Maximization of log likelihood wrt $\Sigma_{t}$ subject to $\Sigma_{t}$ is positive definite

Given the random vector $$y_{t}=A_{t}^{1/2}\epsilon_{t}$$ with $$\epsilon_{t}=N(0,I)$$, where $$A_{t}$$ is a positive definite covariance matrix for every t, I want to minimize the negative log marginal likelihood for every t: $$-\log p(y|X) = \frac{1}{2}y_{t}^T\Sigma_{t}^{-1}y_{t} + \frac{1}{2}\log|\Sigma_{t}|+\frac{n}{2}\log2\pi$$

wrt to $$\Sigma_{t}$$ (notice that, for every t, $$\Sigma_{t}$$ is indeed allowed to change). Setting the gradient to $$0$$, for a generic t, I get:

$$\Sigma_{t}=y_{t}y_{t}^{T}=A_{t}^{1/2}\epsilon_{t}\epsilon_{t}^{T}A_{t}^{1/2T}$$

which clearly is not positive definite because is the outer product of a vector $$y_{t}$$ so it has rank 1.

So I should impose the constraint $$det(\Sigma_{t})>0$$ and solve the constrained optimum with the Lagrangian. However I know that $$E(\epsilon_{t}\epsilon_{t}^{T})=I$$, so that, on average for a generic observation of $$y_{t}y_{t}^{T}$$, I will have have that $$\epsilon_{t} \epsilon_{t}^{T}=I$$ and therefore setting $$\Sigma_{t}=A_{t}^{1/2}A_{t}^{1/2T}$$ will maximize the log-likelihood of a generic observation under $$det(\Sigma_{t})>0$$.

Is there an elegant way of saying so and avoid solving the maximization with the Lagrangian and the constraint $$det(\Sigma_{t})>0$$?

• $\log \det$ is concave, thus $-\log \det$ is convex. How do you know you're maximizing a concave function ? – Gabriel Romon Aug 23 '19 at 21:32
• Gabriel suppose (as edited) I want to minimize the following (edited) function. if I take the derivative of the gradient wrt to $\Sigma$ I now get $yy^{T}$ which is psd, therefore that is a minimum. – JMallin Aug 23 '19 at 22:08
• The minimum you exhibit is local though. – Gabriel Romon Aug 23 '19 at 22:11
• Ok thanks, anyway the problem to me is: how can I show that this is a min st $det \Sigma>0$. As an alternative do you have the derivative of $det(\Sigma)$? But I would prefer to just go without and show that for a generic observation $yy^{T}=A$ Holds so I can take $\Sigma=A$.. many thanks in advance Gabriel – JMallin Aug 23 '19 at 22:22

We know that, assuming the process is $$y_{t}=A_{t}^{1/2}\epsilon_{t}$$ with iid $$\epsilon_{t}=N(0,I)$$ and positive definite $$A_{t}$$ for every t, then $$y_{t}y_{t}^{T}=A_{t}^{1/2}\epsilon_{t}\epsilon_{t}^{T}A^{1/2T}$$ which implies $$E_{t-1}(y_{t}y_{t}^{T})=A_{t}$$, therefore we can write $$y_{t}y_{t}^{T}=A_{t}+U$$ where U is such that $$E_{t-1}(U)=0$$ and $$E(U)=0$$. Therefore the –log likelihood to be minimized wrt $$\Sigma_{t}$$ can be re-written as $$-\log p(y|X) = \sum_{t=1}^{N} \frac{1}{2}y_{t}^T\Sigma_{t}^{-1}y_{t} + \frac{1}{2}\log|\Sigma_{t}|+\frac{n}{2}\log2\pi = \sum_{t=1}^{N} \frac{1}{2}\log|\Sigma_{t}|+\frac{n}{2}\log2\pi + \frac{1}{2}Tr(\Sigma_{t}^{-1}y_{t} y_{t}^{T})$$
Substituting $$y_{t}y_{t}^{T}=A_{t}+U$$ $$-\log p(y|X) = \sum_{t=1}^{N} \frac{1}{2}\log|\Sigma_{t}|+\frac{n}{2}\log2\pi + \frac{1}{2}Tr(\Sigma_{t}^{-1} (A_{t}+U))$$
Notice that $$Tr(\Sigma_{t}^{-1} (A_{t}+U)) = Tr(\Sigma_{t}^{-1} A_{t}+\Sigma_{t}^{-1} U) = Tr(\Sigma_{t}^{-1} A_{t})+Tr(\Sigma_{t}^{-1} U)$$. Notice also that, since Tr is a linear operator and commutes with expectations (see for example this), then the followinhg holds $$E(Tr(\Sigma_{t}^{-1}U))= Tr(E(\Sigma_{t}^{-1})E(U))$$. Assuming that $$E(U)=0$$ holds in the sample (i.e. $$1/N \sum_{t=1}^{N} U_{t} =0 \rightarrow \sum_{t=1}^{N} Tr(\Sigma_{t}^{-1} U)=0$$ with some neglectable rounding in a large well-behaved sample), then we could write a proxy for the -log likelihood as: $$-\log p(y|X) = \sum_{t=1}^{N} \frac{1}{2}\log|\Sigma_{t}|+\frac{n}{2}\log2\pi + \frac{1}{2}Tr(\Sigma_{t}^{-1} A_{t})$$ Taking the gradient wrt a generic $$\Sigma_{t}$$ and setting it equal to 0, we get: $$\Sigma_{t}=A_{t}$$ Where $$A_{t}$$ is pd by assumption. Notice also that taking and additional derivation, the Hessian is $$A_{t}$$ which is pd by assumption, so we have found a minimum for the negative log marginal likelihood, after rounding the expression of the sample negative log likelihood. So we do not need to impose the constraint $$det(\Sigma_{t})>0$$ to show that the choice of $$A_{t}$$ is the choice of $$\Sigma_{t}$$ that minimizes the –log lik subject to the requirement that $$\Sigma_{t}$$ is pd for every t.