Consider a coin-toss experiment where the probability of head showing up is \(p\). In the \(i^{\text{th}}\) coin toss, let \(X_{i} = 1\) if head appears, and \(X_{i} = 0\) if tail appears. Consider
\[ \widehat{p} = \frac{1}{n}\sum_{i=1}^{n} X_{i} \]
where \(n\) is the total number of independent coin tosses.
Which of the following statements is/are correct?
Answer
(A), (C)
Explanation
Each \(X_{i}\) is a Bernoulli variable with \(E[X_{i}] = p\) and \(\operatorname{Var}(X_{i}) = p(1 - p)\).
- By linearity of expectation, \(E[\,\widehat{p}\,] = \frac{1}{n} \cdot np = p\). So A holds and B fails; \(\frac{p}{n}\) forgets that the sum has \(n\) terms.
- The tosses are independent, so the variances add: \(\operatorname{Var}(\widehat{p}) = \frac{1}{n^{2}} \cdot np(1 - p) = \frac{p(1 - p)}{n}\). This shrinks as \(n\) grows, so C holds and D fails.
In short, \(\widehat{p}\) is an unbiased estimator of \(p\) that gets more precise with more tosses.
Answer: A, C