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

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?

• It looks right to me - and the form is also approximately of the form $T(x)e^{T(X)-1}$ since $(1-1/n)^n \to e$, which is what we would intuitively expect. Oct 25, 2018 at 0:58
• This answer seems to confirm your working Oct 25, 2018 at 1:00
• It's been a long day and I should really be watching mindless TV, so maybe I'm missing something, but I think you may have an error. If $\lambda = 5$ and $n = 20,$ then I get an improbable result. Please check. Oct 25, 2018 at 2:13

## 2 Answers

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{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}

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.

• Thanks for clarification. Should I delete simulation? Oct 25, 2018 at 2:59
• I thinks your simulation is pretty cool (+1) - a nice way of showing the OP that his estimator is biased. I would keep it if I were you. In general, it is nice to have a theory answer juxtaposed with a simulation answer.
– Ben
Oct 25, 2018 at 3:02
• Thank you Ben. I will accept your answer once I fully understand it. Is $\mathbb{I}$ your notation for the indicator function?
– Remy
Oct 25, 2018 at 3:34
• @Remy: Yes, that is correct.
– Ben
Oct 25, 2018 at 4:00

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

• Im not sure I follow your answer, though admittedly I am unfamilar with this language. If $a$ is the sample mean, then $a (1-1/n)^{(n*a-1)}$ is the UMVUE he derived, is it not? His derived UMVUE is $\bar{X}(1-1/n)^{n \bar{X}-1}$ Oct 25, 2018 at 2:47
• I took t to be $\sum_i X_i,$ not $\bar X.$ Oct 25, 2018 at 2:50
• I see. He conditioned on $\bar{X}$ so here $t$ is the sample mean, not total.I can see the confusion though since he defines the sum as $T$ first to justify using the sample mean Oct 25, 2018 at 2:53
• At the very least there is a confusion of definition. Note displayed $T(\vec{X})=\sum_{i=1}^n t(X_i)=\sum_{i=1}^n X_i.$ But I think you're right about the conditioning. Oct 25, 2018 at 2:55