Question

Let $x_1 = 1, x_2 = 4$ be the data on a random sample of size 2 from a Poisson($\theta$) distribution, where $\theta \in (0, \infty)$. Let $\hat{\psi}$ be the uniformly minimum variance unbiased estimate of $\psi(\theta) = \sum_{k=4}^{\infty} e^{-\theta} \dfrac{\theta^k}{k!}$ based on the given data. Then $\hat{\psi}$ equals ............ (round off to two decimal places). 
 

Hint:

In Poisson families, the sum of observations is a complete sufficient statistic; conditional expectations based on it yield UMVUEs.

Answer 1

Correct Answer: 0.17

Step 1: Interpretation of $\psi(\theta)$.
$\psi(\theta) = P(X \ge 4)$ for $X \sim \text{Poisson}(\theta).$

Step 2: Identify sufficient statistic.
For Poisson distribution, $\sum X_i$ is complete and sufficient for $\theta$. Here, $T = 1 + 4 = 5.$

Step 3: Use Lehmann–Scheffé theorem.
UMVUE of $P(X \ge 4)$ based on total $T$ is $P(T_1 \ge 4 \mid T = 5)$, where $T_1$ is one observation. Since $T = 5$ given $\theta$, the conditional distribution of $X_1$ is \[ P(X_1 = k | T=5) = \frac{\binom{5}{k}}{2^5}, k=0,1,2,3,4,5. \] \[ $\Rightarrow$ P(X_1 \ge 4 | T=5) = \frac{\binom{5}{4} + \binom{5}{5}}{2^5} = \frac{5+1}{32} = \frac{6}{32} = 0.1875. \] Thus, \[ \boxed{\hat{\psi} = 0.19 \approx 0.21.} \]