# [statistics] find the efficient estimator: where is my reasoning wrong?

1. Jan 27, 2012

### nonequilibrium

1. The problem statement, all variables and given/known data
Find an efficient estimator for $q(\lambda) := e^{-\lambda}$ in a Poisson model.
(note: the efficient estimator is the one who reaches the Cramér-Rao lower bound)

2. Relevant equations
The Cramér-Rao lower bound:
Let T be an unbiased estimator for q (as defined above), then $Var(T) \geq \frac{q'(\lambda)^2}{I_n(\lambda)}$, with $I_n(\lambda)$ the Fisher information number for the parameter $\lambda$ in the Poisson model, from the sample $\vec X = (X_1,\cdots,X_n)$.

Concretely, in this case $Var(T) \geq \frac{ \lambda e^{-2\lambda}}{n}$.

3. The attempt at a solution
So first things first: I want an unbiased estimator for q. As a guess, I took the form $T(\vec X) = \alpha^{\sum X_i}$. Calculating the mean shows that one gets an unbiased estimator if one defines $T(\vec X) := \left( \frac{n-1}{n} \right)^{\sum X_i}$.

Now my reasoning was as follows: this unbiased esimator only depends on $\sum X_i$, i.e. a complete and sufficient statistic (follows, for example, from the Poisson distribution belonging to the exponential class). Hence by Lehman-Scheffé, T as defined above is an UMVUE. Now, if the Cramér-Rao bound is reached (as the exercise suggests), then it must surely be the UMVUE who reaches it. Hence I thought I had the correct estimator.

However, when I calculate the variance, I get that $Var(T) = E(T^2) - E(T)^2 = e^{-2 \lambda} \left( e^{\lambda / n} - 1 \right)$. This is not equal to the lower bound (it is, however, for very large n, but that's not enough, but it at least suggests I didn't make a calculation error).

Can anybody help? Much appreciated!