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If $X_1,X_2,\ldots,X_n\sim \mathrm{Pois}(\lambda)$, find an unbiased estimator of $e^{-2\lambda}$.

I am actually supposed to find the UMVUE of $e^{-2\lambda}$ but I first have to find its unbiased estimator. I tried using the MLE of $\lambda$ which is $\hat{\lambda}:= \frac{1}{n}\sum_{i=1}^n X_i$ but I'm not sure where to go from there. I know that by invariance property that $e^{-2\hat{\lambda}}$ will be the MLE of $e^{-2\lambda}$ but I'm not sure if it is also unbiased.

Did
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Mr. Bromwich I
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  • The title of the question (repeated in the first line of its body) and the rest of the body of the question are asking different questions. Re the question asked in the title, one should at least mention this unbiased estimator of $e^{-2\lambda}$: $$(-1)^{Y_1}$$ – Did Mar 13 '18 at 15:23
  • Of course, since $E(\hat\lambda)=\lambda$, one knows that $E(e^{-2\hat\lambda})>e^{-2\lambda}$, in particular $E(e^{-2\hat\lambda})\ne e^{-2\lambda}$. – Did Mar 13 '18 at 15:26
  • @Did Why does $E(e^{-2\hat{\lambda}}) > e^{-2\lambda}$? – Mr. Bromwich I Mar 14 '18 at 01:21
  • Because $x\mapsto e^{-2x}$ is strictly convex. – Did Mar 14 '18 at 06:34
  • @Mr.BromwichI In case it's not clear, apply Jensen's Inequality. – Clarinetist Mar 14 '18 at 17:18

2 Answers2

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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 $e^{-2\lambda}$.

Clarinetist
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  • Thanks, this helped a lot. WOuld it have been much harder if they wanted (say) an unbiased estimator of $e^{2\lambda}$? – Mr. Bromwich I Mar 13 '18 at 13:14
  • @Mr.BromwichI Yeah, that would not be fun. Maybe it's more obvious than I think it is, but I'm not sure where I would begin with that one. – Clarinetist Mar 13 '18 at 14:33
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    @Mr.BromwichI See my question at https://math.stackexchange.com/questions/2689415/for-what-values-of-k-in-mathbbz-setminus-0-does-there-exist-an-unbiased. – Clarinetist Mar 13 '18 at 14:54
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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 $\hat\rho$.

drhab
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