# Difficulty in obtaining the uniformly minimum variance unbiased estimator of Poisson distribution

I have this problem I know in a Poisson distribution with the parameter theta, the UMVUE is Sample Mean. But I am not sure how do we obtain the UMVUE of theta square. Any help will be appreciated. Thanks in advance.

I think you can use an approach similar to the answer to this CV question about a UMVUE of $\lambda^3$. Not exactly your problem but if you read that answer and understand the logic you can apply it to your problem. The strategy is to argue that $\bar{x}$ is sufficient and complete for $\lambda$ so that any function of $\lambda$ will have a UMVUE that is a function of the the UMVUE of $\lambda$. This is an application of the Lehman-Scheffe theorem.
$\bar{x}^2 - \frac{\bar{x}}{n}$ is the unique UMVUE of $\lambda^2$. This follows from 2 facts:
$\mathbb{E}(\bar{x}^2) = \lambda^2 + \frac{\lambda}{n}$ and that $\mathbb{E}(\bar{x}) =\lambda$. Then a bit of arrangement to arrive at the function of $\bar{x}$ so that you have an unbiased estimator should give you the result.
• The answer cannot be (D) from the question because that is not an unbiased estimator, the second "U" in UMVUE is for unbiased estimator. My answer gives the value of (D) to be $\lambda^2 + \frac{\lambda}{n}$ which is not $\lambda^2$. – Lucas Roberts Dec 28 '17 at 16:48
• To apply Lehmann-Scheffe you need to know: (1) the estimator of the parameter (here $\lambda$) is complete and sufficient. Then Lehmann Scheffe tells you that you can find any function of the parameter by finding the unbiased estimator which is a function of the complete sufficient statistic. Unbiasedness is a linear equation so this makes it fairly easy to determine what the UMVUE will be but this only works if you have a complete + sufficient statistic(s). – Lucas Roberts Dec 28 '17 at 16:51