- Let us assume that $\rho(\mathbf{x})=-\frac{2}{\gamma} \log \mathbb{E}\left[\exp \left(-\frac{\gamma}{2}\left\|\mathbf{x}-\mathbf{x}^{r e f}\right\|_{Q}^{2}\right)\right]$, where $\mathbf{e}_k=\mathbf{x}_k-\mathbf{x}_k^{r e f} \sim N\left(0, \Sigma_{k}\right)$.
- My Questions are:
- How can I get this form? $\rho\left(\mathbf{x}_{k}\right)=\left\|\mathbf{x}_k-\mathbf{x}_k^{r e f}\right\|_{Q}^{2}-\frac{2}{\gamma} \log \mathbb{E}\left[\exp \left\{-\frac{\gamma}{2}\left\|\mathbf{e}_{k}\right\|_{Q}^{2}-\gamma \mathbf{e}_{k}^{T} Q\left(\mathbf{x}_k-\mathbf{x}_k^{r e f}\right)\right\}\right]$
- How can I end up with this formulation? $\rho\left(\mathbf{x}_{k}\right)=\frac{1}{\gamma} \log \operatorname{det}\left(\gamma Q \Sigma_{k}+I\right)+\left\|\mathbf{x}_k-\mathbf{x}_k^{r e f}\right\|_{\left(Q^{-1}+\gamma \Sigma_{k}\right)^{-1}}^{2}$
What I know is that $\rho(\mathbf{x})$ can be approximated using Taylor expansion as follows: $\rho(\mathbf{x}) = \mathbb{E}(\mathbf{e}) - \frac{1}{4}\gamma \mathbb{Var}(\mathbf{e})$
I would be grateful if anyone can help me to get the derivation. It has been proven in this article (See the Appendix) but I am not sure whether it's correct or not.
Many thanks in advance!
First, you need to distinguish $\mathbf x^{ref}$ and $\overline{\mathbf x}$. The first one is a reference value, and the second one is the true value. Both of them are not random!
$$\rho(\mathbf x) = -\frac2\gamma \log\mathbb E\left[\exp \left\{-\frac\gamma2\left\|\mathbf x - \mathbf x^{ref}\right\|_Q^2\right\}\right] = -\frac2\gamma \log\mathbb E\left[\exp \left\{-\frac\gamma2\left(\mathbf x - \mathbf x^{ref}\right)^TQ\left(\mathbf x - \mathbf x^{ref}\right)\right\}\right].$$
Let $\mathbf x_k = \overline{\mathbf x}_k + \mathbf e_k$, where $\mathbf e_k \sim \mathcal N\left(0, \Sigma_k\right)$, replace it $\rho$, and you have:
\begin{align} \rho\left(\mathbf x_k\right) &= -\frac2\gamma\log \mathbb E\left[\exp \left\{-\frac\gamma2\left(\overline{\mathbf x}_k + \mathbf e_k - \mathbf x^{ref}\right)^TQ\left(\overline{\mathbf x}_k + \mathbf e_k - \mathbf x^{ref}\right)\right\}\right]\\ &= -\frac2\gamma\log \mathbb E\left[\exp \left\{-\frac\gamma2\left(\overline{\mathbf x}_k - \mathbf x^{ref}\right)^TQ\left(\overline{\mathbf x}_k - \mathbf x^{ref}\right) -\gamma \mathbf e_k^TQ\left(\overline{\mathbf x}_k - \mathbf x^{ref}\right) -\frac\gamma2\mathbf e_k^TQ\mathbf e_k\right\}\right]\\ &= -\frac2\gamma\log \mathbb E\left[\underbrace{\exp \left\{-\frac\gamma2\left(\overline{\mathbf x}_k - \mathbf x^{ref}\right)^TQ\left(\overline{\mathbf x}_k - \mathbf x^{ref}\right)\right\}}_{\text{constant can go out from the $\mathbb E$}}\exp\left\{-\gamma \mathbf e_k^TQ\left(\overline{\mathbf x}_k - \mathbf x^{ref}\right) -\frac\gamma2\mathbf e_k^TQ\mathbf e_k\right\}\right] \end{align}
and you will have the first formula. To answer the second question let $$\mathbf y_k = \left(\gamma Q + \Sigma_k^{-1}\right)^{-1}Q\left(\overline{\mathbf x}_k - \mathbf x^{ref}\right)$$, compute $$\frac{1}{2}\left(\mathbf e_k - \mathbf y_k\right)^T\left(\gamma Q + \Sigma_k^{-1}\right)\left(\mathbf e_k - \mathbf y_k\right) - \frac12\mathbf y_k^T\left(\gamma Q + \Sigma^{-1}_k\right)\mathbf y_k$$
to obtain
$$\frac\gamma2\mathbf e_k^TQ\mathbf e_k + \frac12\mathbf e_k\Sigma_k^{-1}\mathbf e_k + \gamma \mathbf e_k^TQ\left(\overline{\mathbf x}_k - \mathbf x^{ref}\right)$$
With that you can show: \begin{align} -\frac2\gamma \log \mathbb E\left[\exp \left\{-\gamma \mathbf e_k^TQ\left(\overline{\mathbf x}_k - \mathbf x^{ref}\right) -\frac\gamma2\mathbf e_k^TQ\mathbf e_k\right\}\right] &= -\frac2\gamma \log \left(\frac{1}{\sqrt{(2\pi)^n\det \Sigma_k }}\int_{\mathbb R^n} \exp\left\{\frac{1}{2}\left(\mathbf e_k - \mathbf y_k\right)^T\left(\gamma Q + \Sigma_k^{-1}\right)\left(\mathbf e_k - \mathbf y_k\right) - \frac12\mathbf y_k^T\left(\gamma Q + \Sigma^{-1}_k\right)\mathbf y_k\right\}\mathrm d \mathbf e_k\right) \end{align}
Use the substition $\mathbf z_k = \mathbf e_k -\mathbf y_k$ and you can compute the integral.