UMVUE of $E[X^2]$ where $X_i$ is Poisson $(\lambda)$

Question

I am working on a problem to find the UMVUE of a function.

Let $X_1, X_2, ..., X_n$ be i.i.d. Poison$(\lambda)$.

Find the UMVUE of E[$X^2$].

So far I understand that $$E[X^2]=Var[X]+E[X]^2=\lambda+\lambda^2$$ and thus $$\begin{align} E[\hat{\lambda}_{MLE}^2] & = \bar{X}(1+\bar{X})\\ \end{align}$$ Now the goal is to adjust this so that it is unbiased, so I did $$\begin{align} E[\bar{X}(1+\bar{X})] &= E[\bar{X}]+Var[\bar{X}]+E[\bar{X}]^2\\ &= \lambda +\lambda/n +\lambda^2 \end{align} $$

And then I could not proceed from there.

The notes that I received mentions that we do something from here but it was not explicit and I am not sure how to make this unbiased.

Thank you in advance.

Answer 1

Suppose $X_1, \dots, X_n \overset{\text{iid}}{\sim}\text{Poisson}(\lambda)$.

Then $$L(\lambda) = \prod_{i=1}^{n}\dfrac{e^{-\lambda}\lambda^{x_i}}{x_i!} = \dfrac{e^{-n\lambda}\lambda^{\sum_{i=1}^{n}x_i}}{\prod_{i=1}^{n}x_i!} = \exp\left[\sum_{i=1}^{n}x_i \cdot \log(\lambda)-n\lambda \right]\cdot \dfrac{1}{\prod_{i=1}^{n}x_i!}\text{.}$$ Since the parameter space for $\lambda$, given by $\Theta = (0, \infty)$, contains an open set (we may choose $(0, 1)$ for example), and the Poisson distribution is of the exponential family, it follows that the statistic $T(\mathbf{X})=\sum_{i=1}^{n}X_i$ is sufficient and complete for $\theta$. (See https://stats.stackexchange.com/questions/322381/what-is-exponential-family-criterion-to-test-the-sufficiency-and-completeness-of, for example.)

Now $$\mathbb{E}[T(\mathbf{X})]=n\lambda$$ so that $$g_1(T(\mathbf{X})) = \dfrac{1}{n}T(\mathbf{X})$$ is an unbiased estimator of $\lambda$. Furthermore, $$\mathbb{E}[(T(\mathbf{X})^2]=\text{Var}(T(\mathbf{X}))+(\mathbb{E}[T(\mathbf{X})])^2 = n\lambda+n^2\lambda^2\text{.}$$ It follows that $$\mathbb{E}\left[(T(\mathbf{X}))^2-T(\mathbf{X}) \right] = n^2\lambda^2$$ hence $$g_2(T(\mathbf{X}))=\dfrac{1}{n^2}\left\{[T(\mathbf{X})]^2-T(\mathbf{X}) \right\}$$ is an unbiased estimator of $\lambda^2$. Hence, $$g(T(\mathbf{X})) = g_1(T(\mathbf{X})) + g_2(T(\mathbf{X}))=\dfrac{1}{n}T(\mathbf{X})+\dfrac{1}{n^2}\left\{[T(\mathbf{X})]^2-T(\mathbf{X}) \right\}$$ is an unbiased estimator of $\lambda + \lambda^2 = \mathbb{E}[X^2]$.

By the Lehmann–Scheffé theorem, it follows that $g(T(\mathbf{X}))$ is the UMVUE of $\lambda+\lambda^2$.