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: Find $E(\bar{X})$ and $\text{Var}(\bar{X})$

$$E(\bar{X}) = E\left(\frac{1}{n}\sum_{i=1}^n X_i\right) = \frac{1}{n} \cdot n\lambda = \lambda$$

$$\text{Var}(\bar{X}) = \text{Var}\left(\frac{1}{n}\sum_{i=1}^n X_i\right) = \frac{1}{n^2} \cdot n\lambda = \frac{\lambda}{n}$$

Step 2: Find $E(S^2)$

For any distribution, $S^2$ is an unbiased estimator of the population variance: $$E(S^2) = \text{Var}(X_i) = \lambda$$

Step 3: Find $\text{Var}(S^2)$

For i.i.d. samples: $$\text{Var}(S^2) = \frac{1}{n}\left[\mu_4 - \frac{n-3}{n-1}\sigma^4\right]$$

where $\mu_4 = E[(X_i - \lambda)^4]$ and $\sigma^2 = \lambda$.

For Poisson($\lambda$):

• $\mu_4 = \lambda + 3\lambda^2$
• $\sigma^4 = \lambda^2$

$$\text{Var}(S^2) = \frac{1}{n}\left[\lambda + 3\lambda^2 - \frac{n-3}{n-1}\lambda^2\right] = \frac{1}{n}\left[\lambda + \frac{2\lambda^2}{n-1}\right]$$

Step 4: Evaluate each option

(A) $\text{Var}(\bar{X}) < \text{Var}(S^2)$

$$\text{Var}(\bar{X}) = \frac{\lambda}{n}$$

$$\text{Var}(S^2) = \frac{1}{n}\left[\lambda + \frac{2\lambda^2}{n-1}\right]$$

For large $n$ or $\lambda > 0$: $$\text{Var}(S^2) > \frac{\lambda}{n} = \text{Var}(\bar{X})$$

TRUE 

(B) $\text{Var}(\bar{X}) = \text{Var}(S^2)$

From above, these are not equal. FALSE 

(C) $\text{Var}(\bar{X})$ attains the Cramér-Rao lower bound

For Poisson($\lambda$), the Fisher information is $I(\lambda) = \frac{1}{\lambda}$.

Cramér-Rao lower bound for estimating $\lambda$: $$\text{CRLB} = \frac{1}{nI(\lambda)} = \frac{\lambda}{n}$$

Since $\text{Var}(\bar{X}) = \frac{\lambda}{n}$, it attains the CRLB.

TRUE 

(D) $E(\bar{X}) = E(S^2)$

$$E(\bar{X}) = \lambda$$ $$E(S^2) = \lambda$$

TRUE 

Answer: (A), (C), and (D) are true