Distribution of the square of a random normal variable

Question

If we have $X \sim N(\mu, \sigma^2)$, then what distribution does $X^2$ follow? In the case of the standard normal distribution, this is the chi-squared distribution, and in the case of unit variance - the non-central chi-squared distribution. However, is there a particular distribution for the case of zero mean and non-unit variance, or in the general case? Are the PDF and CDF of closed form?

Answer 1

Let $X \sim N(\mu, \sigma^2)$. Note that $X$ has the same distribution as $\sigma Y$ where $Y \sim N(\mu/\sigma, 1)$. You know that $Y^2$ is non-central chi-squared, so you can write the PDF/CDF of $X^2$ in terms of the PDF/CDF of $Y^2$.

$$F_{X^2}(u) = P(X^2 \le u) = P(Y^2 \le u/\sigma^2) = F_{Y^2}(u/\sigma^2)$$ and $$f_{X^2}(u) = \frac{d}{du} F_{X^2}(u) = \frac{d}{du} F_{Y^2}(u/\sigma^2) = \frac{1}{\sigma^2} f_{Y^2}(u/\sigma^2)$$

That said, the PDF/CDF of chi-squared distributions don't seem to have a nice closed form.

Answer 2

For a general normal distribution, you can use the fact that $$X = \mu + \sigma N$$ where $N$ is a standard normal to get $$X^2 = \mu^2 + 2\sigma \mu N + \sigma^2 N^2$$