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As the title already says I have to show that for $Z=(Z_1,...,Z_n)^T$ it is $$ y^T\text{Cov}(Z)y\geqslant 0~\forall~y\in\mathbb{R}^n. $$

In a book I read $$ y^T\text{Cov}(Z)y=\text{Var}(y^TZ)\geqslant 0 $$ but that's not clear to me, because I would have said that $$ y^T\text{Cov}(Z)y=\text{Cov}(y^TZ) $$ (and that can be negative). So in other words, I do not see why $$ \text{Cov}(y^TZ)=\text{Var}(y^TZ). $$

Symmetric is clear because of $\text{Cov}(Z_i,Z_j)=\text{Cov}(Z_j,Z_i)$.

mathfemi
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1 Answers1

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Indeed $y^T\text{Cov}(Z)y=\text{Var}(y^TZ)$ (variance, not covariance) since $y^TZ$ is a random matrix of size $1\times1$, that is, a real-valued random variable. Thus, $y^T\text{Cov}(Z)y$ is a nonnegative real number for every vector $y$, which proves that the symmetric matrix $\text{Cov}(Z)$ of size $n\times n$ is nonnegative definite.

Did
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