Basic result of expectation of the sample mean

Question

Let $X_1,\dots,X_n$ be idependent and identically distributed random variables with mean $\mu$ and variance $\sigma^2$. Also define $\bar X=\sum_{i=1}^n X_i/n$.

It's well known that $E(\bar X)=\mu$ and $Var(\bar X)=\sigma^2/n=E(\bar X^2)-E(\bar X)^2$.

But, when I directly calculate $E(\bar X^2)$, which is equal to $\sigma^2/n+\mu^2$, I get instead

\begin{align} E(\bar X^2)&=E\bigg[\bigg(\frac{\sum_{i=1}^n X_i}{n}\bigg)^2\bigg]=E\bigg[\frac{\sum_{i,j=1}^n X_iX_j}{n^2}\bigg]\\ &=\frac{\sum_{i=1}^n E(X_i^2)}{n^2}=\frac{E(X_1^2)}{n}=\frac{\mu^2+\sigma^2}{n}. \end{align}

Hence, somethig is wrong in my calculations. Can you point me out which part is flawed?

Thanks

Answer 1

$$\bigg(\frac{1}{n}\sum_{1\leq k \leq n}X_k\bigg)^2=\frac{1}{n^2}\sum_{1\leq k\ \leq n}X_k^2+\frac{1}{n^2}\sum_{k\neq h}X_kX_h$$ Note $E[X_kX_h]=E[X_k]E[X_h]=\mu^2$ by IID-ness. So by taking the expectation: $$\frac{1}{n}E[X_1^2]+\frac{1}{n^2}n(n-1)\mu^2=\frac{1}{n}(\mu^2+\sigma^2)+\mu^2-\frac{1}{n}\mu^2=\frac{\sigma^2}{n}+\mu^2$$

Answer 2

You forgot to include the terms with $X_iX_j$ for $i\neq j$. With this, you get

$$\begin{align*}\mathbb E\left[\sum_{i, j=1}^n X_iX_j\right]&=\sum_{i, j=1}^n \mathbb E\left[X_iX_j\right]\\&= \sum_{i=1}^n \mathbb E\left[X_i^2\right]+\sum_{i, j=1: i\neq j}^n\mathbb E\left[X_i\right]\mathbb E\left[X_j\right]\\&=n(\mu^2+\sigma^2)+(n^2-n)\mu^2\\&=n\sigma^2+n^2\mu^2\end{align*}.$$