Parameter Estimation

Equation (16.6) is useful for parameter estimation. Let the mean $ {\bf y}$ depend on a set of mean parameters $ \phi_1, \phi_2 \ldots$, and let the spectrum values $ \rho_0, \rho_1 \ldots \rho_{N/2+1}$ depend on spectral parameters $ \theta_1, \theta_2 \ldots$. Let the first derivatives of $ \log p({\bf x})$ be denoted by

$\displaystyle D(\theta) = \frac{\partial}{\partial \theta} \log p({\bf x}; \theta).$

We have

$\displaystyle D(\theta) \stackrel{\mbox{\tiny$\Delta$}}{=}\frac{\partial}{\part...
...sum_{k=0}^{N-1} \; \left( \frac{\partial \rho_k}{\partial \theta}\right) T_1(k)$ (16.7)

$\displaystyle D(\phi) \stackrel{\mbox{\tiny$\Delta$}}{=}\frac{\partial}{\partia...
...\rm Re}\left\{ \left(\frac{\partial Y_k}{\partial \phi}\right) T_2(k) \right\},$ (16.8)

where $ E_k = X_k-Y_k$, and $ T_1(k)$, $ T_2(k)$ are defined implicitly.

Using (16.7), the Fisher's information between spectral parameters $ \theta_1$ and $ \theta_2$ equals

$\displaystyle I(\theta_1,\theta_2) = -{\cal E} \left\{
\frac{\partial^2 }{\par...
...1} \; \left( \frac{\partial \rho_k}{\partial \theta_1}\right)
T_1(k) \right\}
$

Before carrying out the derivative with respect to $ \theta_2$, notice that $ T_1(i)$ is zero in expected value. Therefore, the only terms remaining are associated with the derivative of $ T_1(i)$. Note that

$\displaystyle {\cal E} \left\{
\frac{\partial }{ \partial \theta_2} T_1(k)
\r...
...
= \frac{1}{\rho_k^2} \left( \frac{\partial \rho_k}{\partial \theta_2}\right)
$

Therefore,

$\displaystyle I(\theta_1,\theta_2) = \frac{1}{2} \sum_{k=0}^{N-1} \; \left( \fr...
...ght) \frac{1}{\rho_k^2} \left( \frac{\partial \rho_k}{\partial \theta_2}\right)$ (16.9)

Using (16.8), for ``mean" parametrs,

$\displaystyle I(\phi_1,\phi_2) = -{\cal E} \left\{
\frac{\partial^2 }{\partial...
...
\left( \frac{\partial Y_k}{\partial \phi_1}\right)
T_2(k) \right\} \right\}
$

Before carrying out the derivative with respect to $ \phi_2$, notice that $ T_2(k)$ is zero in expected value. Therefore, the only terms remaining are associated with the derivative of $ T_2(k)$. Note that

$\displaystyle {\cal E} \left\{
\frac{\partial }{ \partial \phi_2} T_2(k)
\rig...
...k}{N\rho_k}
= - \frac{1}{N\rho_k} \frac{\partial \bar{Y}_k}{ \partial \phi_2}
$

Therefore,

$\displaystyle I(\phi_1,\phi_2) = \sum_{k=0}^{N-1} \; {\rm Re} \left\{ \left( \f...
...ac{1}{N\rho_k} \left( \frac{\partial \bar{Y}_k}{\partial \phi_2}\right)\right\}$ (16.10)

Baggenstoss 2017-05-19