Let $X$ be a $\text{Poisson} (\theta)$ random variable. Unbiased estimator

Question

Let $X$ be a $\text{Poisson} (\theta)$ random variable. Show that $(-1)^X$ is an unbiased estimator for $e^{-2 \theta}$ This is a fairly bad estimator for a number of reasons - so this exercise helps show why unbiasedness is not the most important criterion for an estimator.

Can someone please help me??? Pleaseee

Answer 1

Let us derive this in detail, to address your confusion.

By definition, an estimator $\hat{\mu}$ is an unbiased estimator for some quantity $\mu$ if $\mathbb{E}[\hat{\mu}] = \mu$.

So what we need to show is that $\mathbb{E}[(-1)^X] = e^{-2\theta}$.

Writing the definition of expectation, $$ \mathbb{E}[(-1)^X] = \sum_{n=0}^\infty (-1)^n \mathbb{P}\{X=n\} = \sum_{n=0}^\infty (-1)^n \frac{e^{-\theta}\theta^n}{n!} = e^{-\theta}\sum_{n=0}^\infty \frac{(-\theta)^n}{n!} $$ where we used the fact that $X\sim\operatorname{Poisson}(\theta)$ for the second equality.

To conclude, we recall the definition of exponential: for any $x\in\mathbb{R}$, $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$. Thus, $$ \mathbb{E}[(-1)^X] = e^{-\theta}\sum_{n=0}^\infty \frac{(-\theta)^n}{n!} = e^{-\theta}e^{-\theta} = e^{-2\theta} $$ concluding the proof.