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]