Can off-diagonal elements in a covariance matrix ever be greater than the diagonal elements? Can off-diagonal elements in a variance-covariance matrix ever be greater than the diagonal elements?
 A: *

*No. Consider the example with two variables $X_1$ and $X_2$.

Suppose $$Cov(X_1,X_2) > max\{Var(X_1),Var(X_2)\}$$
Then it must be that $$Corr(X_1,X_2) > max\biggl\{ \sqrt{\frac{Var(X_1)}{Var(X_2)}}, \sqrt{\frac{Var(X_2)}{Var(X_1)}}\biggr\}$$
Suppose that $Var(X_1) > Var(X_2)$, then
$$Corr(X_1,X_2) > \sqrt{\frac{Var(X_1)}{Var(X_2)}}$$
But this is impossible since
$$1 \geq Corr(X_1,X_2) > \sqrt{\frac{Var(X_1)}{Var(X_2)}} > 1$$
The same would hold if you suppose that $Var(X_1) < Var(X_2)$.


*No. By the same reasoning you can have

$$Cov(X_1,X_2) > Var(X_1)$$
since it is possible to have
$$1 \geq Corr(X_1,X_2) > \sqrt{\frac{Var(X_1)}{Var(X_2)}} > 0$$
if $Var(X_1) < Var(X_2)$.
A: No, it cannot be greater than the maximum diagonal entry. Define the centered random variables $\overline{X}_i=X_i-E[X_i]$ for all $i=1,2,\ldots,N$.
Let $E[\overline{X}_{\ell}^2]\geq E[\overline{X}_j^2]$ for all $j=1,2,\ldots,N$, then
$$0\leq E[\left(\overline{X}_{i}-\overline{X}_k\right)^2]=E[\overline{X}_{i}^2]+E[\overline{X}_k^2]-2E[\overline{X}_{i} \overline{X}_k]\leq 2E[\overline{X}_{\ell}^2]-2E[\overline{X}_{i} \overline{X}_k], $$
and, therefore, $E[\overline{X}_{\ell}^2]\geq E[\overline{X}_{i} \overline{X}_k]$ for all $i,k$.
Yes, it can be greater than the smallest diagonal entry (a counter-example has been provided in the comments).
