CRLB of Amplitude Estimate in Gamma Noise

**ronbeck** · November 11th 2012, 03:26 PM

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 signal s in additive gamma noise w with N independent elements w_n
$\utilde{\mathbf{x}} = A\mathbf{s} + \utilde{\mathbf{w}}$

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

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

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

Let $x_n - As_n = y_n$ . So,

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

$= \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$

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

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

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

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

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

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

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

**chiro** · November 11th 2012, 11:44 PM

Hey ronbeck.

I'm wondering if you could outline the calculation for the 2nd derivative of the log-likelihood: I'm sure it's correct but I'm wondering where a lot of the simplifications come from explicitly.

If xn - Asn ~ Gamma (which it looks to be from definition), then you should have no problem finding E[1/w^2].

**ronbeck** · November 12th 2012, 08:11 AM

Hi Chiro,

I missed the '+' sign for $L(\mathbf{x};A)$ . Anyway, here is the calculation:

The Gamma distribution is defined as: (Assuming independent noise elements $w_n$ in $\mathbf{w}$ )

$f_{\utilde{\mathbf{x}}}(\mathbf{x}) &= \displaystyle\prod_{n=0}^{N-1}\frac{1}{\Gamma(k_n)\theta_n^{k_n}} (x_n-As_n)^{k_n-1}e^{-\frac{(x_n-As_n)}{\theta_n}} \quad \mbox{for $x_n \ge As_n$}$

So the log-likelihood function is the 'log' of the above:

$$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}}$

$\implies \frac{\partial L(\mathbf{x};A)}{\partial A} = \displaystyle\sum_{n=0}^{N-1} 0 + (k_n-1) \left[\dfrac{\partial\ln(x_n-As_n) }{\partial A}\right] - \dfrac{1}{\theta_n}\dfrac{\partial (x_n-As_n)} {\partial A}$

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

$\implies \frac{\partial^2 L(\mathbf{x};A)}{\partial A^2} = \displaystyle\sum_{n=0}^{N-1} -(k_n-1) \dfrac{\partial}{\partial A}\left(\dfrac{s_n}{x_n-As_n}\right) + 0$

$\implies \frac{\partial^2 L(\mathbf{x};A)}{\partial A^2} = \displaystyle\sum_{n=0}^{N-1} -(k_n-1) \left(\dfrac{s_n}{x_n-As_n}\right)^2$

I just wanted to see if I am doing the CRLB calculations right.

Thanks

**chiro** · November 12th 2012, 05:39 PM

Yeah the CRLB calculations look right to me, assuming that particular distribution (Gamma distribution) for the residual noise elements.

One way to check these kinds of things is to use simulation if you ever need some kind confirmation (but don't have anyone at your disposal to check your work), but in terms of the mathematics it looks good.

**ronbeck** · November 13th 2012, 03:24 AM

Thanks Chiro. I will try simulations. But an efficient estimator is needed to check the CRLB? It's a catch-22 kind of situation unless there is some simulation method that can check the CRLB without really estimating A.

Another concern is, if $k_n < 2 \quad \forall\, n$ then the CRLB will be negative? This doesn't make sense.

Thanks

**chiro** · November 13th 2012, 01:53 PM

I was just thinking that there may be a problem:

If sn is dependent on w and x then you can't factorize the sn outside of the expectation sign and do what you did.

You are essentially finding E[1/W^2] but the sn dependent on w which means that if you want to take expectations, you will need include the sn term and transform it if taking expectations with respect to some random variable (like W ~ Gamma). So instead of finding the normal mean you are finding the mean of a function of 1/W^2 but since Gamma is defined over the whole real positive line it means that you are going to get singularity blow-ups for various values of kn which means some won't make sense to be used.

So the PDF has to "vanish" quick enough to go to zero when you go to zero for this random variable to have a mean that makes any sense numerically.

Thread: CRLB of Amplitude Estimate in Gamma Noise

LinkBack

Thread Tools

Display

CRLB of Amplitude Estimate in Gamma Noise

Re: CRLB of Amplitude Estimate in Gamma Noise

Re: CRLB of Amplitude Estimate in Gamma Noise

Re: CRLB of Amplitude Estimate in Gamma Noise

Re: CRLB of Amplitude Estimate in Gamma Noise

Re: CRLB of Amplitude Estimate in Gamma Noise

Similar Math Help Forum Discussions

CRLB mistake (Gamma)

CRLB for a NB distribution

Difference between Gaussian noise and white Gaussian noise?

Find a lower estimate and an upper estimate

[SOLVED] complex numbers amplitude and phase - complex amplitude?

Search Tags