Sample from $\mathrm{Poi}(\lambda)$. Is the following an unbiased estimator of $1/\lambda$?

Question

I simply don't know how to go about answering this question. I've done a good few other questions about point estimation, but I really don't know where I'm going with this one:

Thanks for the help!

EDIT: My question is regarding the unbiasedness section, but any input on the second part would also be great.

Answer 1

Recall that if $X,Y\sim\mathrm{Pois}(\lambda)$ are independent, then $X+Y\sim\mathrm{Pois}(2\lambda)$. So $$\sum_{i=1}^n X_i\sim\mathrm{Pois}(n\lambda).$$ We can use the law of the unconscious statistician to compute the expectation of the given estimator: $$ \begin{align*} \mathbb E\left[\frac n{\sum_{i=1}^n X_i + 1}\right] &= \sum_{k=0}^\infty \frac n{k+1}\mathbb P\left(\sum_{i=1}^n X_i=k\right)\\ &= \sum_{k=0}^\infty \frac n{k+1}\cdot\frac{e^{-n\lambda}(n\lambda)^k}{k!}\\ &= ne^{-n\lambda}(n\lambda)^{-1}\sum_{k=0}^\infty\frac{(n\lambda)^{k+1}}{(k+1)!}\\ &= \frac{e^{-n\lambda}}\lambda\left(e^{n\lambda}-1\right)\\ &= \frac1\lambda(1 - e^{-n\lambda}). \end{align*} $$ Since this expectation is not equal to $\frac1\lambda$, we see that the estimator is biased. However, as @André Nicolas pointed out, it is asymptotically unbiased, as $$\lim_{n\to\infty}\frac1\lambda(1-e^{-n\lambda})=\frac1\lambda.$$

Answer 2

What would go wrong if $1$ had not been added in the denominator is that the probability that the denominator is $0$ would be positive so the expected value would be $\infty\ne 1/\lambda$.

The problem is simply to find the expected value: \begin{align} & \operatorname{E}\left( \frac n {1+\sum_{i=1}^n X_i} \right) = \sum_{x=0}^\infty \frac n {1+x} \Pr\left( \sum_{i=1}^n X_i = x \right) = n\sum_{x=0}^\infty \frac 1 {1+x} \cdot \frac{e^{-n\lambda} (n\lambda)^x}{x!} \\[10pt] = {} & n \sum_{x=0}^\infty \frac{e^{-n\lambda} (n\lambda)^x }{(x+1)!} = \frac n {n\lambda} \sum_{x=0}^\infty \frac{e^{-n\lambda} (n\lambda)^{x+1} }{(x+1)!} = \frac 1 \lambda e^{-n\lambda} \underbrace{\sum_{x=1}^\infty \frac{(n\lambda)^x}{x!}}_{\text{One term is missing.}} \end{align} In the last equality, a substitution was done: the new $x$ was the old $x+1$, and as the old $x$ goes from $0$ to $\infty$, the new $x$ goes from $1$ to $\infty$; thus the sum is missing the first term in a sum that evaluates to $e^{n\lambda}$. The last sum is therefore $e^{n\lambda}-1$.