Let's assume $x\left(n\right)\:=\:b^n+w\left(n\right)$
where $w\left(n\right)$ has a normal distribution of $w\left(n\right)\in N\left(0,\sigma ^2\right)$
I need to estimate $b$ by finding the CRLB (Cramer-Rao Lower Bound) as well as the variance.
Here's my work: $p\left(x;\theta \right)\:=\:\frac{1}{\left(2\pi \:\sigma \:^2\right)^{\frac{N}{2}}}e^{-\frac{1}{2\sigma ^2}\sum _{n=0}^{N-1}\left(x\left(n\right)-b^n\right)^2}$
$\frac{d}{db}\left(ln\left(p\left(x;\theta \:\right)\:\right)\right)=\:\frac{-ln\left(b\right)}{\sigma \:^2}\sum _{n=0}^{N-1}\left(x\left(n\right)-b^n\right)$
and
$\frac{d^2}{db^2}\left(ln\left(p\left(x;\theta \:\:\right)\:\right)\right)=\:\frac{\left(\frac{1}{r}+ln^2\left(b\right)\right)}{\sigma \:^2\left(1-b\right)}-\sum \:_{n=0}^{N-1}\frac{\left(x\left(n\right)\right)}{\sigma \:^2r}\:=\:I\left(\theta \right)$
where the $Var(b) = \frac{1}{I\left(\theta \:\right)}\:$
Did I do this write? am confused on how to approach this problem. This is a confusing subject and would like some help to clarify this please.
$f(x(1),...,x(N)|b) = a \times exp\left(\frac{-\sum_n (x(n)-b^n)^2}{2\sigma^2}\right)$.
$\frac{d ln(f(x(1),...,x(N)|b))}{db} = \frac{d\frac{(-\sum_n (x(n)-b^n)^2)}{2\sigma^2}}{b}$ $= \frac{\sum_n nb^{n-1}(x(n)-b^n)}{\sigma^2}$
$\frac{d^2 ln(f(x(1),...,x(N)|b))}{db^2} = \frac{d\frac{\sum_n n(b^{n-1}x(n)-b^{2n-1})}{\sigma^2}}{db} = \frac{\sum_n n((n-1)b^{n-2}x(n)-(2n-1)b^{2n-2})}{\sigma^2} $
$-E(\frac{d^2 ln(f(x(1),...,x(N)|b))}{db^2}) = -\frac{\sum_n n((n-1)b^{n-2}b^n-(2n-1)b^{2n-2})}{\sigma^2} =\frac{\sum_n n^2 b^{2n-2}}{\sigma^2} $.
So the Cramer-Rao bound is : $Var(b_{est}) \geq \frac{\sigma^2}{\sum_n n^2 b^{2n-2}}$