Calculate expected value $\mathbb{E}[(\frac{X}{n}-p)^2$] with $X$ is binomial $(n,p)$

Question

I have to calculate the expected value $\mathbb{E}[(\frac{X}{n}-p)^2] = \frac{pq}{n}$, but everytime i try to solve it my answer is $\frac{p}{n} - p^2$, which is wrong.

What i did: Let X be binomial distributed.

$ \mathbb{E}[(\frac{X}{n}-p)^2] =\mathbb{E}[(\frac{X}{n})^2-2(\frac{X} {n}p)+p^2] \\ =\mathbb{E}[(\frac{X}{n})^2]+(-2)\mathbb{E}[\frac{X}{n}p]+(\mathbb{E}[X])^2 \\ = \mathbb{E}[(\frac{X}{n})^2] - 2p\mathbb{E}[\frac{X}{n}]+p^2 \\= \mathbb{E}[(\frac{X}{n})^2] - 2p\frac{1}{n}\mathbb{E}[X]+p^2\\= \mathbb{E}[(\frac{X}{n})^2] - 2p\frac{1}{n}np+p^2 \\= \mathbb{E}[(\frac{X}{n})^2] - p^2 \\ = \frac{p}{n}-p^2$

Answer 1

Use your derivation until you arrived at this equality: $$\mathbb{E}[\Big(\frac{X}{n}-p\Big)^2]=\ldots=\mathbb{E}[\Big(\frac{X}{n}\Big)^2]-p^2=\frac{1}{n^2}\mathbb{E}[X^2]-p^2$$ Now insert $$\mathbb{E}[X^2]=\mathrm{Var}[X]+(\mathbb{E}[X])^2=np(1-p)+(np)^2\tag{$\ast$}$$ to obtain $$\mathbb{E}[\Big(\frac{X}{n}-p\Big)^2]=\frac{p(1-p)}{n}=\frac{pq}{n}$$ Edit: If you don't know the formula given in $(\ast)$, then you have to calculate \begin{align} \mathbb{E}[X^2]&=\sum_{k=0}^n k^2P(X=k)\\ &=\sum_{k=1}^n [k(k-1)+k]P(X=k)\\ &=\sum_{k=1}^n k(k-1)P(X=k)+\sum_{k=1}^n kP(X=k)\\ &=\sum_{k=2}^n k(k-1)P(X=k)+\sum_{k=1}^n kP(X=k)\\ &=n(n-1)p^2+np \end{align}

Answer 2

$ \mathbb{E}[(\frac{X}{n}-p)^2] =\mathbb{E}[(\frac{X}{n})^2-2(\frac{X} {n}p)+p^2] \\ =\mathbb{E}[(\frac{X}{n})^2]+(-2)\mathbb{E}[\frac{X}{n}p]+(\mathbb{E}[X])^2 \\ = \mathbb{E}[(\frac{X}{n})^2] - 2p\mathbb{E}[\frac{X}{n}]+p^2 \\= \mathbb{E}[(\frac{X}{n})^2] - 2p\frac{1}{n}\mathbb{E}[X]+p^2\\ $

$\mathbb{E}[(X)^2] -(\mathbb{E}[(X)])^2 = Var(X)$

$\mathbb{E}[(X)^2]= Var(X)+(\mathbb{E}[(X)])^2$

$\mathbb{E}[(X)^2] = np(1-p) + n^2p^2$

($Var(X) = np(1-p)$)

$(\mathbb{E}[(\frac{X}{n})^2] = \frac{1}{n^2}(\mathbb{E}[(X)^2)$

$ \mathbb{E}[(\frac{X}{n}-p)^2]=\frac{np(1-p)}{n^2} + \frac{n^2p^2}{n^2}- 2p\frac{np}{n}+p^2 \\= \frac{np(1-p)}{n^2} = \frac{pq}{n}$