PDF form

Let $ {\bf x}$ be the magnitude-squared of the DFT bins of a size-$ N_t$ DFT. Then, $ N=N_t/2+1$. Let $ {\bf x}$ have mean $ \bar{{\bf x}}$. Then, the distribution of the zero and Nyquist frequency bins ($ i=1,N$) follows the Chi-squared distribution with 1 degree of freedom scaled by $ \bar{x}_i$,

$\displaystyle p(x_i) = \frac{1}{\bar{x}_i\sqrt{2\pi}}\; (x_i/\bar{x}_i)^{-1/2} \; \exp\left\{
-\frac{x_i}{2\bar{x}_i}\right\}, \;\; i=1, \;i=N.
$

DFT bins 1 through $ N-1$ are complex so have the Chi-squared distribution with 2 degrees of freedom scaled by $ \bar{x}_i/2$, or equivalently the exponential distribution with mean $ \bar{x}_i$:

$\displaystyle p(x_i) = \frac{1}{\bar{x}_i}\; \exp\left\{-\frac{x_i}{\bar{x}_i}\right\}, \;\; i\in [2,N-1].
$

The joint PDF of $ p({\bf x};\bar{{\bf x}})$ is obtained from the product of the above bin densities. Interestingly, maximizing $ p({\bf x};\bar{{\bf x}})$ over $ \bar{{\bf x}}$ will also maximize the PDF of the input data to the DFT. Let $ {\bf u}$ be the length-$ N_t$ input to the DFT, then

$\displaystyle \log p({\bf u}; \bar{{\bf x}}) = - \frac{1}{2} \sum_{i=1}^N \; \left\{ \log 2\pi \bar{x}_i + \frac{\vert U_i\vert^2}{\bar{x}_i} \right\},$ (5.6)

where $ U_i$ are the DFT bins. Therefore, we can replace $ \vert U_i\vert^2$ with $ x_i$:

$\displaystyle \log p({\bf u}; \bar{{\bf x}}) = - \frac{1}{2} \sum_{i=1}^N \; \left\{ \log 2\pi \bar{x}_i + \frac{x_i}{\bar{x}_i} \right\},$ (5.7)

It is mathematically simpler to maximize (5.7) instead of $ p({\bf x};\bar{{\bf x}})$.

Although (5.7) is a function of $ \bar{{\bf x}}$, we need to evaluate it for a particular value of the parameters $ {\bf a}$ and $ \sigma^2$. In Section 5.2.5, we showed how to compute $ \bar{{\bf x}}$ from the AR parameters.

Baggenstoss 2017-05-19