Convolution of multivariate Gaussian

Question

Let $f: \mathbb R^n \rightarrow \mathbb R$ be a multivariate Gaussian and $g: \mathbb R^n \rightarrow \mathbb R$ a multivariate dirac-function, namely with $i,j,m,n\in\mathbb N,m<n$ and $x=(x_1,\ldots,x_n)$ and $a,b\in\mathbb R$:

\begin{align} & f(x)= \frac {1}{\sqrt {(2\pi )^k|{\boldsymbol {\Sigma }}|}} \cdot \exp \left(-{\frac {1}{2}}({\mathbf {x} }-{\boldsymbol {\mu }})^{\mathrm {T} }{\boldsymbol {\Sigma }}^{-1}({\mathbf {x} }-{\boldsymbol {\mu }})\right) \\[10pt] & g(x) = \begin{cases} \lim\limits_{a\rightarrow0,\ b\rightarrow\infty} \quad \dfrac{1}{a^m\cdot b^{n-m}} & |x_i|\le\frac a2,1\le i\le m,\quad |x_j|\le\frac b2,m<j\le n \\[6pt] \quad 0 & \text{otherwise.} \end{cases} \end{align}

So bascially $g$ is a boxcar function where I have some of the borders shrink in some other expand, so that I get sort of a tube in 2D. If I have not done any mistakes the volume of $g$ should be $1.$ Let us further assume, the simplification that $\Sigma$ is a multiple of the identify matrix.

I would like to convolve these two to $h(x) = (f ∗ g)(x)$. If we would have a regular dirac function with $\lim\limits_{a\rightarrow0,\ b\rightarrow0}$, then the result would be the $h(x)=f(x)$, however in my case this is not the case. Here, we get for the partially evaluated function

$$ h(x\mid\ x_i=c_i, m<i) = \text{constant} $$

where all $c_i\in\mathbb R$. Now, my question: Is

$$ h(x\mid\ x_i=c_i, i\le m) \overset?\propto \exp \left(-{\frac {1}{2}}({\mathbf {x} }-{\boldsymbol {\mu }})^{\mathrm {T} }{\boldsymbol {\Sigma^\star }}^{-1}({\mathbf {x} }-{\boldsymbol {\mu }})\right) $$

a multivariate Gaussian (of dimension $m$)? How would $\Sigma^\star$ would be scaled compared to $\Sigma$?

Answer 1

EDIT: I simplified the notation a bit to make the answer clearer.

I assume your limiting function $g$ should actually be $\prod_{i=1}^m \delta(x_i)$ ( the normalization by $1\over b$ will cause the entire convolution to vanish, so I treat it as a confusion). In that case, the convolution $$ \int _{-\infty}^{\infty}f(x-\xi)g(\xi)\,d\xi $$ becomes $$ {1\over{\sqrt {(2\pi )^k|{\boldsymbol {\Sigma }}|}}} \int _{-\infty}^{\infty} \exp \left(-{\frac {1}{2}}({\mathbf {\xi} }-{\boldsymbol {\mu }})^{\mathrm {T} }{\boldsymbol {\Sigma }}^{-1}({\mathbf {\xi} }-{\boldsymbol {\mu }})\right) \prod_{i=1}^m \delta(x_i-\xi_i)\prod_{j=1}^{m} d\xi_j\prod_{k=m+1}^{n} d\xi_k $$

At this point some simplification can be made by noticing that the $\xi$ are integration variables, and we can "swallow" the $\mu$ into them, integrating over $\xi-\mu$ instead. This leaves us with $$ {1\over{\sqrt {(2\pi )^k|{\boldsymbol {\Sigma }}|}}} \int _{-\infty}^{\infty} \exp \left(-{\frac {1}{2}}({\mathbf {\xi} })^{\mathrm {T} }{\boldsymbol {\Sigma }}^{-1}({\mathbf {\xi} })\right) \prod_{i=1}^m \delta(x_i-\xi_i-\mu_i)\prod_{j=1}^{m} d\xi_j\prod_{k=m+1}^{n} d\xi_k $$

if we split the inverse covariance $\Sigma^{-1} $ into 4 parts: $\Sigma^{-1}_{<<};\Sigma^{-1}_{>>};\Sigma^{-1}_{><};\Sigma^{-1}_{<>}$ where the subscripts determine whether the indices are $i\le m;i>m$ respectively, we see that the expression in the exponent can be split into $$({\mathbf {\xi} })^{\mathrm {T} }{\boldsymbol {\Sigma }}^{-1}({\mathbf {\xi} }) = ({\mathbf {\xi}_{<} })^{\mathrm {T} } {\boldsymbol {\Sigma }}^{-1}_{<<} ({\mathbf {\xi}_{<} })+ ({\mathbf {\xi}_{>} })^{\mathrm {T} } {\boldsymbol {\Sigma }}^{-1}_{>>} ({\mathbf {\xi}_{>} })+ 2({\mathbf {\xi_{<}} })^{\mathrm {T} } {\boldsymbol {\Sigma }}^{-1}_{<>} ({\mathbf {\xi}_{>} })$$

The $\delta$'s enforce $\xi=x-\mu$ on the $_<$ indices, turning the expression in the exponent to $$ ({\mathbf {x}_{<} }-{\boldsymbol {\mu }_{<}})^{\mathrm {T} } {\boldsymbol {\Sigma }}^{-1}_{<<} ({\mathbf {x}_{<} }-{\boldsymbol {\mu }}_{<})+ ({\mathbf {\xi}_{>} })^{\mathrm {T} } {\boldsymbol {\Sigma }}^{-1}_{>>} ({\mathbf {\xi}_{>} })+ 2({\mathbf {x_{<}} }-{\boldsymbol {\mu }_{<}})^{\mathrm {T} } {\boldsymbol {\Sigma }}^{-1}_{<>} ({\mathbf {\xi}_{>} }) $$ and we are left with an integral over the $\xi_>$ terms: $$ {1\over{\sqrt {(2\pi )^k|{\boldsymbol {\Sigma }}|}}} \left(\exp(-{1\over2}({\mathbf {x}_{<} }-{\boldsymbol {\mu }_{<}})^{\mathrm {T} } {\boldsymbol {\Sigma }}^{-1}_{<<} ({\mathbf {x}_{<} }-{\boldsymbol {\mu }}_{<}) \right)\int \prod d{\xi}_> \exp\left(-{1\over2} \left(({\mathbf {\xi}_{>} })^{\mathrm {T} } {\boldsymbol {\Sigma }}^{-1}_{>>} ({\mathbf {\xi}_{>} })+ 2({\mathbf {x_{<}} }-{\boldsymbol {\mu }_{<}})^{\mathrm {T} } {\boldsymbol {\Sigma }}^{-1}_{<>} ({\mathbf {\xi}_{>} }) \right)\right) $$

which can be calculated by completing the square.

Answer 2

To add to the answer of user619894:

---

First we rewrite

\begin{align} ({\mathbf {\xi}_{>} }-{\boldsymbol {\mu }_{>}})^{\mathrm {T} } {\boldsymbol {\Sigma }}^{-1}_{>>} ({\mathbf {\xi}_{>} }-{\boldsymbol {\mu }_{>}})+ 2({\mathbf {x_{<}} }-{\boldsymbol {\mu }_{<}})^{\mathrm {T} } {\boldsymbol {\Sigma }}^{-1}_{<>} ({\mathbf {\xi}_{>} }-{\boldsymbol {\mu }_{>}}) \\ \end{align}

with

\begin{align} a &:= (\mathbf\xi_>-\boldsymbol\mu_>) \\ M &:= \boldsymbol\Sigma_{>>}^{-1} \\ b &:= -(({\mathbf {x_{<}} }-{\boldsymbol {\mu }_{<}})^{\mathrm {T} } {\boldsymbol {\Sigma }}^{-1}_{<>})^{\mathrm {T} } = -(\boldsymbol\Sigma_{<>}^{-1})^T \cdot (\mathbf x_<-\boldsymbol\mu_<) \end{align}

to

\begin{align} & a^T \cdot M \cdot a - 2 \cdot b^T \cdot a \\ =& (a-M^{-1}\cdot b)^T \cdot M \cdot (a-M^{-1}\cdot b) - b^T\cdot M^{-1} \cdot b \\ \end{align}

thanks to http://gregorygundersen.com/blog/2019/09/18/completing-the-square/ which becomes

\begin{align} &( (\mathbf\xi_>-\boldsymbol\mu_>) + \boldsymbol\Sigma_{>>} \cdot (\boldsymbol\Sigma_{<>}^{-1})^T \cdot (\mathbf x_<-\boldsymbol\mu_<) )^T \\ &\cdot \boldsymbol\Sigma_{>>}^{-1} \cdot \\ &( (\mathbf\xi_>-\boldsymbol\mu_>) + \boldsymbol\Sigma_{>>} \cdot (\boldsymbol\Sigma_{<>}^{-1})^T \cdot (\mathbf x_<-\boldsymbol\mu_<) ) \\ &- (\mathbf x_<-\boldsymbol\mu_<)^T \cdot (\boldsymbol\Sigma_{<>}^{-1})^T \cdot \boldsymbol\Sigma_{>>} \cdot (\boldsymbol\Sigma_{<>}^{-1})^T \cdot (\mathbf x_<-\boldsymbol\mu_<) \end{align}