1
$\begingroup$

My attempt: We know, since $X\sim Pois(\theta)$ that $\mathbb{P}_{\theta}(X=x)=e^{-\theta}\theta^{x}/x!$. A given tip is that we must recall that $e^{x}=\sum^{\infty}_{k=0}\frac{x^{k}}{k!}$. I know that once we find an unbiased estimator that is a function of our complete sufficient statistic $X$, that this estimator must then automatically be UMVU. However, I'm not sure how to approach this question by even finding an expression for an unbiased estimator.

Question: How to approach/solve this exercise?

Thanks!

$\endgroup$
1
$\begingroup$

We need to find some function $g(k)$ s.t. $\mathbb E[g(X)]=e^{-3\theta}$ for all $\theta>0$. Write this expectation: $$ \mathbb E[g(X)]=\sum_{k=0}^\infty g(k)\dfrac{\theta^k}{k!}e^{-\theta}=e^{-3\theta}. $$ Multiply both parts by $e^\theta$ and get $$ \sum_{k=0}^\infty g(k)\dfrac{\theta^k}{k!}=e^{-2\theta} = \sum_{k=0}^\infty \frac{(-2)^k\theta^k}{k!}. $$ Since these sums are equal for each $\theta>0$, we get $g(k)=(-2)^k$ and $$g(X)=(-2)^X$$ is unique unbiased estimator for $e^{-3\theta}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.