integral of squared exponential function and multivariate normal product

Question

I found in a paper that the multi-dimensional integral of a squared exponential correlation function times a multivariate normal has an analytical expression as follows. \begin{equation} \int\exp{\left\{-(\theta-c)^T\Omega_t (\theta-c)\right\}}(2\pi)^{-k/2}|V_\theta|^{-\frac{1}{2}}\exp{\left\{-\frac{1}{2}(\theta-m_\theta)^TV_\theta^{-1} (\theta-m_\theta)\right\}} d\theta= \end{equation} \begin{equation} =|I+2V_\theta\Omega_t|^{-1/2}\exp{\left\{-(m_\theta-c)^T(2V_\theta+\Omega_t^{-1})^{-1} (m_\theta-c)\right\}} \end{equation} where $c$ is a constant, $\Omega_t$ is a $k$-dimensional diagonal matrix with the inverse of the length scales of the correlation function and $m_\theta$ and $V_\theta$ are the mean and covariance of the MVN distribution, respectively. I would like to see a proof or have a hint as to how the analytic expression is as shown above. I've searched and found the following related posts

Sum of two quadratic forms

Integral of product of two normal distribution densities

https://en.wikipedia.org/wiki/Gaussian_integral#n-dimensional_and_functional_generalization

Thank you in advance.

Answer 1

I was able to find a breakthrough to crack it. Essentially I've used the equation shown in this answer https://math.stackexchange.com/a/2996439/608359, and realised that the matrix in the numerator can be written much more simply as $(\Gamma + \Sigma)^{-1}$. Similarly, the determinants in the denominator can also be simplified in a similar manner. Therefore, it is necessary to convert the correlation function to a MVN distribution, and then to apply the simplified formula

$$\int_{\mathbb{R}^D} \rho_{\mu, \Sigma}(x)\cdot\rho_{\mathbf{0},\Gamma}(x)\,dx=\frac{\exp\left(-\frac 12 (\mu^T(\Gamma+\Sigma)^{-1}\mu) \right)} {\sqrt{(2\pi)^D |\Sigma||\Gamma\Sigma^{-1}+I|}}$$