# Cramér–Rao Lower Bound for a Function of a Parameter

• Mar 26th 2010, 06:36 AM
alan4cult
Cramér–Rao Lower Bound for a Function of a Parameter
I'm trying to compute the Cramér–Rao Lower Bound for a function of the parameter $\lambda$ of a Poisson distribution.

The function is $g (\lambda) = e^{-\lambda}$.
Let $\hat{g}$ be an estimate of $g(\lambda)$

I'm using the formula:

$Var (\hat{g}) \ge \dfrac{(g'(\lambda))^2}{nI(\lambda)}$

Now from estimating $\lambda$ previously, I know that $nI(\lambda) = \dfrac {n}{\lambda}$

And
$g'(\lambda) = -e^{-\lambda}$
$(g'(\lambda))^2 = e^{-2\lambda}$

So
$Var (\hat{g}) \ge \dfrac{e^{-2\lambda}}{\dfrac {n}{\lambda}}$

$Var (\hat{g}) \ge \dfrac{\lambda e^{-2\lambda}}{n}$

Is this correct because I tried it another way and got?
$Var (\hat{g}) \ge \dfrac{e^{-\lambda}(1-e^{-\lambda})}{n}$
• Mar 26th 2010, 07:13 AM
CaptainBlack
Quote:

Originally Posted by alan4cult
I'm trying to compute the Cramér–Rao Lower Bound for a function of the parameter $\lambda$ of a Poisson distribution.

The function is $g (\lambda) = e^{-\lambda}$.
Let $\hat{g}$ be an estimate of $g(\lambda)$

I'm using the formula:

$Var (\hat{g}) \ge \dfrac{(g'(\lambda))^2}{nI(\lambda)}$

Now from estimating $\lambda$ previously, I know that $nI(\lambda) = \dfrac {n}{\lambda}$

And
$g'(\lambda) = -e^{-\lambda}$
$(g'(\lambda))^2 = e^{-2\lambda}$

So
$Var (\hat{g}) \ge \dfrac{e^{-2\lambda}}{\dfrac {n}{\lambda}}$

$Var (\hat{g}) \ge \dfrac{\lambda e^{-2\lambda}}{n}$

Is this correct because I tried it another way and got?
$Var (\hat{g}) \ge \dfrac{e^{-\lambda}(1-e^{-\lambda})}{n}$

Well I'm not going to work this out but you first form looks wrong as it has the same dimensions as $\lambda$ (lets say $\text{[T]}^{-1}$) but it should be a pure number. The second form is a pure number which is what we should expect.

CB
• Mar 26th 2010, 08:30 AM
alan4cult

Can you see what's wrong with my end result by looking at the calculations that led me to it? Did I differentiate incorrectly?

The derivative of g with respect to $\lambda$ is $-e^{-2\lambda}$

And then I've just used that in the formula?
Do you think I'm using the formula incorrectly/out of context?

About the dimensions thing you're talking about, what does that mean? I haven't encountered dimensions before?
• Mar 27th 2010, 07:21 AM
alan4cult
Anyone got any more ideas?