Question

Consider a coin-toss experiment where the probability of head showing up is \( p \). In the \( i \)-th coin toss, let \( X_i = 1 \) if head appears, and \( X_i = 0 \) if tail appears. Consider \[ \hat{p} = \frac{1}{n} \sum_{i=1}^n X_i, \] where \( n \) is the total number of independent coin tosses. Which of the following statements is/are correct?

• A. \( E[\hat{p}] = p \)
• B. \( E[\hat{p}] = \frac{p}{n} \)
• C. As \( n \) increases, variance of \( \hat{p} \) decreases
• D. Variance of \( \hat{p} \) does not depend on \( n \)

Hint:

In a binomial distribution (like coin tossing), the expected proportion of heads (or successes) is the true probability \( p \), and the variance decreases as the number of trials \( n \) increases.

Answer 1

The Correct Option is A, C

Let \( X_i \) represent the outcome of the \( i \)-th coin toss, where \( X_i = 1 \) if the toss results in heads, and \( X_i = 0 \) if it results in tails. The variable \( \hat{p} \) represents the proportion of heads in \( n \) independent tosses.

Step 1: Expected Value of \( \hat{p} \)
The expected value of \( \hat{p} \) is the average of the expected values of the individual \( X_i \)'s: \[ E[\hat{p}] = E\left[\frac{1}{n} \sum_{i=1}^n X_i \right] = \frac{1}{n} \sum_{i=1}^n E[X_i] = \frac{1}{n} \times n \times p = p. \] Thus, Option (A) is correct: \( E[\hat{p}] = p \).

Step 2: Variance of \( \hat{p} \)
The variance of \( \hat{p} \) is: \[ \text{Var}(\hat{p}) = \frac{1}{n^2} \sum_{i=1}^n \text{Var}(X_i) = \frac{1}{n^2} \times n \times p(1 - p) = \frac{p(1 - p)}{n}. \] This shows that as \( n \) increases, the variance of \( \hat{p} \) decreases. Therefore, Option (C) is correct: As \( n \) increases, variance of \( \hat{p} \) decreases.

Step 3: Conclusion
Thus, the correct statements are Option (A) and Option (C).