Question-54

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: \[ \widehat{p} =\cfrac{1}{n}\sum X_{i} \] where \(n\) is the total number of independent coin tosses. Which of the following statements is/are correct?

• \(E[\widehat{p}] =p\)

• \(E[\widehat{p}] =\cfrac{p}{n}\)

• As \(n\) increases, the variance of \(\widehat{p}\) decreases

• Variance of \(\widehat{p}\) does not depend on \(n\)

NoteAnswer

• \(E[\widehat{p}] =p\)

• \(E[\widehat{p}] =\cfrac{p}{n}\)

• As \(n\) increases, the variance of \(\widehat{p}\) decreases

• Variance of \(\widehat{p}\) does not depend on \(n\)

NoteSolution

From the linearity of expectation, we have \(E[\widehat{p}] =\cfrac{1}{n}\sum E[ X_{i}] =p\). Note that \(X_{i} \sim \text{Bernoulli}( p)\), hence \(E[ X_{i}] =p\). The variance is:

\[ \sigma _{\widehat{p}}^{2} =\cfrac{1}{n^{2}}\sum \sigma _{X_{i}}^{2} =\cfrac{p( 1-p)}{n} \]

We see that the variance decreases as \(n\) increases.