Stuck on Variance proof

Question

Let $(X_{k})_{k \in\ \mathbb{N}}$ be a sequence of square integrable and i.i.d random variables. Define $\bar{X}_{n} = \frac{1}{n}\sum^{n}_{i=1}X_{i}$. Then we have $Var(\bar{X}_{n}) = \mathbb{E}[(\bar{X}_{n} - \mathbb{E}[X_{1}])^{2}] = \frac{1}{n}Var[X_{1}]$. The proof is an exercise.

I am stuck here because of this $X_{1}$ constant, which I know if I take the expectation of, is just a constant. Here is what I have tried:

I started by using the usual variance derivation formula: $Var(X) = \mathbb{E}[({X} - \mathbb{E}[X])^{2}]= \mathbb{E}[X^{2}] - (\mathbb{E}[X])^{2}$

Plugging in the relevant values yielded me: $\mathbb{E}[\bar{X}_{n}^{2} - 2 \bar{X}_{n}\mathbb{E}[X_{1}] + (\mathbb{E}[X_{1}])^{2}]$

By the property of expectation of a constant, this got me: $\mathbb{E}[\bar{X}_{n}^{2} - 2X_{1} \bar{X}_{n} + X_{1}^{2}]$

However, I am not sure where to got next, this $X_{1}$ has really thrown me off here.

Answer 1

Use the properties of variance: \begin{align} \operatorname{Var}(\bar{X}_n)&=\frac{1}{n^2}\operatorname{Var}\!\left(\sum_{1\le i\le n}{X}_i\right)=\frac{1}{n^2}\sum_{1\le i,j \le n}\operatorname{Cov}(X_i,X_j) \\ &=\frac{1}{n^2}\sum_{1\le i\le n}\operatorname{Var}(X_i)+\frac{1}{n^2}\sum_{i\ne j}\operatorname{Cov}(X_i,X_j)=\ldots \end{align}

Answer 2

Let $\mu=E[X_{1}]$. Note that $E[\bar{X}_{n}]=\frac{1}{n}\sum_{k=1}^{n}E[X_{k}]=\frac{1}{n}\sum_{k=1}^{n}\mu=\mu$. Define $Y_{i}=X_{i}-\mu$. Note that $Y_{1},Y_{2},\ldots$ are square-integrable and i.i.d.. Moreover, $E[Y_{i}]=0$. If $i\neq j$, we have $E[Y_{i}Y_{j}]=E[Y_{i}]E[Y_{j}]=0$. $E[Y_{i}^{2}]=E[Y_{1}^{2}]=Var(X_{1}).$

Observe that \begin{eqnarray*} \bar{X}_{n}-\mu & = & \frac{1}{n}\sum_{k=1}^{n}(X_{k}-\mu)\\ & = & \frac{1}{n}\sum_{k=1}^{n}Y_{k}. \end{eqnarray*} Therefore, \begin{eqnarray*} \left(\bar{X}_{n}-\mu\right)^{2} & = & \frac{1}{n^{2}}\sum_{i=1}^{n}\sum_{j=1}^{n}Y_{i}Y_{j}\\ & = & \frac{1}{n^{2}}\left\{ \sum_{i=1}^{n}Y_{i}^{2}+\sum_{(i,j),i\neq j}Y_{i}Y_{j}\right\} . \end{eqnarray*} Taking expectation on both sides yields, \begin{eqnarray*} Var(\bar{X}_{n}) & = & E\left[\left(\bar{X}_{n}-\mu\right)^{2}\right]\\ & = & \frac{1}{n^{2}}\left\{ \sum_{i=1}^{n}E[Y_{i}^{2}]+\sum_{(i,j),i\neq j}E[Y_{i}Y_{j}]\right\} \\ & = & \frac{1}{n^{2}}\left\{ \sum_{i=1}^{n}Var(X_{1})+0\right\} \\ & = & \frac{1}{n}Var(X_{1}). \end{eqnarray*}