Can I get $\mathrm{Cov}(X_1, X_2)$ in this case?

Question

I know the values of:

1. $\mathrm{Cov}(X_1,Z_1)=M_1$,
2. $\mathrm{Cov}(X_1,Z_2)=M_1*A$,
3. $\mathrm{Cov}(X_2,Z_1)=M_2*B$,
4. $\mathrm{Cov}(X_2,Z_2)=M_2$

Is it possible to get the value of $\mathrm{Cov}(X_1,X_2)$ in this case?

Answer 1

This is somewhat tedious to disprove.

Consider $X_{1},X_{2},\bar{X}_{2},Z_{1},Z_{2}$ such that

$Z_{1}=Z_{2}$ , hence $A=B$ , $Z_{1}\in\left\{ -2,-1,1,2\right\} $ where each value has a likelihood of 0.25.

$Z_{1}=-2\Rightarrow X_{1}=1$

$Z_{1}=-1\Rightarrow X_{1}=-2$

$Z_{1}=1\Rightarrow X_{1}=2$

$Z_{1}=2\Rightarrow X_{1}=-1$

Obviously $\mathbb{E}\left[X_{1}\right]=\mathbb{E}\left[Z_{1}\right]=0 . X_{1}Z_{1}$ takes the value 2 in 2 cases and -2 in 2 cases, so the expectation is 0 as well. There we get 0 covariance. So $M_{1}=0$ .

Now lets define $X_{2}=X_{1}$ , so their correlation is one and covariance 2.5 (or sth. like that) and obviously $M_{2}=0$ . Lets define $\bar{X}_{2}=0$ almost surely, so $\bar{M}_{2}=0$ as well, but the covariance with $X_{1}$ is $0$ but more importantly different from the covariance of $X_1$ and $X_2$ .