let $X\sim Bin(n,p)$ and $\hat{p} =\frac{X}{n}$
a) Find a constant c such that $E[c\hat{p}(1-\hat{p})]=p(1-p)$
My work: $$ \begin{align} cE[\hat{p}(1-\hat{p})] &=E[\frac{X}{n}]-E[\frac{X^2}{n^2}]\\ &= \frac{1}{n}E[X]-\frac{1}{n^2}E[X^2] \\ \end{align} $$ And I continue with $E[X] = p$ and $E[X^2] = Var(X)+E[X]^2$ but this gives me $$ = \frac{p}{n} - \frac{1}{n^2}np(1-p)-\frac{p^2}{n^4} $$ Which doesnt seem righ. Am I doing something wrong?
Since $X$ is a binomial random variable, $E[X]=np$ and $Var(X)=np(1-p)$.
Hence, $$\frac{1}{n}E[X]-\frac{1}{n^2}E[X^{2}]=\frac{np}{n}-\frac{np(1-p)+n^{2}p^{2}}{n^2}$$ Simplifying, $$=p-p^{2}-\frac{1}{n}p(1-p)=\left(1-\frac{1}{n}\right)p(1-p)$$
So we simply let $c=\frac{1}{1-\frac{1}{n}}=\frac{n}{n-1}$