Finding UMVUE of $\theta e^{-\theta}$ where $X_i\sim\text{Pois}(\theta)$

Question

Suppose $X_1, X_2, . . . , X_n$ are i.i.d Poisson ($\theta$) random variables, where $\theta\in(0,\infty)$. Give the UMVUE of $\theta e^{-\theta}$

I found a similar problem here.

I have that the Poisson distribution is an exponential family where

$$f(x\mid\theta)=\frac{\theta^x e^{-\theta}}{x!}=\left(\frac{I_{(0,1,...,)}(x)}{x!}\right)e^{-\theta}\text{exp}\left(x\cdot\text{log}(\theta)\right)$$

where $w(\theta)=\text{log}(\theta)$ and $t(x)=x$. Since

$$\{{w(\theta):\theta\in\Theta\}}=\{{\text{log}(\theta):\theta\in(0,\infty)\}}=(-\infty,\infty)$$

contains an open set in $\mathbb{R}$

$$T(\vec{X})=\sum_{i=1}^n t(X_i)=\sum_{i=1}^n X_i$$

is a complete statistic (Complete Statistics in the Exponential Family Theorem) and is also sufficient (Factorization Theorem). Hence $\bar{X}=\frac{T}{n}$ is sufficient for $\theta$, and hence, for $\theta e^{-\theta}$.

I tried to Rao-Blackwellize the unbiased estimator $\bar{X}$. For all possible values of $T$ we have that since $\theta e^{-\theta}=\mathsf P(X_i=1)$ then

$$\begin{align*} \mathsf P(X_1=1\mid\bar{X}=t) &=\frac{\mathsf P(X_1=1,\sum_{i=2}^n {X_i} =nt-1)}{\mathsf P(\bar{X} = t)}\\\\ &=\frac{\frac{e^{-\theta}\theta}{1}\cdot\frac{e^{-(n-1)\theta}((n-1)\theta)^{nt-1}}{nt-1!}}{\frac{e^{-n\theta}(n\theta)^{nt}}{nt!}}\\\\ &=t\cdot\left(1-\frac{1}{n}\right)^{nt-1} \end{align*}$$

Since $\mathsf E(X_1)=\theta$, then $X_1$ is an unbiased estimator of $\theta$, and so it follows from the Lehmann–Scheffé theorem that $t\cdot\left(1-\frac{1}{n}\right)^{nt-1}$ is the UMVUE.

Is this a valid solution? Were my justifications correct and sufficient?

Answer 1

The Poisson distribution is a one-parameter exponential family distribution, with natural sufficient statistic given by the sample total $T(\mathbf{x}) = \sum_{i=1}^n x_i$. The canonical form is:

$$p(\mathbf{x}|\theta) = \exp \Big( \ln (\theta) T(\mathbf{x}) - n\theta \Big) \cdot h(\mathbf{x}) \quad \quad \quad h(\mathbf{x}) = \coprod_{i=1}^n x_i! $$

From this form it is easy to establish that $T$ is a complete sufficient statistic for the parameter $\theta$. So the Lehmann–Scheffé theorem means that for any $g(\theta)$ there is only one unbiased estimator of this quantity that is a function of $T$, and this is the is UMVUE of $g(\theta)$. One way to find this estimator (the method you are using) is via the Rao-Blackwell theorem --- start with an arbitrary unbiased estimator of $g(\theta)$ and then condition on the complete sufficient statistic to get the unique unbiased estimator that is a function of $T$.

Using Rao-Blackwell to find the UMVUE: In your case you want to find the UMVUE of:

$$g(\theta) \equiv \theta \exp (-\theta).$$

Using the initial estimator $\hat{g}_*(\mathbf{X}) \equiv \mathbb{I}(X_1=1)$ you can confirm that,

$$\mathbb{E}(\hat{g}_*(\mathbf{X})) = \mathbb{E}(\mathbb{I}(X_1=1)) = \mathbb{P}(X_1=1) = \theta \exp(-\theta) = g(\theta),$$

so this is indeed an unbiased estimator. Hence, the unique UMVUE obtained from the Rao-Blackwell technique is:

$$\begin{equation} \begin{aligned} \hat{g}(\mathbf{X}) &\equiv \mathbb{E}(\mathbb{I}(X_1=1) | T(\mathbf{X}) = t) \\[6pt] &= \mathbb{P}(X_1=1 | T(\mathbf{X}) = t) \\[6pt] &= \mathbb{P} \Big( X_1=1 \Big| \sum_{i=1}^n X_i = t \Big) \\[6pt] &= \frac{\mathbb{P} \Big( X_1=1 \Big) \mathbb{P} \Big( \sum_{i=2}^n X_i = t-1 \Big)}{\mathbb{P} \Big( \sum_{i=1}^n X_i = t \Big)} \\[6pt] &= \frac{\text{Pois}(1| \theta) \cdot \text{Pois}(t-1| (n-1)\theta)}{\text{Pois}(t| n\theta)} \\[6pt] &= \frac{t!}{(t-1)!} \cdot \frac{ \theta \exp(-\theta) \cdot ((n-1) \theta)^{t-1} \exp(-(n-1)\theta)}{(n \theta)^t \exp(-n\theta)} \\[6pt] &= t \cdot \frac{ (n-1)^{t-1}}{n^t} \\[6pt] &= \frac{t}{n} \Big( 1- \frac{1}{n} \Big)^{t-1} \\[6pt] \end{aligned} \end{equation}$$

Your answer has a slight error where you have conflated the sample mean and the sample total, but most of your working is correct. As $n \rightarrow \infty$ we have $(1-\tfrac{1}{n})^n \rightarrow \exp(-1)$ and $t/n \rightarrow \theta$, so taking these asymptotic results together we can also confirm consistency of the estimator:

$$\hat{g}(\mathbf{X}) = \frac{t}{n} \Big[ \Big( 1- \frac{1}{n} \Big)^n \Big] ^{\frac{t}{n} - \frac{1}{n}} \rightarrow \theta [ \exp (-1) ]^\theta = \theta \exp (-\theta) = g(\theta).$$

This latter demonstration is heuristic, but it gives a nice check on the working. It is interesting here that you get an estimator that is a finite approximation to the exponential function of interest.

Answer 2

Here is a simulation in R that I did using a the average of $n = 20$ observations, where $\lambda = 5.$ The parameter par is $P(X = 1) = \lambda e^{-\lambda}.$ The estimate par.fcn, which tries to estimate $P(X = 1)$ merely as a function of the average, is biased. My version of your UMVUE for $P(X=1)$ using a function of average a (instead of total) seems to work OK.

set.seed(2018);  m = 10^5; n = 20; lam=5; par=dpois(1, lam)
x = rpois(m*n, lam);  MAT=matrix(x, nrow=m)  # each row a sample of size 20
a = rowMeans(MAT)

lam.umvue = a;  lam;  mean(lam.umvue);  sd(lam.umvue)
[1] 5             # exact lambda
[1] 5.000788      # mean est of lambda
[1] 0.4989791     # aprx SD of est

par.fcn = exp(-lam.umvue)*lam.umvue;  par;  mean(par.fcn);  sd(par.fcn)
[1] 0.03368973    # exact P(X=1)
[1] 0.03620296    # slightly biased
[1] 0.01444379
sqrt(mean((par.fcn - par)^2))
[1] 0.01466074    # aprx root mean square error (rmse) of par.fun

par.umvue = a*(1-1/n)^(n*a - 1);  par;  mean(par.umvue);   sd(par.umvue)
[1] 0.03368973    # exact P(X=1)
[1] 0.03365454    # mean est of P(X=1), seems unbiased
[1] 0.01388531
sqrt(mean((par.umvue - par)^2))
[1] 0.01388528    # aprx rmse of umvue of P(X=1);  smaller than rmse of par.fun