# Poisson: finding UMVUE for $\lambda + \lambda^2$ [closed]

Let $$X_1,..., X_n$$ be iid sample from the Poisson distribution with parameter $$\lambda$$. Find the UMVUE of $$\lambda + \lambda^2$$.

I know $$T := \sum\limits_{i=1}^n X_i$$ is complete and sufficient for $$\lambda$$. Also, $$T/n$$ is the MLE of $$\lambda$$. By the invariance principle of the MLE, $$T/n + T^2 / n^2$$ is the MLE for $$\lambda + \lambda^2$$.

I tried following the approach for computing UMVUE of $$\lambda^3$$ from an earlier post, but could not compute the expectation that gives $$\lambda + \lambda^2$$.

• I think the question is specific: finding the UMVUE, which I assume is somehow linked with the concept of finding the expectation that gives $\lambda + \lambda^2$. Commented Mar 22, 2023 at 4:49
• OK, from the 2nd paragragh, you almost reached the answer. Only a minor adjustment is needed to get the unbiased estimator from what you wrote. Commented Mar 22, 2023 at 4:52
• In this case, $E[X_i]^2$ is the only expectation I see that yields $\lambda + \lambda^2$. Since $X_i$ is an unbiased estimator, ${X_i}^2$ is unbiased as well I assume. But when computing the bias, what should be the reference parameter: $\lambda$ or $\lambda + \lambda^2$? I think I have some gaps in logic and sequence here, so could you post it as an answer? Commented Mar 22, 2023 at 4:58

There is no need to bring MLE into the discussion, all you need is $$T \sim \text{Poisson}(n\lambda)$$ hence \begin{align} E(T) = n\lambda, \; E(T^2) = n\lambda + n^2\lambda^2. \end{align} From this it is easy to see $$n^{-2}T^2 + (n^{-1} - n^{-2})T$$ is an unbiased estimator of $$\lambda + \lambda^2$$. Now use Lehmann-Scheffe theorem to conclude.