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

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

with PDF

$\displaystyle 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,

$\displaystyle \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

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

and the variance of $ t$ is

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

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

$\displaystyle \log p(t; 1,\sigma^2)= -\log(\sigma^2) -\frac{1}{2} \log(2\pi)-\left({t\over 2\sigma^2} \right) -\frac{1}{2}\log \left({t\over \sigma^2} \right).$ (16.1)

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

$\displaystyle \log p(t; 2,\sigma^2) = -\log (2 \sigma^2) - {t \over 2 \sigma^2}.$ (16.2)



Baggenstoss 2017-05-19