When $cov(Z_s,Z_u)=\sigma^2_{\max(s,u)}$ for normal random variables? Let $Z_1, Z_2, \dots, Z_n$ be normal random variables with mean zero and $Var(Z_i)=\sigma_i^2$.
Under what condition $cov(Z_s,Z_u)=\sigma^2_{\max(s,u)}$?
I will only consider the $n=2$ case for a counter-example, as my linear algebra is not strong. But a covariance matrix M must be positive semi-definite, i.e. $z^T M z \geq 0$ for all column vectors $z$ of length $n$. As you have defined it, the covariance matrix would be of the form $M = (a,b;b,b)$, for some $a,b &gt; 0$. Let $z = (x;y)$. Then to be positive semi-definite, $ax^2 + 2bxy + by^2 = ax^2 + b(2xy + y^2)\geq 0$ for all $x,y$ (this expression can be obtained by standard matrix operations). Only $b(2xy + y^2)$ can be negative, so we need $ax^2 \geq -b(2xy + y^2)$. For any $y, a, b$ there should be a suitable $x$ which breaks this inequality, so we have shown that this matrix cannot be a covariance matrix.
I will only consider the $n=2$ case for a counter-example, as my linear algebra is not strong. But a covariance matrix M must be positive semi-definite, i.e. $z^T M z \geq 0$ for all column vectors $z$ of length $n$. As you have defined it, the covariance matrix would be of the form $M = (a,b;b,b)$, for some $a,b &gt; 0$. Let $z = (x;y)$. Then to be positive semi-definite, $ax^2 + 2bxy + by^2 = ax^2 + b(2xy + y^2)\geq 0$ for all $x,y$ (this expression can be obtained by standard matrix operations). Only $b(2xy + y^2)$ can be negative, so we need $ax^2 \geq -b(2xy + y^2)$. For any $y, a, b$ there should be a suitable $x$ which breaks this inequality, so we have shown that this matrix cannot be a covariance matrix.