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 $\lambda + \lambda^2$.

  • $\begingroup$ I think the question is specific: finding the UMVUE, which I assume is somehow linked with the concept of finding the expectation that gives $\lambda + \lambda^2$. $\endgroup$ Commented Mar 22, 2023 at 4:49
  • $\begingroup$ 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. $\endgroup$
    – Zhanxiong
    Commented Mar 22, 2023 at 4:52
  • $\begingroup$ 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? $\endgroup$ Commented Mar 22, 2023 at 4:58

1 Answer 1


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.


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