# Showing the multivariate normal is log-concave?

I'm trying to show that $\log p(x) = -\frac{1}{2}(x-\mu)^{T} \Sigma^{-1}(x-\mu)$ is concave. How would I go about this in $\mathbb{R}^n$? I've tried taking derivatives but I'm getting stuck once I get to $$\nabla\log p(x) = - \det[ (x-\mu)^{T}\Sigma^{-1}(x-\mu) ] * \Sigma^{-1}(x-\mu) * [(x-\mu)^{T}\Sigma^{-1}(x-\mu)]^{-1}.$$ I still would need to take another gradient with respect to $x\in \mathbb{R}^n$ so this takes me nowhere. Can someone help?

Btw, $\mu$ is the mean and $\Sigma$ is the covariance matrix.

@ Superman: for simplicity, I'll write $l=\log p(x)$ and $A=\Sigma^{-1}$. Then $$\nabla l=\frac{\partial}{\partial x}\left[-\frac{1}{2}(x-\mu)^{T}A(x-\mu)\right]=-\frac{1}{2}(A+A^T)(x-\mu)=-A(x-\mu)$$ where the last equality is due to the symmetry of $A$. It follows that the Hessian of $l$ is $$H=\frac{\partial}{\partial x^T}\nabla l =\frac{\partial}{\partial x^T}-A(x-\mu)=-A$$ which is negative definite. So $l$ is concave.
p.s. We have used a result that if $X$ is symmetric and positive definite, then $X^{-1}$ is symmetric and positive definite. This follows from $$(X^{-1})^T=(X^T)^{-1}=X^{-1},\\ v^TX^{-1}v=v^T X^{-1}X X^{-1}v=(X^{-1}v)^TX(X^{-1}v).$$
• How did you get $\nabla l = -\frac{1}{2}(A+A^T)(x-\mu)$? Sep 18, 2014 at 16:54