# On integrating a "gaussian-like" integral

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!

• Might you have intended $(2\pi)^{n/2}$ instead of $(2\pi)^{1/2}$? If so, then what you have if you omit the $\mathbf x$ just before the $d\mathbf x$ is a probability density (provided I'm right in understanding that you meant $\Sigma$ is a positive-definite symmetric matrix). Then the integral would just be the expected value, which is the vector $\mu$. Commented Nov 15, 2013 at 23:05
• You don't seem to have expressed the determinant of the matrix as a function of $d_i$, $i=1,\dots,n$. Commented Nov 15, 2013 at 23:10
• @lcv That's correct, I omitted that, but I think it's rather obvisous. By the way, I will add it to the original post. Thanks! Commented Nov 15, 2013 at 23:10
• @MichaelHardy you're right. It's a typo, I will fix it. But, in that case, are you sure that $\mu$ is the right answer? What you state is rational, but can anyone else confirm? I'm confused! Thanks, anyway! Commented Nov 15, 2013 at 23:13
• @MichaelHardy, I am not sure I understand what you are saying about the determinant of the matrix? What am I supposed to do with the $d$'s? Thanks. Commented Nov 15, 2013 at 23:16

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.
• The integral as written above evaluates to $0$. Somewhere you pulled out a separate constant term depending on $\mu$. If we'd had $(y_k+\mu_k)$, then the integral would evaluate to $\mu_k$ because the part before $y_k\,dy_k$ is a density function if the value of the constant is right. So you should end up with $\mu$ as the value of the integral you started with. Commented Nov 15, 2013 at 23:33
• So, the original integral $I$ is equal to the mean vector $\mu$, right? That simple, right? And it can also be explained via the Gaussian distribution and it's expected value? Commented Nov 15, 2013 at 23:39
• Correct. This is a way of proving that the expected value of a distribution with that density function is what its....expected....to be. Similarly if on integrates the $n\times n$ matrix $(\mathbf{x-\mu})(\mathbf{x-\mu})^T$ times that density function, one should get $\Sigma$. Commented Nov 15, 2013 at 23:44