I'm reading through the following paper: https://arxiv.org/pdf/0704.1751.pdf

I'm stuck in the middle of page 8, at the statement:

$$E[||S(X)-S^*(X)||^2] = J(X) - J(X^*)$$

Where $S(X)$ is the score of the random vector $X$, $S^*(X)$ is the score of the gaussian r.v. with the same covariance as $X$, and $J(X)$ is the Fisher information of $X$ defined to be $\text{tr}(\text{Cov}(S(X), S(X)))$.

What I have so far is:

\begin{align} E[||S(X)-S^*(X)||^2] &= E[||S(X)||^2 - 2(S(X)\cdot S^*(X)) + ||S^*(X)||^2]\\ &= E[||S(X)||^2] - 2E[S(X)\cdot S^*(X)] + E[||S^*(X)||^2]\\ &= J(X) - 2E[S(X)\cdot S^*(X)] + J(X^*) \end{align}

From this point it suffices to show that: $$E[S(X)\cdot S^*(X)] = J(X^*)$$

It can be shown that the score of the associated Gaussian is $-K^{-1}x$, where $K=\text{Cov}(X),$ so in other words, we look to show: $$E[S(X)\cdot (-K^{-1}X)] = E[(K^{-1}X)\cdot (K^{-1}X)]$$

How can this be shown? Or is there an alternative way to show this?

Thanks.