# $X_1, X_2, \cdots, X_n$ : i.i.d. $\sim \text{Bernoulli}(p)$. Then $\bar{x}$ is an unbiased estimator of $p$.

Let $$X_1, X_2, \cdots, X_n$$ be i.i.d. $$\sim \text{Bernoulli}(p)$$. Then $$\bar{x}$$ is an unbiased estimator of $$p$$.

How should I approach for this types of problems. Some hint will also help me.

• – Lord Shark the Unknown Dec 20 '18 at 6:49
• Actually independence is redundant here. – drhab Dec 20 '18 at 9:20

You know that $$E(X) = p$$ or, for any $$i$$, $$E(X_i) = p$$.

So,

$$E(\bar X) = E\Bigl(\frac {\sum_{i=1}^n X_i}{n}\Bigl)$$ $$=\frac{1}{n}\Big(E{\sum_{i=1}^n X_i}\Big)$$ $$=\frac{1}{n}\Big({\sum_{i=1}^n E(X_i)}\Big)$$ $$=\frac{1}{n}(np)$$ $$=p$$

By the definition of Unbiased Estimator, $$\bar X$$ is an unbiased estimator of $$p$$.

If you define $$\bar{x}$$ as the sample mean, i.e. $$\bar{x} = \frac{1}{n}\sum_{i=1}^n X_i$$ Take expectations on both sides $$E \bar{x} = E \frac{1}{n}\sum_{i=1}^n X_i$$ Since expectation is a linear operator, then $$E \bar{x} = \frac{1}{n}\sum_{i=1}^n E X_i$$ But since $$E X_i = p$$, then $$E \bar{x} = \frac{1}{n}\sum_{i=1}^n p$$ Since $$p$$ is a constant, then we could extract it as $$E \bar{x} = \frac{p}{n}\sum_{i=1}^n 1$$ Now since $$\sum_{i=1}^n 1 = n$$, we get $$E \bar{x} = \frac{p}{n}n = p = \mu$$ where $$\mu = p$$ is the true mean of the Bernoulli distribution, hence we say that $$\bar{x}$$ is an unbiased estimator of $$p$$, since in average, it gives us $$p$$.

Use the fact that $$X_i$$ are Identical. And the linearity property of Expectation.

$$E(\bar{X})=E(X_1)=p$$