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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

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    Try calculating $\operatorname E\left[(-1)^X\right]$. – 2017-02-21
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    Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – 2017-02-21
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    @TheoreticalEconomist, I don't know how to compute nor show that it is an unbiased estimator . I really need lots of help. – 2017-02-21
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    @Kat Do you know what it means for an estimator to be unbiased? If you do, then you would know why I made my initial suggestion. If you don't, a good place to start trying to answer that question would be looking up what an unbiased estimator is. (Of course, we can just tell you what it is, but I think you'll learn better if you find the meaning yourself.) – 2017-02-21
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    @Clement C : Thank you soo much. – 2017-02-25

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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.