Hello everyone,

This is regarding a research problem I am working on. Trying to see if I am doing any mistake in the calculation of the CRLB.

Consider the case of a signalsin additive gamma noisewwith N independent elements w_{n}

$\displaystyle \utilde{\mathbf{x}} = A\mathbf{s} + \utilde{\mathbf{w}} $

I am trying to find the CRLB for $\displaystyle \hat{A}$. I am skipping a few trivial steps...

The log-likelihood function forxis - $\displaystyle L(\mathbf{x};A) = \displaystyle\sum_{n=0}^{N-1}\ln\left[\frac{1}{\Gamma(k_n)\theta_n^{k_n}}\right] (k_n-1) \ln(x_n-As_n)-\frac{(x_n-As_n)}{\theta_n}\right] $

So now, $\displaystyle E\left[ \frac{\partial^2 L(\mathbf{x};A)}{\partial A^2}\right] &= E\left[ \displaystyle\sum_{n=0}^{N-1} -(k_n-1) \left(\dfrac{s_n}{x_n-As_n}\right)^2 \right] = \displaystyle\sum_{n=0}^{N-1} -(k_n-1)s_n^2\, E\left[\dfrac{1}{ \left(x_n-As_n\right)^2} \right] $

Let $\displaystyle x_n - As_n = y_n$. So,

$\displaystyle E\left[\dfrac{1}{ y_n^2} \right] &= \int_0^\infty \dfrac{1}{y_n^2} \frac{1}{\Gamma(k_n)\theta_n^{k_n}} y_n^{k_n-1} e^{-\dfrac{y_n}{\theta_n}} dy\notag $

$\displaystyle = \frac{1}{\Gamma(k_n)\theta_n^{k_n}} \int_0^\infty y_n^{k_n-3} e^{-\dfrac{y_n}{\theta_n}}dy_n \quad \mbox{, Substituting $y_n = \theta z$}\notag$

$\displaystyle = \frac{1}{\Gamma(k_n)\theta_n^{k_n}} \int_0^\infty (\theta z)^{k_n-3} e^{-z} \theta dz $

$\displaystyle = \frac{\theta_n^{k_n-2}}{\Gamma(k_n)\theta_n^{k_n}} \int_0^\infty z^{k_n-3} e^{-z} dz$

$\displaystyle = \frac{\Gamma(k_n-2)}{\Gamma(k_n)\theta_n^{2}}$

$\displaystyle = \frac{1}{(k_n-1)(k_n-2)\theta_n^2}$

The Fisher Information is,

$\displaystyle E\left[ \frac{\partial^2 L(\mathbf{x};A)}{\partial A^2}\right] = \displaystyle\sum_{n=0}^{N-1} -\frac{s_n^2} {(k_n-2)\theta_n^2}$

Hence CRLB is,

$\displaystyle var(\hat{A}) \ge \dfrac{1}{-E\left[ \frac{\partial^2 l_A(\mathbf{x})}{\partial A^2}\right]} = \dfrac{1}{\displaystyle\sum_{n=0}^{N-1} \frac{s_n^2} {(k_n-2)\theta_n^2}} $

Does this look right? I am a little unsure about the substitution $\displaystyle y_n = x_n - As_n $.

TIA for the help.