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I'm trying to analytically find the inverse of the covariance matrix generated by the exponential covariance function (also known as Ornstein-Uhlenbeck kernel) in $\mathbb{R}$, that is

$K_{ij} = k(x_i, x_j) = \exp(-\frac{|x_i - x_j|}{\theta})$

Where $x_i \in \mathbb{R}$ and that $K$ is an $n\times n$ matrix.

Here is what I've thought about doing:

Since $K$ is a covariance matrix, I know it is positive semidefinite and so is $K^{-1}$. We can do a singular value decomposition on $K = UDU^*$ and $K^{-1} = VCV^*$ where $U, V$ are unitary matrices and $D, C$ are diagonal matrices. But that I haven't been able to develop the idea further.

What would be right approach to get an analytical expression for $K^{-1}$?

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  • $\begingroup$ I worked out the math. It is a quite complicated expression but the main idea is to assume that the inverse is a symmetric tridiagonal matrix (which it is) and find the values for each of the entries. $\endgroup$ – Ana Echavarria Dec 9 '16 at 17:11
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I'll keep a $\sigma$ out front for completeness. Okay so assume $C(x,x') = \sigma^2 \exp(-|x-x'|/T)$, and $x \in \mathbb{R}$. (a more complicated but similar approach works for $x \in \mathbb{R}^n$ too)

The kernel associates a function $\hat{e}(x')$ with $e(x)$ via,

1) $e(x) = \int_{x_1}^{x_2} dx' C(x,x')\hat{e}(x')$.

Notice, if $\phi(x) = \sigma^2 \exp(-|x|/T)$ then two things:

2) $\partial_x \phi(x) = -\frac{1}{T}sign(x)\phi(x)$ and

3) $\partial_x^2 \phi(x) = -\frac{1}{T^2}\phi(x) - \frac{2\sigma^2}{T}\delta(x)$

therefore plugging this in,

4) $\partial_x e(x) = \frac{1}{-T}\int_{x_1}^{x_2} dx' sign(x-x')C(x,x')\hat{e}(x')$

and,

5) $\partial_x^2 e(x) = \frac{1}{T^2} - \frac{2 \sigma^2}{T} \hat{e}(x)$

using 1) and 4) shows that at $x=x_1$ and $x = x_2$ that $\partial_x e(x)$ and $e(x)$ are not independent.

6) $\partial_x e(x_1) = \frac{1}{T} e(x_1)$ and likewise for $x_2$.

Eq 5) then gives,

7) $\hat{e}(x) = \frac{1}{2\sigma^2 T} e(x) - \frac{T}{2\sigma^2} \partial_x^2 e(x)$

$C^{-1}(x,x')$ has domain of definition thus satifying relations 6) and accosiates via eq 7). Taking the integral representation,

8) $\hat{e}(x) = \int_{x_1}^{x_2} dx' C^{-1}(x,x')e(x')$

it then follows that,

$C^{-1}(x,x') = \frac{1}{2\sigma} (\frac{1}{T} \delta(x-x') - T \delta^{(2)}(x-x'))$

where we use $\int dx' \delta^{(n)}(x') f(x') = (-1)^n \partial_x^n f(x)$.

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We first write the kernel in the following form: $$exp(-||x-y||)=exp(\min\{x,y\})exp(-\max\{x,y\})=:p(\min\{x,y\})q(\max\{x,y\})$$

Then the inverse matrix is a symmetric tridiagonal matrix with entries: $$[K^{-1}]_{i,i}=\frac{p(x_{i+1})q(x_{i-1})-p(x_{i-1})q(x_{i+1})}{[p(x_{i+1})q(x_{i})-p(x_{i})q(x_{i+1})][p(x_{i})q(x_{i-1})-p(x_{i-1})q(x_{i})]} $$ $$[K^{-1}]_{i,i-1}=[K^{-1}]_{i-1,i}=\frac{-1}{p(x_{i})q(x_{i-1})-p(x_{i-1})q(x_{i})} $$ which is easier and more general. you can refer to the following paper for details: https://ldingaa.github.io/files/ScalableSK.pdf

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