Why is $E[(\sum^n_{m=1}X_m)^2]=n(\sigma^2+\mu^2)+(n^2-n)\mu^2$?

Question

Let $\hat \mu=\frac{1}{n} \sum\limits^n_{m=1}X_m $, and $X_m$ is iid , that is, $E[X_m]=\mu$, $V[X_m]=\sigma ^2$.

Now, I have calculated the $\hat \sigma^2=\left(\frac{1}{n}\sum\limits^n_{m=1}X_m^2\right)-\hat \mu ^2$, so

\begin{align} E\left[\hat \sigma^2\right]& =E\left[\left(\frac{1}{n}\sum\limits^n_{m=1}X_m^2\right)-\hat \mu ^2\right] \\ & =E\left[\frac{1}{n}\sum\limits^n_{m=1}X_m^2\right]-E\left[\hat \mu ^2\right]\\ & =\frac{1}{n}\sum\limits^n_{m=1}E\left[X_m^2\right]-E\left[\frac{1}{n^2}\left(\sum\limits^n_{m=1}X_m\right)^2\right] \\ & =\frac{1}{n}\sum\limits^n_{m=1}(\sigma^2+\mu^2)-\frac{1}{n^2}(n(\sigma^2+\mu^2)+(n^2-n)\mu^2). \\ \end{align}

However, I can't understand why is $E\left[\left(\sum\limits^n_{m=1}X_m\right)^2\right]=n(\sigma^2+\mu^2)+(n^2-n)\mu^2$, can anyone explain it to me?

Answer 1

$$\mathbb{E}\left(\sum_j X_j \right)^2 =\sum_j \mathbb{E} (X_j^2) +\sum_{i\neq j } \mathbb{E}( X_i )\mathbb{E} (X_j )$$ but $$\mathbb{E} (X^2 ) = \mbox{Var} (X) +(\mathbb{E} (X_j ))^2 =\sigma^2 +\mu^2$$ hence $$\mathbb{E}\left(\sum_j X_j \right)^2 =\sum_j \mathbb{E} (X_j^2) +\sum_{i\neq j } \mathbb{E}( X_i )\mathbb{E} (X_j )=\sum_j (\sigma^2 +\mu^2 ) +\sum_{i\neq j } \mu\cdot \mu =n(\sigma^2 +\mu^2 ) +n(n-1) \mu^2$$