Trying to prove the sample mean $\hat{p} = \bar{X}$ is an efficient estimator for the Bernoulli parameter $p$ Consider a random sample $X_1, X_2, \dots, X_n$ from a Bernoulli distribution with parameter $p$. We want to show that the estimator $\hat{p} = \bar{X}$ is efficient. #### **1) Unbiasedness** First, we check if the estimator is unbiased. $\hat{p} = \bar{X} \quad \text{is unbiased } \checkmark$ $E[\hat{p}] = E[\bar{X}] = E[X_1] = p$ #### **2) Cramer-Rao Inequality & Efficiency** To show efficiency, we check if the variance of the estimator achieves the Cramer-Rao Lower Bound (CRLB). The Cramer-Rao inequality states: $Var(\hat{p}) \ge \frac{1}{n \cdot I(p)}$ where $I(p)$ is the Fisher information for a single observation. For a Bernoulli distribution, the Fisher information is: $I(p) = \frac{1}{p(1 - p)}$ **Actual Variance of the Estimator:** $Var(\hat{p}) = Var(\bar{X}) = \frac{Var(X_1)}{n} = \frac{p(1 - p)}{n}$ Note: $MSE(\hat{p}, p) = E[(\hat{p} - p)^2] = E[(\bar{X} - p)^2]$ **Cramer-Rao Lower Bound (CRLB):** $\frac{1}{n \cdot I(p)} = \frac{1}{n \cdot \frac{1}{p(1 - p)}} = \frac{p(1 - p)}{n}$ **Comparison:** Comparing the actual variance to the lower bound: $\frac{p(1 - p)}{n} = \frac{p(1 - p)}{n}$ #### **Conclusion** Since the estimator is unbiased and its variance is equal to the Cramer-Rao Lower Bound: $\Rightarrow (1), (2) \Rightarrow \hat{p} \text{ is an efficient estimator for } p$