Parameter Estimation

Equation (17.6) 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

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

We have

$$D(\theta) \stackrel{\mbox{\tiny$\Delta$}}{=}\frac{\partial}{\part... ...sum_{k=0}^{N-1} \; \left( \frac{\partial \rho_k}{\partial \theta}\right) T_1(k)$$ (17.7)

$$D(\phi) \stackrel{\mbox{\tiny$\Delta$}}{=}\frac{\partial}{\partia... ...\rm Re}\left\{ \left(\frac{\partial Y_k}{\partial \phi}\right) T_2(k) \right\},$$ (17.8)

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

Using (17.7), the Fisher's information between spectral parameters $\theta_1$ and $\theta_2$ equals

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

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

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

Therefore,

$$I(\theta_1,\theta_2) = \frac{1}{2} \sum_{k=0}^{N-1} \; \left( \fr... ...ght) \frac{1}{\rho_k^2} \left( \frac{\partial \rho_k}{\partial \theta_2}\right)$$ (17.9)

Using (17.8), for “mean" parametrs,

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

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

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

Therefore,

$$I(\phi_1,\phi_2) = \sum_{k=0}^{N-1} \; {\rm Re} \left\{ \left( \f... ...ac{1}{N\rho_k} \left( \frac{\partial \bar{Y}_k}{\partial \phi_2}\right)\right\}$$ (17.10)