Question

Let $X_1, X_2, \ldots, X_n$ be i.i.d. Poisson($\lambda$) random variables, where $\lambda > 0$. Define \[ \bar{X} = \frac{1}{n}\sum_{i=1}^n X_i \text{and} S^2 = \frac{1}{n - 1}\sum_{i=1}^n (X_i - \bar{X})^2. \] Then which of the following statements is/are TRUE?

• A. $\mathrm{Var}(\bar{X}) < \mathrm{Var}(S^2)$
• B. $\mathrm{Var}(\bar{X}) = \mathrm{Var}(S^2)$
• C. $\mathrm{Var}(\bar{X})$ attains the Cramer–Rao lower bound
• D. $E(\bar{X}) = E(S^2)$

Hint:

For the Poisson distribution, the sample mean is both unbiased and efficient—it achieves the Cramer–Rao lower bound.

Answer 1

The Correct Option is A, C, D

Step 1: Recall properties of Poisson distribution. 
For $X_i \sim \text{Poisson}(\lambda)$, \[ E(X_i) = \lambda, \mathrm{Var}(X_i) = \lambda. \]

Step 2: Compute $E(\bar{X})$ and $\mathrm{Var}(\bar{X})$.
\[ E(\bar{X}) = \lambda, \mathrm{Var}(\bar{X}) = \frac{\lambda}{n}. \]

Step 3: Compute $E(S^2)$. 
For Poisson, $S^2$ is an unbiased estimator of $\lambda$: \[ E(S^2) = \lambda. \] Hence, (D) is true.

Step 4: Cramer–Rao lower bound. 
For a Poisson distribution, \[ \text{CRLB} = \frac{\lambda}{n}. \] Since $\mathrm{Var}(\bar{X}) = \frac{\lambda}{n}$, the sample mean $\bar{X}$ attains the CRLB. Hence, (C) is true.

Step 5: Conclusion. 
\[ \boxed{(C) \text{ and } (D) \text{ are correct.}} \]