How can I find the distribution of a stochastic variable X^2 if X is normal standard distributed?

Question

I am considering a stochastic variable X that is standard normal distributed i.e. $$ F_X(x) = \int_{-\infty}^x\frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}dt $$ How do I find out the distribution of $X^2$?

/Gus

Answer 1

Let $Y=X^2$. We presumably want to find the density function of $Y$. First we (sort of) find the cumulative distribution function $F_Y(y)$ of $Y$. Note that $Y$ is never negative. For $y\gt 0$, we have $$F_Y(y)=\Pr(Y\le y)=\Pr(X^2\le y)=\Pr(-\sqrt{y}\le X\le \sqrt{y}).\tag{1}$$ It follows that $$F_Y(y)=\int_{-\sqrt{y}}^{\sqrt{y}}\frac{1}{\sqrt{2\pi}}e^{-t^2/2}\,dt.\tag{2}$$ We could integrate. However, if all we want is to find the density function $f_Y(y)$, we can use the Fundamental Theorem of Calculus to differentiate the integral in (2). Using the Chain Rule, we get $$f_Y(y)=\frac{2}{2\sqrt{y}}\cdot\frac{1}{\sqrt{2\pi}}e^{-y/2}$$ (for $y\gt 0$).