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nonequilibrium
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Homework Statement
Find an efficient estimator for [itex]q(\lambda) := e^{-\lambda}[/itex] in a Poisson model.
(note: the efficient estimator is the one who reaches the Cramér-Rao lower bound)
Homework Equations
The Cramér-Rao lower bound:
Let T be an unbiased estimator for q (as defined above), then [itex]Var(T) \geq \frac{q'(\lambda)^2}{I_n(\lambda)}[/itex], with [itex]I_n(\lambda)[/itex] the Fisher information number for the parameter [itex]\lambda[/itex] in the Poisson model, from the sample [itex]\vec X = (X_1,\cdots,X_n)[/itex].
Concretely, in this case [itex]Var(T) \geq \frac{ \lambda e^{-2\lambda}}{n} [/itex].
The Attempt at a Solution
So first things first: I want an unbiased estimator for q. As a guess, I took the form [itex]T(\vec X) = \alpha^{\sum X_i}[/itex]. Calculating the mean shows that one gets an unbiased estimator if one defines [itex]T(\vec X) := \left( \frac{n-1}{n} \right)^{\sum X_i}[/itex].
Now my reasoning was as follows: this unbiased esimator only depends on [itex]\sum X_i[/itex], 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 [itex]Var(T) = E(T^2) - E(T)^2 = e^{-2 \lambda} \left( e^{\lambda / n} - 1 \right)[/itex]. 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!
ADDENDUM: While writing the red text above, I realized it also depends on n explicitly. Is this important? I don't think so.
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