On integrating a "gaussian-like" integral

Question

Let the following "gaussian-like" integral:

$$ I = \int_{\Re^n} \! \frac{1}{(2\pi)^{n/2}|\Sigma|^{1/2}} \exp \left\{ -\frac{1}{2} (\mathbf{x}-\mathbf{\mu})^T\Sigma^{-1}(\mathbf{x}-\mathbf{\mu}) \right\} \mathbf{x} \,\mathbf{d}\mathbf{x}, $$

where $\mathbf{x}=(x_1,\dots,x_n)^T$, $\mathbf{\mu} = (\mu_1,\dots,\mu_n)^T\in\Re^n$, and $\Sigma\in\mathbb{S}_{++}^{n}$.

Our main goal is to evaluate the above integral. To this end, let $\mathbf{x}-\mathbf{\mu}=S\mathbf{y}$, where $S$ is an $n \times n$ orthogonal matrix ($S^T=S^{-1}$) with determinant equal to $1$. Using this change of variable, the quadratic form shown in the integral written as:

$$ -\frac{1}{2}(\mathbf{x}-\mathbf{\mu})^T\Sigma^{-1}(\mathbf{x}-\mathbf{\mu})= -\frac{1}{2}\mathbf{y}^T(S^T\Sigma^{-1}S)\mathbf{y}= -\frac{1}{2}\mathbf{y}^T(S^{-1}\Sigma^{-1}S)\mathbf{y}= -\frac{1}{2}\mathbf{y}^TD\mathbf{y}, $$

where $D=\operatorname{diag}\{d_1,\dots,d_n\}$. As a result it is rewritten as follows:

$$ -\frac{1}{2}(\mathbf{x}-\mathbf{\mu})^T\Sigma^{-1}(\mathbf{x}-\mathbf{\mu})= -\frac{1}{2}\sum_{k=1}^{n} d_i y_i^2 $$

Moreover, $\mathbf{x}-\mathbf{\mu}=S\mathbf{y} \Rightarrow \mathbf{x}=S\mathbf{y}+\mathbf{\mu}=[\mathbf{s_1}\:\dots\:\mathbf{s_n}]\mathbf{y}+\mathbf{\mu}=(\mathbf{s_1}\cdot\mathbf{y}+\mu_1,\dots, \mathbf{s_n}\cdot\mathbf{y}+\mu_n)^T,$ where $\mathbf{s}_j$ is the $j$-th column of matrix $S$.

Using the above results, the original integral can be rewritten as follows:

$$ I = (I_1,\dots,I_n)^T, $$

where the $j$-th element of $I$ is given by:

$$ I_j = \int_{\Re^n} \! \frac{1}{(2\pi)^{n/2}|\Sigma|^{1/2}} \exp \left\{ -\frac{1}{2}\sum_{k=1}^{n} d_i y_i^2 \right\} (\mathbf{s}_j\cdot\mathbf{y}+\mu_j) \,\mathbf{d}\mathbf{y}\\ = \int_{\Re^n} \! \frac{1}{(2\pi)^{n/2}|\Sigma|^{1/2}} \exp \left\{ -\frac{1}{2}\sum_{k=1}^{n} d_i y_i^2 \right\} \mathbf{s}_j\cdot\mathbf{y} \,\mathbf{d}\mathbf{y}\\ + \int_{\Re^n} \! \frac{1}{(2\pi)^{n/2}|\Sigma|^{1/2}} \exp \left\{ -\frac{1}{2}\sum_{k=1}^{n} d_i y_i^2 \right\} \mu_j \,\mathbf{d}\mathbf{y} \Rightarrow\\ I_j = \int_{\Re^n} \! \frac{1}{(2\pi)^{n/2}|\Sigma|^{1/2}} \exp \left\{ -\frac{1}{2}\sum_{k=1}^{n} d_i y_i^2 \right\} \mathbf{s}_j\cdot\mathbf{y} \,\mathbf{d}\mathbf{y} + \mu_j $$

If we write the dot product $\mathbf{s}_j\cdot\mathbf{y}$ as

$$ \mathbf{s}_j\cdot\mathbf{y} = \sum_{r=1}^{n} s_{jr}y_r, $$

then the integral $I_j$ is given by:

$$ I_j = \int_{\Re^n} \! \frac{1}{(2\pi)^{n/2}|\Sigma|^{1/2}} \left(\sum_{r=1}^{n} s_{jr}y_r\right) \exp \left\{ -\frac{1}{2}\sum_{k=1}^{n} d_i y_i^2 \right\} \,\mathbf{d}\mathbf{y} + \mu_j $$

I would like to ask, first, whether the whole approach above is correct or not(if so, please correct me), and, second, how could I evaluate the last integral, $I_j$. Does it converge, like the gaussian integral over $\Re^n$?

Thanks in advance! Every useful comment will be extremely appretiated!

Answer 1

Consider a vector whose $j$th component is \begin{align} & \phantom{={}}\text{constant}\cdot\int_{\mathbb R^n} \frac{1}{\sqrt{d_1\cdots d_n}} \exp\left(\frac{-1}{2} \sum_{k=1}^n d_k y_k^2 \right) y_j\,dy_1\cdots dy_n \\[12pt] & = c\int_{\mathbb R^n} \prod_{k=1}^n\left(\frac{1}{\sqrt{d_k}} \exp\left(\frac{-1}{2} d_ky_k^2\right)\right) y_j\,dy_1 \cdots dy_k \\[12pt] & = c\prod_{k=1}^n \int_{\mathbb R} \frac{1}{\sqrt{d_k}} \exp\left(\frac{-1}{2} d_ky_k^2\right) y_k\, dy_k. \end{align}

So it's reducible to integrals over $\mathbb R^1$, and if you're thinking about this particular problem, you probably know how to evaluate these particular integrals.

It can often happen that the purpose of diagonalizing a matrix is to reduce a problem involving a vector in $n$-space to $n$ problems involving scalars.