The random variable \(X\) has the binomial distribution \({\text{B}}(n,{\text{ }}p)\), where \(n > 1\).

Show that

(a)     \(\frac{X}{n}\) is an unbiased estimator for \(p\);

(b)     \({\left( {\frac{X}{n}} \right)^2}\) is not an unbiased estimator for \({p^2}\);

(c)     \(\frac{{X(X - 1)}}{{n(n - 1)}}\) is an unbiased estimator for \({p^2}\).

(a)     \({\text{E}}\left( {\frac{X}{n}} \right) = \frac{1}{n}{\text{E}}(X)\)     M1

\( = \frac{1}{n} \times np = p\)     AG

therefore unbiased     AG

[2 marks]

(b)     \({\text{E}}\left[ {{{\left( {\frac{X}{n}} \right)}^2}} \right] = \frac{1}{{{n^2}}}\left( {{\text{Var}}(X) + {{[{\text{E}}(X)]}^2}} \right)\)     M1A1

\( = \frac{1}{{{n^2}}}\left( {np(1 - p) + {n^2}{p^2}} \right)\)     A1

\( \ne {p^2}\)     A1

therefore not unbiased     AG

[4 marks]

(c)     \({\text{E}}\left[ {\left( {\frac{{X(X - 1)}}{{n(n - 1)}}} \right)} \right] = \frac{{{\text{E}}({X^2}) - {\text{E}}(X)}}{{n(n - 1)}}\)     M1

\( = \frac{{np(1 - p) + {n^2}{p^2} - np}}{{n(n - 1)}}\)     A1

\( = \frac{{n{p^2}(n - 1)}}{{n(n - 1)}}\)     A1

\( = {p^2}\)

therefore unbiased     AG

[3 marks]