Is this an unbiased estimator?

Question

I'm working in the following problem:

Let X be a sample of size = 1 from a Poisson distribution with parameter $\lambda$, and let $h(\lambda) = e^{-3\lambda}$.

a.) Check if $T = (-2)^X$ is an unbiased estimator of $h(\lambda)$.

b.) Give some bad property of the above estimator, which is unrelated to the question of bias.

To start part (a) I wanted to show that $E[T] = h(\lambda)$.

Using $E \left[g(x) \right] = \sum g(x)p(x)$:

$E \left[ (-2)^X \right] = \sum (-2)^x \dfrac{\lambda^xe^{-\lambda}}{x!}$

The part I'm stuck on are the bounds for the summation. Above, it was mentioned that $n = 1$ so I'm inclined to just sum for $x = 1$ and get: $-2\lambda e^{-\lambda}$. But maybe I should do 0 to 1? Something else? Maybe I need a different approach entirely to this problem?

Regardless, (assuming my math is correct which it may not be) it appears that there will be no way to reach a $e^{-3\lambda}$ term because of the $e^{-\lambda}$ from the Poisson equation.

Also, I'm at a loss for part (b) too, so any advice there would be appreciated.

Thanks in advance.

Answer 1

Hint: for the first part use the infinite series representation of $e^{x}$, namely

$$e^{x} = \sum_{i=0}^{\infty} \frac{x^{i}}{i!}$$

and you will see why $(-2)^{X} $ is unbiased for $e^{-3 \lambda}$. The limits of the summation are not related to the sample size but to the range of your random variable, which for the Poisson case is the nonnegative integers. Specifically,

\begin{align} E[T] = \sum_{x=0}^{\infty} (-2)^{x} \frac{e^{-\lambda} \lambda^{x}}{x!} = e^{-\lambda} \sum_{x=0}^\infty \frac{(-2 \lambda)^{x}}{x!} \end{align}

and you should be able to continue from here.

For the second part, ask yourself what values your estimator assumes and what values $h(\lambda)$ assumes. Do you see any problem there?

Hope this helps.