Unbaised estimator of $\ e^{- 2 \lambda } $ is t(x) = $ \ (-1)^x $ Prove that
Unbaised estimator of $\ e^{- 2 \lambda  } $ is t(x) =  $ \ (-1)^x $  ,
when x follows poisson distribution with parameter $\lambda $
Sorry to ask the direct question but I need a starting hint to solve the question , actually last time I studied statistical inference was 1 year back , I would be great if anyone could just give some starting steps .

I'll provide a step by step answer. In case you want to try it yourself, please stop at each point and try from there.

*

*Define what you are looking for. What is an unbiased estimator?


An unbiased estimator, $t(x)$, of a quantity $\theta$ (which in this case is $\theta = e^{-2\lambda}$), is such that:
$$  \mathbb{E}t(x) = \theta = e^{-2\lambda} $$



*Plug-in information available to start solving the problem


we know $t(x)=(-1)^{x}$ and $x\sim \text{Poisson}(\lambda)$. Then we want to compute the expected value of this quantity:
$$ \mathbb{E}\left[(-1)^x\right]. $$
Recall that expected value (for a discrete random variable) is obtained by summing all possible values the random variable can take multiplied by their probability.
Recall further that if $x\sim \text{Poisson}(\lambda)$ then $P(x=i)=\frac{\lambda^{i}e^{-\lambda}}{i!}$. Thus, just using definitions:
$$  $$
$$ \mathbb{E}\left[(-1)^x\right] = \sum_{i=0}^{\infty} (-1)^{i}P(x=i) $$



*Solve the above.


Using the Poisson law, we have
$$ \sum_{i=0}^{\infty} (-1)^{i}P(x=i) = \sum_{i=0}^{\infty} (-1)^{i}\frac{\lambda^{i}e^{-\lambda}}{i!}=e^{-\lambda}-\lambda e^{-\lambda}+\lambda^{2} \frac{e^{-\lambda}}{2!}-\lambda^{3} \frac{e^{-\lambda}}{3!}+\ldots$$



*What is the right-hand side term equal to?


Note that we can further simplify by taking common factor $e^{-\lambda}$
$$ e^{-\lambda}\left(1-\lambda+ \frac{\lambda^{2}}{2!}- \frac{\lambda^{3}}{3!}+\ldots\right) $$
Recall Taylor series? You can use them to rewrite $e^{-\lambda}$ around $0$ (which is actually the definition of exponential):
$$ e^{-\lambda}= e^{-0} +\frac{\partial e^{-\lambda}}{\partial\lambda}|_{\lambda=0}\frac{(\lambda-0)}{1!}+\frac{\partial^{2} e^{-\lambda}}{\partial\lambda^{2}}|_{\lambda=0}\frac{(\lambda-0)^{2}}{2!} \ldots$$
solving the derivatives, you see that the above is:
$$ e^{-\lambda}= 1 -\lambda+\frac{\lambda^{2}}{2!}-\frac{\lambda^{3}}{3!} \ldots$$
So, we use this fact to rewrite:
$$ e^{-\lambda}\left(1-\lambda+ \frac{\lambda^{2}}{2!}- \frac{\lambda^{3}}{3!}+\ldots\right)=e^{-\lambda}e^{-\lambda}=e^{-2\lambda}. $$

Remember where we started from?
$$  \mathbb{E}t(x) = e^{-\lambda}\left(1-\lambda+ \frac{\lambda^{2}}{2!}- \frac{\lambda^{3}}{3!}+\ldots\right)=e^{-\lambda}e^{-\lambda}=e^{-2\lambda} $$
Done!
