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.


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 $X_1,...,X_n$.

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    $\begingroup$ Are you trying to ask if $$ \sum_i \left(E(X_i) - \sum_j E(X_j)\right)^2 =\sum_{i,j} Cov(X_i,X_j)$$ or some variant of this? $\endgroup$ – Calvin Khor Jul 12 '18 at 19:45
  • $\begingroup$ @CalvinKhor no... $\endgroup$ – user571299 Jul 12 '18 at 19:52
  • $\begingroup$ Well as written, the answer below is right. Once you take an expectation, there is no more randomness. The expectation of a random variable is not random. Since variance is defined with expectations, the same is also true for variance. Maybe you want to consider conditional expectations? $\endgroup$ – Calvin Khor Jul 12 '18 at 19:53
  • $\begingroup$ @CalvinKhor but $Cov(X_i,X_j)$ is not a constant. $\endgroup$ – user571299 Jul 12 '18 at 20:00
  • $\begingroup$ It is a constant. For each index $i$ and $j$. That is why I introduced a discrete sum which can be interpreted as making the index uniformly random $\endgroup$ – Calvin Khor Jul 12 '18 at 20:02

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 $.


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