I am trying to finish a problem, my method requires to prove variance of mean equals mean of covariance, but I have trouble proving it. Is it correct? Or more condition needed?
Now I use $$Var(X)=E(X^2)-E(X)^2$$ $$Cov(X_i,X_j)=E(X_iX_j)-E(X_i)E(X_j)$$ but fail to show whether$$Var(E(X_i))=E(Cov(X_i,X_j))$$is true.
id.est.$$\frac{1}{n}\sum_{i=1}^{n}{(E(X_i)-E(X))^2}=\frac{1}{n^2}\sum_{i=1}^{n}\sum_{j=1}^{n}{Cov(X_i,X_j)}$$
It seems that this problem assumes$$E(X_i)=E(X)$$and$$Var(X_i)=Var(X)$$
$X_i$ is a finite set consisting of $n$ real numbers for each $i \in \left\{1,...,n\right\}$.
$X$ means the set of all $n^2$ real numbers from $X_1,...,X_n$.
Counterexample. Suppose $n=2$, $X_1 = (1,0)$ and $X_2 = (1,0)$, so $X = (1,0,1,0)$. Then $X_1X_2 = (1,0) = X_1$ by multiplying the entries termwise, and under the interpretation that $\mathbb E((x_1,\dots,x_n)) = \frac{1}{n}\sum_{i=1}^n x_i$,
$$ \mathbb E (X_1) = \mathbb E(X_2) = \mathbb E(X) =1/2,$$ $$ \operatorname{Var}X_1 = \operatorname{Var}X_2 = 1/4, $$ $$\operatorname{Cov}(X_1,X_2) = \operatorname{Cov}(X_2,X_1) = E(X_1 X_2)-E(X_1)E(X_2) = 1/4, $$
So $\text{LHS} = 0$ but $\text{RHS} =1/4 $.