Is covariance matrix of multivariate normal distribution always positive semi-definite? Can it be negative definite? What will happen if the covariance matrix is negative definite???
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To show that it is positive semi definite we need to show that $z^T\Sigma z \ge 0$ for all $z$: \begin{align*} z^T\Sigma z &= \sum_{i=1}\sum_{j=1} (\Sigma_{ij}z_iz_j) \\ &= \sum_{i=1} \sum_{j=1} (E[(X_i - EX_i)(X_j - EX_j)])z_iz_j\\ &= E \Bigg[ \sum_i \sum_j (X_i - EX_i)(X_j - EX_j) z_iz_j \Bigg] \ge 0\\ \end{align*} So for any random vector $X$, the covariance matrix is positive semidefinite.
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Thank you. That helps a lot. But I'm still wondering what will happen if it's negative definite. Does it mean that the variance would be negative?(which of course doesn't make any sense) – user10386405 Oct 06 '20 at 08:33