Distribution of squares of standard Normal variates

(I'm not entirely sure about the validity, but it shows the general idea.)

If $X$ has p.d.f. $\mathrm{f}(x)$, then

$$\ensuremath{\mathrm{Pr}(X \in [a,b])}= \int_{a}^{b} \mathrm{f}(x) \d{x}$$

To find the probability of a function (assuming single-valued, for now) of $X$ being in a specific range, we find the limits within which $X$ would lie, i.e.

$$\ensuremath{\mathrm{Pr}(\mathrm{g}(X) \in [a,b])}= \ensuremath{\mathrm{Pr}(X \in [\mathrm{g}^{-1}(a), \mathrm{g}^{-1}(b)])}$$

(where .) Hence,

$$\begin{eqnarray*} \ensuremath{\mathrm{Pr}(\mathrm{g}(X) \in [a,b])}& = & \int_{\... ... & \int_a^b \mathrm{f}(\mathrm{g}^{-1}(u)) \d{\mathrm{g}^{-1}(u)}\end{eqnarray*}$$

upon substituting . If , then

$$\ensuremath{\mathrm{Pr}(Z^2 = x)}= \ensuremath{\mathrm{Pr}(Z... ...uremath{\mathrm{Pr}(Z = -x)}= 2 \ensuremath{\mathrm{Pr}(Z = x)}$$

(since the Normal distribution is symmetrical). Similarly,

$$\ensuremath{\mathrm{Pr}(Z^2 \in [a,b])}= 2 \ensuremath{\mathrm{Pr}(Z \in [\sqrt{a}, \sqrt{b}])}$$

From the above,

$$\begin{eqnarray*} \ensuremath{\mathrm{Pr}(Z^2 \in [a,b])}& = & \frac{2}{\sqrt{2\... ... \int_a^b u^{\frac{1}{2} - 1} \ensuremath{\mathrm{e}^{-u/2}}\d{u}\end{eqnarray*}$$

We recognise that this is the p.d.f. of the chi-squared distribution with . Hence

$$Z \ensuremath{\sim \mathrm{\chi^2}(1)}$$