Sylvester's determinant theorem states that for matrices $A\in\mathbb{R}^{n\times d}, B\in\mathbb{R}^{d\times n}$:
\begin{equation} \det(I_{n}+AB)=\det(I_d+BA) \end{equation}
In my case I consider $A$ and $B$ to be pos. def., so that by extension I may consider the $\log\circ\det$ and it is a concave function:
\begin{equation} \log\circ\det(I_{n}+AB)=\log\circ\det(I_d+BA) \end{equation}
Because of the concavity I am interested to compute Jensen's inequality (so I need to pass in the expectation). From computer experiments I obtain the following:
\begin{align} \mathbb{E}[\log\circ\det(I_{n}+AB)]&=\mathbb{E}[\log\circ\det(I_d+BA)]\quad\text{($\mathbb{E}$[Sylvester's])}\\ \log\circ\det(I_{n}+\mathbb{E}[AB])&\neq\log\circ\det(I_d+\mathbb{E}[BA])\quad\text{(Why is it not "="?)} \end{align}
I am not sure why that inequality isn't an equality?
Overall the inequality chain is satisfied:
\begin{align} \mathbb{E}[\log\circ\det(I_{n}+AB)]=\mathbb{E}[\log\circ\det(I_d+BA)]\leq \log\circ\det(I_{n}+\mathbb{E}[AB]) \end{align}
And
\begin{align} \mathbb{E}[\log\circ\det(I_{n}+AB)]=\mathbb{E}[\log\circ\det(I_d+BA)]\leq \log\circ\det(I_{n}+\mathbb{E}[BA]) \end{align}
So that Jensen's inequality is satisfied as expected...
Ultimately I am unsure of why this statement is true:
\begin{align} \log\circ\det(I_{n}+\mathbb{E}[AB])&\neq\log\circ\det(I_d+\mathbb{E}[BA]) \end{align}
Could someone offer an explanation for this?
The logarithmic function and Jensen's inequality are irrelevant. The key issue here is that log-determinant is not an additive function, so that you cannot obtain $$ \log\det(I+E[AB])=\log\det(I+E[BA]) $$ from the equality $$ E[\log\det(I+AB)]=E[\log\det(I+BA)]. $$
However, we do have $\log\det(I+E[AB])=\log\det(I+E[BA])$ when $A$ and $B$ are sampled from a certain set of symmetric matrices of the same sizes (this is the case when $n=d$ and $A,B$ are positive definite matrices of the same sizes). This is simply because the determinant of a matrix is equal to the determinant of its transpose: \begin{aligned} \log\det(I+E[BA]) &=\log\det\left((I+E[BA])^\top\right)\\ &=\log\det(I+E[(BA)^\top])\\ &=\log\det(I+E[AB]). \end{aligned}