What is wrong with my proof of Law of Large Numbers? Need help in deriving variance for LLN, but am getting something wrong here, can you help point out where I'm going wrong?
Thanks a ton!
For i.i.d random variables $X_1,X_2,\ldots,X_n$ with mean $\mu$ and variance $\sigma^2$.
Let $X=\frac{1}{n}\displaystyle\sum_{i=1}^nX_i$ and it's mean be $\mu$(from LLN) then
\begin{align}
\operatorname{var}(X)& =E[(X-\mu)^2]\\[10pt]
& =E[X^2]-\mu^2\\[10pt]
& =E\left[\left(\frac{1}{n}\sum_{i=1}^nX_i\right)^2\right]-\mu^2\\[10pt]
& =E\left[\frac{1}{n^2} \left( \sum_{i=1}^nX_i\right)^2\right]-\mu^2\\[10pt]
& =\frac{1}{n^2}E\left[\left(\sum_{i=1}^nX_i\right)^2\right]-\mu^2\\[10pt]
& =\frac{1}{n^2}E\left[\sum_{i=1}^nX_i^2+\sum_{i,j=1,i\neq j}^{i,j=n} X_iX_j\right]-\mu^2\\[10pt]
& =\frac{1}{n^2}\left(\sum_{i=1}^nE[X_i^2]+\sum_{i,j=1,i\neq j}^{i,j=n}E[X_iX_j]\right)-\mu^2\\[10pt]
& =\frac{1}{n^2}(n\sigma^2+(n^2-n)\mu^2-n^2\mu^2)\\[10pt]
& =\frac{1}{n^2}(n\sigma^2-n\mu^2)\\[10pt]
& =\frac{\sigma^2-\mu^2}{n}
\end{align}

You say "from LLN", but you are not using LLN anywhere.
Moreover, it's much easier to do this:
$$
\text{Var}\left(\frac{1}{n}\sum X_{i}\right)=\frac{1}{n^{2}}\sum\text{Var}\left(X_{i}\right)=\frac{n\sigma^{2}}{n^{2}}=\frac{\sigma^{2}}{n}.
$$
