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I am trying to understand intuitively how a kernel works in a Gaussian process. I know that GP are distributions over functions, in short you have the model $y = f(x)+\epsilon$ and the $f(x)$ follows a Gaussian process

$GP(m(x),K(x,x^{\prime}))$.

After deriving the conditional distribution we arrive at the predictive equations (with noise):

$\hat{f}=K(x,x^{\prime})(K(x,x)+\sigma^{2}I)^{-1}y$ for the mean

and

$cov(f)=K(x^{\prime},x^{\prime})-K(x^{\prime},x)(K(x,x)+\sigma^{2}I)^{-1}K(x,x^{\prime})$ for the covariance.

The covariance kernel is the SE (Squared Exponential):

$k_{f}(x_{i},x_{j}) = \sigma^{2}exp(-\frac{1}{2\ell^{2}}\sum_{j=1}^{q}(x_{i,j}-x_{k,j})^{2})$

I know that a kernel, when we have a vector with inputs $x$ and size $(n)$, has dimensions $(n\times n)$. But how it works in multidimensional environment? For example, the covariance above $cov(f)$ with one dimensional inputs $(n)$ and for a single test point $(p = 1)$ will have the form:

$(p \times p) - (p \times n)(n \times n)(n \times p)$ which gives a scalar (the variance).

I can't comprehend how the above equation will work when we accept as input a matrix $(n \times d)$ size instead.

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  • $\begingroup$ You might find this question helpful. Also the Kernel cookbook is a great resource for more standard kernels. $\endgroup$ – combo Aug 22 '17 at 15:48
  • $\begingroup$ What is $d$? (The dimension of what?) Also, looks like the question is asking about some issue with multidimensional something rather than "understand intuitively how a kernel works"? $\endgroup$ – Juho Kokkala Aug 23 '17 at 5:07
  • $\begingroup$ I have been clear on that. "We accept as input a matrix with dimensions $(n x d)$, so is the dimensions of matrix $x$. Now the question asks how the kernel works in a multi-dimension, in which case you are right. I do not state that explicitly but rather I have put forward an example, as the calculation of covariance function depends on the kernel. So, in that example, if we have a multidimensional input $x$, how do we calculate the covariance function? Thanks $\endgroup$ – Jespar Aug 23 '17 at 5:36
  • $\begingroup$ Scrap that, I just realised that I also state "how the kernel works in a multidimensional environment". And then I provide an example. $\endgroup$ – Jespar Aug 23 '17 at 5:41
  • $\begingroup$ Please read the question carefully, "one dimensional inputs (n)". (p) is the size of the test point. I do state that. So you have as input a one dimensional vector with size n and a scalar test point. I do not know how to be more clear on that. Yes, the question asks "how it works in a multi dimensional environment". $\endgroup$ – Jespar Aug 23 '17 at 5:55
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Notation / setting

We are considering a GP regression model: \begin{equation} y_i = f(x_i) + \epsilon_i \end{equation} where $y_i\in \mathbb{R}$,$x_i \in \mathbb{R}^d$, $f$ a Gaussian process (whose realizations are functions $f:\mathbb{R}^d\rightarrow \mathbb{R}$), \begin{equation} f \sim \mathrm{GP}(m(x_i), \kappa(x_i,x_j)). \end{equation} $n$ datapoints $(y_1,x_1), (y_2,x_2), (y_3,x_3),\ldots, (y_n,x_n)$ are given. (I use $\kappa$ to distinguish the function from the matrices $K(\cdot,\cdot)$ that contain values of $\kappa$ evaluated at certain points. The question denotes both by $K$)

How to handle $d$-dimensional inputs

The question covers computing the posterior predictive distribution for a test point (or $p$ test points) in the case $d=1$ and asks how to extend to the general $d=2,3,\ldots$.

Answer: nothing changes and the formulas from the one-dimensional case work as well in this case. Note that $m$ is then a function from $\mathbb{R}^d$ to $\mathbb{R}$ and $\kappa$ a function from $\mathbb{R}^d \times \mathbb{R}^d$ to $\mathbb{R}$.

So, for example the matrix denoted by $K(x,x)$ in the question is a $n\times n$ matrix for which $K(x,x)_{i,j} = \kappa(x_i, x_j)$ ($x_i$ and $x_j$ are $d$-dimensional but since $\kappa$ maps two $d$-dimensional vectors to a scalar, $\kappa(x_i, x_j)$ is a scalar. Similarly for $K(x,x')$ and $K(x',x')$ where $x'$ are the test points.

Thus, the dimensions of the matrices in the predictive covariance equation are $(p\times p) - (p \times n)\,(n \times n)\,(n \times p)$ independent of whether the elements of the matrices are obtained by evaluating a function $\kappa(\cdot,\cdot)$ whose arguments are $1$-dimensional of a function $\kappa(\cdot, \cdot)$ whose arguments are $d$-dimensional. In fact, the inputs could even be in some space other than $\mathbb{R}^d$ (such as if we have a categorical predictor) as long as a positive-definite covariance function can be defined.

An extra remark about the SE kernel appearing in the question

The question mentions the SE kernel \begin{equation} k_{f}(x_{i},x_{j}) = \sigma^{2}\exp\!\Big(-\frac{1}{2\ell^{2}}\sum_{j=1}^{q}(x_{i,j}-x_{k,j})^{2}\Big) \end{equation} Note that this is already a function from $\mathbb{R}^q \times \mathbb{R}^q$ to $\mathbb{R}$ (with scalar inputs there would no "$x_{i,j}$ and $x_{k,j}$" for different values of $j$. And $q$ should be $d$ if $d$ is the dimension of inputs.

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    $\begingroup$ Especially for the last part, about the kernel. Very helpful indeed! I should delete my previous 'answer' $\endgroup$ – Jespar Aug 23 '17 at 18:20

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