# Expected value of log-likelihood and KL divergence

Background:

Let $$x_t = Ax_{t-1} + w_t$$ be a discrete linear time invariant system where:

• $$x_t \in \mathbb{R}^d$$ for all time samples $$t$$ corresponds to the state vector
• $$A\in \mathbb{R}^{d\times d}$$ is the system matrix
• $$w$$ is noise vector (i.i.d. from a Gaussian with zero mean and identity covariance)

In this reference the authors try to put low bounds on the number of samples needed to estimate the matrix $$A$$. They say they can consider a change-of-measurement argument (I am unfamiliar with) and use some $$A'\neq A$$ such as to compute the log-likelihood:

\begin{align} L_t = \log \left(\frac{f_A(x_1,\ldots,x_t)}{f_{A'}(x_1,\ldots,x_t)}\right) \end{align} where $$f_A = \prod_{s=1}^t f_A(x_s|x_{s-1})$$ is the density of the state distribution at time $$t$$ given that at time $$t-1$$ the state was $$x_{t-1}$$ and this density is Gaussian: $$\nu_{A,x} = \mathcal{Ax, Id}\\ \nu_{A',x} = \mathcal{Ax', Id}\\$$

Confusion:

I am struggling to understand the following derivation as presented in their paper.

\begin{aligned} &\mathbb{E}_{A}\left[L_{t}\right]=\mathbb{E}_{A}\left[\sum_{s=1}^{t} \mathbb{E}_{A}\left[\log \left(\frac{f_{A}\left(x_{s} \mid x_{s-1}\right)}{f_{A^{\prime}}\left(x_{s} \mid x_{s-1}\right)}\right) \mid \mathcal{F}_{s-1}\right]\right] \\ &\quad=\mathbb{E}_{A}\left[\sum_{s=1}^{t} \mathrm{KL}\left(\nu_{A, x_{s-1}}\,, \nu_{A^{\prime}, x_{s-1}}\right)\right] \\ &\quad=\mathbb{E}_{A}\left[\sum_{s=0}^{t-1} \frac{1}{2} x_{s}^{\top}\left(A-A^{\prime}\right)^{\top}\left(A-A^{\prime}\right) x_{s}\right] \end{aligned}

Specifically:

• In the first equality where does the second expectation value come from?
• Why the inner expectation in the first equality is the same as the KL divergence of the two distributions?
• More generally, what is the meaning and intuition of taking the expectation w.r.t. $$A$$ of either the function $$L_t$$?