chi-squared 1

Defintion

If \(Z\) is a standard normal random variable, the distribution of \(U = Z^2\) is chi-squared with 1 degree of freedom, denoted \(\chi^2_1\).

Let \(Z\) be a standard normal random variable with probability density function (pdf) has pdf: \[ f_Z(z) = \frac{1}{\sqrt{2\pi}} e^{-z^2/2} \] and cumulative distribution function (CDF) \(F_Z(z)\). The variable \(U = Z^2\) is non-negative.

The cumulative distribution function of \(U\) is given by: \[ F_U(u) = P(U \leq u) = P(Z^2 \leq u) \]

Since \(Z\) is standard normal, we have: \[ P(Z^2 \leq u) = P(-\sqrt{u} \leq Z \leq \sqrt{u}) \]

Expressing this in terms of the standard normal CDF: \[ F_U(u) = F_Z(\sqrt{u}) - F_Z(-\sqrt{u}) \]

The probability density function (pdf) of \(U\) is then obtained by taking the derivative: \[ f_U(u) = \frac{d}{du} \left[F_Z(\sqrt{u}) - F_Z(-\sqrt{u})\right] \]

Since we do not have an explicit expression for \(F_Z(z)\) it would appear that we are stuck at this point, but we can still get the corresponding \(f_U(u)\) by simple differentiation: \[ \begin{align} f_U(u) &=\frac{d}{du} F_Z(\sqrt{u}) - \frac{d}{du} F_Z(-\sqrt{u})\\ &= \frac{1}{2}u^{-1/2}f_z(\sqrt{u}) - (-\frac{1}{2})u^{-1/2}f_z(-\sqrt{u})\\ &= \frac{1}{2}u^{-1/2} \frac{1}{\sqrt{2\pi}} e^{-(\sqrt{u})^2/2} + \frac{1}{2}u^{-1/2}\frac{1}{\sqrt{2\pi}} e^{-(-\sqrt{u})^2/2}\\ &= \frac{1}{\sqrt{2\pi u}}e^{-u/2} \end{align} \]

Using the chain rule, the pdf for \(U\) is found to be: \[ f_U(u) = \frac{1}{\sqrt{2\pi u}} e^{-u/2} \]

Compare this with the probability density function (pdf) of a chi-squared distribution with \(k\) degrees of freedom, denoted \(\chi^2_k\), given by: \[ f(x; k) = \frac{1}{2^{k/2} \Gamma(k/2)} x^{(k/2) - 1} e^{-x/2}, \quad x > 0 \] where \(\Gamma(\cdot)\) is the gamma function. By comparing the two pdfs, it is evident that \(U\) follows a chi-squared distribution with 1 degree of freedom (\(\chi^2_1\)).

\[ f_{\chi^2_1}(u) = \frac{1}{2^{\frac{1}{2}} \Gamma\left(\frac{1}{2}\right)} u^{-\frac{1}{2}} e^{-\frac{u}{2}} \]

This follows from the Gaussian Integral

\[ \int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}. \] This follows from the fact that we know the pdf of a Normal distribution with a \(\mu=0\) and \(\sigma^2 = 1/2\) should integrate to one. \[ \begin{align} 1 &= \int _{-\infty }^{\infty } \frac {1}{\sigma {\sqrt {2\pi }}}e^{-{\frac {1}{2}}\left({\frac {x-\mu }{\sigma }}\right)^{2}} \\ 1 &= \int _{-\infty }^{\infty } \frac {1}{ {\sqrt {\pi }}}e^{-x^{2}}\\ \sqrt {\pi } &= \int _{-\infty }^{\infty } e^{-x^{2}} \end{align} \] Note that its MGF is thus equal to \((1 − 2t)^{−1/2}\). See result here