Finding an unbiased estimator of $e^{-2\lambda}$ for Poisson distribution

Solution 1:

I'm going to assume that $X_1, \dots, X_n$ are independent.

I remember seeing this question (or a very similar one) on my Master's qualifying exam, and being rather angry that I couldn't figure out this seemingly simple question out.

Let $\mathbf{I}$ be the indicator function with $\mathbf{I}(A) = 1$ if the statement $A$ is true, and $0$ otherwise. Since $X_1, X_2$ are independent, it follows that $Y = X_1 + X_2 \sim \text{Poisson}(2\lambda)$.

Hence, the PMF of $Y$ is $$f_{Y}(y) = \dfrac{e^{-2\lambda}(2\lambda)^{y}}{y!}$$ for $y = 0, 1, \dots$.

Observe that $f_{Y}(0) = e^{-2\lambda}$.

Recall that the expected value of the indicator function based on an event is the probability of the event. Hence, $$\mathbf{I}(Y = 0) = \mathbf{I}(X_1 + X_2 = 0)$$ is an unbiased estimator of $e^{-2\lambda}$.

Solution 2:

If $X_i$ and $X_j$ are independent and $i\neq j$ then:$$\mathsf E\mathbf1_{X_i=0}\mathbf1_{X_j=0}=\mathsf E\mathbf1_{X_i=0}\mathsf E\mathbf1_{X_j=0}=e^{-\lambda}e^{-\lambda}=e^{-2\lambda}$$

So a nice unbiased estimator of $e^{-2\lambda}$ is: $$\hat\rho=\binom{n}2^{-1}\sum_{1\leq i<j\leq n}\mathbf1_{X_i=0}\mathbf1_{X_j=0}$$

Nice exercise: find the variance of $\hat\rho$.