Convolution of multivariate distributions wrt linear subspace

Question

Is it possible to perform analytically the following integral? $$ \int d^n a_\alpha \exp \left[ -\frac{1}{2} (z_i -a_{\alpha}g_{\alpha i})C^{-1}_{ij} (z_j -a_{\beta}g_{\beta j}) \right] \exp \left[ -\frac{1}{2} (x_i -a_{\alpha}g_{\alpha i})C^{-1}_{ij} (x_j -a_{\beta}g_{\beta j}) \right] $$ where $C_{ij}$ is the symmetric $N$ X $N$ covariance matrix, $g_{\alpha i}$ is a $n$ X $N$ matrix, repeated indexes are summed over, and $n\le N$.

If $N=n$ and $\det g \neq 0$, then one can change variables and the integral becomes a simple convolution (and one sums the covariance matrixes and means). And if $n<N$?

Answer 1

Not with this notation. If $M^2$ is a positive definite matrix then $$\int_{\mathbb{R}^n} e^{-\pi (\mu- G x)^T M^2 (\mu- G x) } e^{-2 i \pi \langle x, \xi\rangle} d^n x$$ is directly related to the Fourier transform of the Gaussian, thanks to the change of variable $$y = M (\mu -G x), \qquad d^n y = |\det MG|d^n x$$

By analytic continuation, the obtained formula stays true for $\xi \in \mathbb{C}^n$.

Note $e^{-\pi t^2}$ and hence $e^{-\pi \|x\|^2}$ is its own Fourier transform.

Answer 2

I assume that the $a_\alpha$-dependence is explicit, otherwise I don't think there exist an analytic formula.

In this case I guess you can open up the products in the exponentials as follows:

$$\int d^n a_\alpha \exp \left[-\frac{1}{2}z_iC^{-1}_{ij}z_j -\frac{1}{2}x_iC^{-1}_{ij}x_j - \frac{1}{2}A_{\alpha\beta}a_\alpha a_\beta + B_\alpha a_\alpha\right]$$

where I have defined

$$A_{\alpha\beta} \equiv 2g_{\alpha i}C^{-1}_{ij}g_{\beta j}$$

$$B_\beta \equiv z_iC^{-1}_{ij}g_{\beta j} + x_iC^{-1}_{ij}g_{\beta j}$$

which you should be able to compute. I have used the symmetry of the covariance matrix in order to define $B_\alpha$. Finally, use the Gaussian integral and obtain

$$\exp \left[-\frac{1}{2}z_iC^{-1}_{ij}z_j -\frac{1}{2}x_iC^{-1}_{ij}x_j\right]\sqrt{\frac{(2\pi)^n}{\det A}} \exp\left[ \frac{1}{2}B_\alpha A^{-1}_{\alpha\beta}B_\beta\right]$$