# Finding UMVU stimator of $\frac{\lambda^3}{3!}e^{-\lambda}$ function for Poisson distribution $(Poiss(\lambda))$ sample

Let's say that we want to find Unbiased Minimum Variance estimator of $\lambda^2$ for a sample $X=(X_1,\dots,X_n)$ from Poisson distribution. We can consider $\overline{X}^2$ and notice that

$$\mathbb{E}(\overline{X}^2) = Var(\overline{X}) + \left(\mathbb{E}(\overline{X})\right)^2 = \frac{1}{n^2} \cdot n \cdot Var(X_1) + \lambda^2 = \lambda^2 + \frac{\lambda}{n} = \lambda^2 + \mathbb{E}\left(\frac{1}{n}\overline{X} \right)$$

Which gives us that $\overline{X}^2 + \frac{1}{n}\overline{X}$ is what we are searching for: because it's a function of $\overline{X}$, which for exponential families is sufficient, and in this case also complete, statistic, then from Lehmann-Scheffe theorem we conclude that it's UMVU estimator.

Now, let's consider a function like: $$g(\lambda) = \frac{\lambda^3}{3!}e^{-\lambda}.$$

Which is actually a probability $P(X_i=3)$. What can we do with this, to find an UMVU estimator of this one? I was wondering to start, just like above, from considering

$$T(X)= \frac{\overline{X}^3}{3!}e^{-\overline{X}},$$

and eventually adjust it with some other components, but trying to find expected value of this almost made me cry. I was thinking about transforming it somehow and using Basu theorem make things easier, but not sure how.

Any advice how to do this? Or maybe I can find the UMVUE easier here?

Start from $$g_n = \mathcal{I}\{X_1 = 3\},$$ as an unbiased estimator. Then, using Rao-Blackwell, compute $$g_n^{RB} = \mathbb{E}[g_n|\sum _{i=1}^n X_i =t].$$ Note that $g_n^{RB}$ is an unbiased estimator and function of the complete minimal sufficient statistic $\sum_{i=1}^n X_i$. Thus, by Lehmann-Scheffe, it is a UMVUE.
Namely, \begin{align} g_n^{RB} &= \mathbb{E}[g_n|\sum _{i=1}^n X_i =t]\\ &= \frac{\mathbb{P}(X_1 = 3) \mathbb{P}( \sum _{i=2}^n X_i =t - 3)}{\mathbb{P}( \sum _{i=1}^n X_i =t)}\\ & =\frac{e^{-\lambda}\lambda^3/3! \times e^{-\lambda (n-1)}\lambda^{t-3} (n-1)^{t-3} / (t-3)!}{e^{-\lambda n}\lambda^t n^t/t!} \\ & = \left( \frac{n-1}{n} \right)^{t} \binom{t}{3}(n-1)^{-3} \\ & = \left( 1 - \frac{1}{n} \right)^{n\bar{x}_n} \binom{n\bar{x}_n}{3}(n-1)^{-3}. \end{align} Now, for validation, you can use the continuous mapping theorem and the WLLN. Note that $\bar{X}_n \xrightarrow{p} \lambda$, same is true for $\frac{1}{n-1}\sum X_i \xrightarrow{p} \lambda$, and note that $(1-1/n)^n \xrightarrow{} e^{-1}$. Hence, combining it all, the estimator converges to $g(\lambda)$.
• What does $\mathcal{I}$ in $g_n$ stands for? Fisher Information? – Kusavil Feb 6 '18 at 21:13
• $\mathcal{I}$ - indicator function, $g_n$ - an estimator of $g(\lambda)$ – V. Vancak Feb 6 '18 at 21:15
• So we have: $\mathbb{P}(X_1 = 3)=\frac{\lambda^3}{3!}e^{-\lambda}$, and because for $X\sim Poiss(\lambda)$ we have $\sum_{i=1}^n \sim Poiss(n \lambda)$, thus we have $\mathbb{P}( \sum _{i=1}^n X_i =t) = \frac{(n \lambda)^t}{t!}e^{-n \lambda}$, then similarly $\mathbb{P}(\sum_{i=1}^n X_i =t-3)= \frac{(n \lambda)^{t-3} }{(t-3)!} e^{-n \lambda}$, which gives me $\frac{\mathbb{P}(X_1 = 3) \mathbb{P}( \sum _{i=1}^n X_i =t - 3)}{\mathbb{P}( \sum _{i=1}^n X_i =t)} = \binom{t}{3}e^{-\lambda}(1-\frac{1}{n})^3(n-1)^{-3}$. How to I transform this into your form? – Kusavil Feb 6 '18 at 21:46
• Also, what does $t$ stands for? Where do we know $t$ from? – Kusavil Feb 6 '18 at 21:47