i.d, we can rewrite the expectation as:
$$
E[\hat{p}] = \frac{1}{n}\sum_{i=1}^n E[Y_i]
$$
We know that $E[Y_i] = p$ for a Bernoulli random variable, so:
$$
E[\hat{p}] = \frac{1}{n}\sum_{i=1}^n p = \frac{1}{n} \cdot np = p
$$
Thus, $\hat{p}$ is an unbiased
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