Chi-Square

Let $x_i$ be a zero mean Gaussian with variance $\sigma^2$. The Chi-square random variable with $N$ degrees of freedom, also written $\chi^2(N,\sigma^2)$, is

$$t = \sum_{i=1}^N \; x_i^2,$$

with PDF

$$p(t; N,\sigma^2) = \frac{1}{\sigma^2} \; \Gamma^{-1}\left({N\ove... ...\over \sigma^2} \right)^{N/2-1} \; \exp\left\{ -{t \over 2 \sigma^2} \right\}$$

or in log form,

$$\log p(t; N,\sigma^2) = -\log \sigma^2 - \log \Gamma\left({N\over... ...og 2 + (N/2-1) \log \left( {t\over \sigma^2} \right) \; -{t \over 2 \sigma^2}.$$

The mean of $t$ is

$$\mu = {\cal E}\{t\}=N\sigma^2,$$

and the variance of $t$ is

$${\cal E}\{(t-\mu)^2\}=2N\sigma^4.$$

If $N=1$, the mean of $t$ is $\sigma^2$, the variance of $t$ is $2 \sigma^4$, and

$$\log p(t; 1,\sigma^2)= -\log(\sigma^2) -\frac{1}{2} \log(2\pi)-\left({t\over 2\sigma^2} \right) -\frac{1}{2}\log \left({t\over \sigma^2} \right).$$ (17.1)

If $N=2$, $t$ is an exponential RVs and with mean $2 \sigma^2$, the variance of $t$ is $4 \sigma^4$, and

$$\log p(t; 2,\sigma^2) = -\log (2 \sigma^2) - {t \over 2 \sigma^2}.$$ (17.2)