Question

Let $0,1,1,2,0$ be five observations of a random variable $X$ which follows a Poisson distribution with parameter $\theta>0$. Let the minimum variance unbiased estimate of $P(X\le 1)$, based on this data, be $\alpha$. Then $5^{4}\alpha$ (in integer) is equal to $\;__________$

Hint:

For Poisson samples, obtain UMVUEs by conditioning on the complete sufficient statistic $T=\sum X_i$ and using Poisson–multinomial splitting.

Answer 1

Correct Answer: 512

For i.i.d. Poisson$(\theta)$, the sufficient and complete statistic is the total $T=\sum_{i=1}^{5}X_i\sim \text{Poisson}(5\theta)$.
For any one observation, $g(\theta)=P(X\le 1)=e^{-\theta}(1+\theta)$. The UMVUE is $h(T)=\mathbb{E}\big[\mathbf{1}\{X_1\le 1\}\mid T\big]$. Given $T=t$, Poisson splitting gives $X_1\mid T=t\sim\text{Binomial}(t,1/5)$, hence \[ h(t)=P(X_1\le 1\mid T=t)=\left(\frac45\right)^{t}+ \frac{t}{5}\left(\frac45\right)^{t-1}. \] With the observed data, $t=\sum(0,1,1,2,0)=4$; therefore \[ \alpha=h(4)=\left(\frac45\right)^{4}+\frac{4}{5}\left(\frac45\right)^{3} =\frac{256}{625}+\frac{256}{625}=\frac{512}{625}. \] Thus, \[ 5^{4}\alpha=625\times\frac{512}{625}= 512. \]