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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}\end{displaymath}

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),

(where \( \mathrm{g}^{-1}(\mathrm{g}(x)) = x \).) Hence,

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

upon substituting \( u = \mathrm{g}(x) \). If \( Z \sim \mathrm{N}(0,1) \), then

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

(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}])}\end{displaymath}

From the above,

\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 \( \nu = 1 \). Hence

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

Alexander Frolkin 2001-02-01