Let $(X_n)_{n∈N}$ be a sequence of i.i.d. Bernoulli random variables with parameter $p ∈ [0, 1]$. For $n ∈ N$, let $X_n = \sum_{i=1}^n X_i/n$.
Prove that $X_n$ converges to $p$ in $L2$?
To converge in $L^2$ $lim_{n->\inf}E(|X_n-X|^2)=0$
In my case: $lim_{n->\inf}E(|X_n-p|^2)=E(X_n^2+p^2-2pX_n)=E(X_n^2)+p^2-2pE(X_n)=\frac{E((\sum_{i=1}^n X_i)^2)}{n^2}+p^2-2p^2=E[n X_i^2+\frac{n(n-1)}{2}X_iX_j]/n^2-p^2=\frac{np+p^2\frac{n(n-1)}{2}}{n^2}-p^2=np(1-p-np)$
Can you help me?
An alternative derivation without as much algebra, using the fact that $\mathbf E[X_n] = p$, and properties of variances:
\begin{align*} \mathbf E [ |X_n -p|^2] & = \mathbf E [ (X_n - \mathbf E[X_n] )^2]\\ & = \text{Var}(X_n) \\ & = \frac1{n^2} \text{Var} \left( \sum_{k=1}^n X_k \right) \\ & = \frac1{n^2} n \text{Var} \left( X_1 \right) \\ & = \frac1n \text{Var}(X_1). \end{align*}
Which converges to $0$ as $n \rightarrow \infty$. Note that we did not need to calculate / manipulate the variance term of a binomial distribution directly.
In particular this proof works for the sum of any independent random variables which have finite mean.
The error in your proof comes in because you need not have the factor of $\frac12$, when expanding the sum i.e.
\begin{align*} \mathbf E \left[ \left( \sum_{k=1}^n X_k \right)^2 \right] &= \sum_{k=1}^n \mathbf E[X_k] + \sum_{1 \leq k \neq j \leq n} \mathbf E[X_k] \mathbf E[X_j] \\ & = n p + \sum_{1 \leq k \neq j \leq n} p^2 \\ & = np + n(n-1)p^2. \end{align*} On correcting for this your proof then gives the expected result.