 2020 at 12:39
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    $\begingroup$ Does this answer your question? Distribution of a quadratic form, non-central chi-squared distribution $\endgroup$
    – Xi'an
    Nov 17, 2020 at 13:30

3 Answers 3

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Graphs in R per @whuber's Comment:

enter image description here

set.seed(1117)
par(mfrow=c(1,3))

w = rnorm(10^6, 150, 15)
summary(w); sd(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  77.08  139.86  150.01  150.02  160.15  221.71 
[1] 1.000814

hdr1 = "W ~ NORM(150, 15)"
hist(w, prob=T, br=30, col="skyblue2", main=hdr1)
 curve(dnorm(x, 150, 15), add=T, col="red")

.

x = w/15
summary(x); sd(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  5.138   9.324  10.001  10.001  10.676  14.781 
[1] 1.000814

hdr2 = "X = W/15 ~ NORM(10, 1)"
hist(x, prob=T, br=40, col="skyblue2", main=hdr2)
 curve(dnorm(x, 10, 1), add=T, col="red")

See Wikipedia on non-central chi-squared distribution. Notice that the mean of $m = 10^6$ observations from $Y \sim \mathsf{Chisq}(\nu=1,\lambda=10^2)$ is consistent with $E(Y) = \nu+\lambda=101.$

y = x^2
summary(y); sd(y)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  26.40   86.94  100.01  101.02  113.99  218.48 
[1] 20.07186

hdr3 = "Y ~ CHISQ(DF=1, NCP=100)"
hist(y, prob=T, br=30, col="skyblue2", main=hdr3)
 curve(dchisq(x,1,100), add=T, col="red")

par(mfrow=c(1,1)) 
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There is a really good proof of this on https://online.stat.psu.edu/stat414/lesson/16/16.5. To prove the theorem, it is needed to show that the p.d.f. of the random variable is the same as the p.d.f. of a chi-square random variable with 1 degree of freedom

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    $\begingroup$ If the link goes dead this will not prove very helpful. Can you abstract the main contents of the link and edit them into your question? $\endgroup$
    – mdewey
    Nov 17, 2020 at 14:36
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Ok so I think I can answer this now based on the suggestions in the comments.

We have $X \sim \mathcal{N}(\mu, \sigma^2) = \sigma \mathcal{N}(\frac{\mu}{\sigma}, 1)$.

So $X/\sigma \sim \mathcal{N}(\frac{\mu}{\sigma},1)$ and thus $X^2/\sigma^2$ is is distributed according to $\chi_{1,\frac{\mu^2}{\sigma^2}}^2$, a noncentral chi-square distribution where the subscripts indicate we have $1$ degree of freedom and a noncentrality parameter $\frac{\mu^2}{\sigma^2}$, respectively.

Then $X^2 \sim \sigma^2 \chi_{1,\frac{\mu^2}{\sigma^2}}^2$. The pdf of $\sigma^2 \chi_{1,\frac{\mu^2}{\sigma^2}}^2$ is $$ f_{\sigma^2 \chi_{1,\frac{\mu^2}{\sigma^2}}^2}(x) = \frac{1}{\sigma^2}f_{\chi_{1,\frac{\mu^2}{\sigma^2}}^2}\bigg(\frac{x}{\sigma^2}\bigg), $$ where $f_{\sigma^2 \chi_{1,\lambda}^2}(x)$ is defined as $$ f_{\sigma^2 \chi_{1,\lambda}^2}(x) = \frac{1}{2}e^{-(x+\lambda)}\bigg(\frac{x}{\lambda}\bigg)^{-1/4}I_{-1/2}(\sqrt{\lambda x}). $$

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    $\begingroup$ When you divide by $\sigma$ you need to consider what that does to the mean $\endgroup$ Nov 17, 2020 at 22:45
  • $\begingroup$ Thanks, I've edited it to properly account for the division by $\sigma$. $\endgroup$
    – Bertus101
    Nov 18, 2020 at 9:54
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    $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – quester
    Nov 18, 2020 at 10:27
  • $\begingroup$ Thanks, I overlooked the fact that the normality parameter was based on the sum of means squared (I mistakenly thought it was just sum of means). $\endgroup$
    – Bertus101
    Nov 18, 2020 at 15:33

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