I could not understand and don't see how the author derived equation (16). The Covariance is with respect to a pdf $\mu_k$ for the random vector $u$. The variable $\xi$ is a ratio of two densities and is equal to $ \frac{\mu_{k+1}}{\mu_k}$, both $\mu_{k+1}$ and $\mu_k$ are densities from the exponential family. The function $h_k(u,v) = \frac{\phi(u,v)}{\mu_{k+1}} = \frac{h_{k-1}}{\xi_{k}}$. And $\phi$ is some given function.
1 Answer
Observe that \begin{array} = Cov\left[\log(\xi), \log\left(\frac{h_k^2}{\xi}\right)\right] \\ = Cov\left[\log(\xi) - \log(h_k) +\log(h_k), 2\log(h_k) - \log(\xi) \right] \\ = Cov\left[\log(\xi) - \log(h_k), \log(h_k) - \log(\xi) \right] + Cov\left[ \log(h_k), \log(h_k) \right] \\ = Var\left[\log(h_k)\right] - Var\left[ \log(h_k) - \log(\xi) \right] \end{array}
So it really is just rewriting the expression.