# Any closed-form solution to this integral (multivariate exponential)?

Here is the probability density function (unnormalized) of a covariance matrix: (from a Bayesian perspective):

$$f(\boldsymbol{V})\propto \det(\boldsymbol{V})^{-\frac{N+J+1}{2}}\int_{\mathbb{R}^{K}}\exp \left [-\frac{1}{2}\sum_{i=1}^{N}(\boldsymbol{y}_{i}-\boldsymbol{X}_{i}\boldsymbol{\beta})^{\top}\boldsymbol{V}^{-1}(\boldsymbol{y}_{i}-\boldsymbol{X}_{i}\boldsymbol{\beta}) \right]d\boldsymbol{\beta}$$

where $$\boldsymbol{\beta}$$ is a $$K \times 1$$ vector,

$$\boldsymbol{y}_i$$ is a $$J \times 1$$ vector,

$$\boldsymbol{X}_i$$ is a $$J \times K$$ matrix,

$$\boldsymbol{V}$$ is a $$J \times J$$ covariance matrix, and

$$N>K>J$$. $$\space \mathbb{R}^{K}$$ is a $$K$$-dimensional real space.

The first $$J$$ columns of $$\boldsymbol{X}_i$$ are an identity matrix $$\boldsymbol{I}_{J \times J}$$. This is a multi-dimensional linear regression model.

My question is:

Is the integral $$\int_{\mathbb{R}^{K}}\exp \left [-\frac{1}{2}\sum_{i=1}^{N}(\boldsymbol{y}_{i}-\boldsymbol{X}_{i}\boldsymbol{\beta})^{\top}\boldsymbol{V}^{-1}(\boldsymbol{y}_{i}-\boldsymbol{X}_{i}\boldsymbol{\beta}) \right]d\boldsymbol{\beta}$$ analytically computable when $$\boldsymbol{V}$$ is fixed? It seems like a multivariate exponential integral. When $$K=J=1$$, this is a non-standard univariate inverse chi-squared distribution with $$N-1$$ degrees of freedom, as proven in Gelman's Bayesian Data Analysis (third edition) pp. 64-66.

• This is a standard multivariate Normal integral.
– whuber
Commented Sep 13, 2023 at 14:38
• @whuber I tried it, but the fact that the integrated variable is $\boldsymbol{\beta}$ seems to make it difficult. Commented Sep 14, 2023 at 13:49
• The sum is a sum of quadratic functions of $\beta$ and therefore is a quadratic function of $\beta.$ That means you can express it in the form $-\beta^\prime \Sigma \beta/2 + \tau^\prime \beta + u$ for some square symmetric matrix $\Sigma,$ vector $\tau,$ and number $u.$ In other words, it is a Gaussian function. You know how to integrate those.
– whuber
Commented Sep 14, 2023 at 14:20

Thanks for the hints by @whuber. Let me answer this question myself.

The quadratic term can be written as $$\sum_{i=1}^{N}(\boldsymbol{y}_{i}-\boldsymbol{X}_{i}\boldsymbol{\beta})^{\top} \boldsymbol{V}^{-1}(\boldsymbol{y}_{i}-\boldsymbol{X}_{i}\boldsymbol{\beta}) =\sum_{i=1}^{N}\boldsymbol{y}_{i}^{\top}\boldsymbol{V}^{-1}\boldsymbol{y}_{i} -2\boldsymbol{\beta}^{\top}\left(\sum_{i=1}^{N}\boldsymbol{X}_{i}^{\top} \boldsymbol{V}^{-1}\boldsymbol{y}_{i}\right) +\boldsymbol{\beta}^{\top}\left(\sum_{i=1}^{N}\boldsymbol{X}_{i}^{\top} \boldsymbol{V}^{-1}\boldsymbol{X}_{i}\right)\boldsymbol{\beta}$$

Let $$\boldsymbol{A}=\sum_{i=1}^{N}\boldsymbol{X}_{i}^{\top}\boldsymbol{V}^{-1}\boldsymbol{X}_{i}$$, $$\boldsymbol{b}=\sum_{i=1}^{N}\boldsymbol{X}_{i}^{\top}\boldsymbol{V}^{-1}\boldsymbol{y}_{i}$$, $$c=\sum_{i=1}^{N}\boldsymbol{y}_{i}^{\top}\boldsymbol{V}^{-1}\boldsymbol{y}_{i}$$. Here, $$\boldsymbol{A}$$ is a $$K \times K$$ symmetric positive definite matrix, $$\boldsymbol{b}$$ is a $$K\times 1$$ vector, and $$c$$ is a scalar.

Now the integral term is

$$\exp\left(-\frac{1}{2}c\right)\int_{\mathbb{R}^{K}} \exp\left(-\frac{1}{2}\boldsymbol{\beta}^{\top}\boldsymbol{A}\boldsymbol{\beta}+ \boldsymbol{b}^{\top}\boldsymbol{\beta}\right)d\boldsymbol{\beta}$$

Applying the Gaussian integral formula (n-dimensional with linear term) from Wikipedia, the integral term is

$$\exp\left(-\frac{1}{2}c\right) \exp\left(\frac{1}{2}\boldsymbol{b}^{\top}\boldsymbol{A}^{-1}\boldsymbol{b}\right) \sqrt{\frac{(2\pi)^{K}}{\det(\boldsymbol{A})}}$$