3
$\begingroup$

I've been studying GPs with Rasmussens book as well as a few of my old probability favorites. I am confused by the results we're getting (doing also a lot of practice) when sampling from our Gaussian Process.

Conditional on the data $ D_n = [{X_n, Y_n}] $ for $n$ datapoints the predictive posterior distribution for a new datapoint $x$ is $GP(x) = Y(x) \sim N(\mu(x), \Sigma(x))$ Where $\mu(x)$ is linearly dependent on the $y_n$.

I do find most of it intuitive, except for the "knots" see the screenshot from Rasmussen's book enter image description here.

At first, I thought that it was just illustrative but I was able to get the same "shapes" (though not as pretty as my plotting skills are limited) with arbitrary data, though also not consistently, just most of the time, at times I don't get the knots everywhere, especially with many points.

If I understand correctly since the covariance at those points is 1, for any realization, at those points $Y(x_1) = Y(x_2)$ if $x_1 = x_2$ But only for that sampled function. Why are all the sampled functions getting the same value?

I do understand that this intuitively is the point to the whole exercise. We sample from possible functions in the $GP$ that map $x$, to $y$. I am just confused as on how to see it from the math.

This is probably quite the silly question and I am missing something here, I apologize in advance and thank you for your patience!

$\endgroup$
2
  • 1
    $\begingroup$ The value of your kernel at $0$ is $0.$ If you were to include a "nugget effect" -- a jump at $0$ -- the interpolators would no longer honor the data. $\endgroup$
    – whuber
    Sep 8, 2022 at 12:47
  • $\begingroup$ @whuber thanks a lot for commenting, can you elaborate? Unfortunately I am not able to follow $\endgroup$
    – Oliver
    Sep 8, 2022 at 13:03

2 Answers 2

1
$\begingroup$

Short answer: The reason that the sampled functions from the predictive posterior distribution perfectly interpolate the data is that they are sampled from a distribution that is conditioned on the data points. In other words, the data points are not just data points, they are also points that the function must go through.

Let's think about what we're doing here. We have a bunch of data points (x_i, y_i) and we want to use a Gaussian process to interpolate between them. First, we need to choose a kernel function. This is a function that gives the covariance between any two points. For example, the RBF kernel is given by

k(x, x') = exp(-||x - x'||^2 / (2 * sigma^2))

where ||x - x'|| is the Euclidean distance between x and x'. sigma is a parameter that controls the "smoothness" of the function.

Next, we need to choose a mean function. This is a function that gives the expected value of the function at any point. For example, we could choose the mean function to be the constant function m(x) = 0.

Now, we can use these to define a Gaussian process. This is a collection of random variables, any one of which is a function from x to y, that have the following properties:

  1. The mean function of the Gaussian process is m(x).
  2. The covariance function of the Gaussian process is k(x, x').

So, what does this all mean? It means that, for any two points x and x', the expected value of the function at those points is m(x) and m(x'), and the covariance between the function values at those points is k(x, x').

Now, let's think about what happens when we sample from a Gaussian process. We're essentially choosing a random function from the set of all possible functions that could have generated our data. So, if we have two points x and x' that are close together, then the function we sample is likely to have values that are close together at those points.

This is why the sampled functions from a Gaussian process perfectly interpolate the data. Because the data points are close together, the sampled functions are likely to have values that are close together at those points. And because the mean function is the same at all points, the sampled functions will have the same value at those points.

I hope that makes sense!

$\endgroup$
2
  • $\begingroup$ Re "the function we sample is likely to have values that are close together at those points:" that is the case only provided $\lim_{x\to x^\prime}k(x,x^\prime)=0.$ $\endgroup$
    – whuber
    Sep 12, 2022 at 15:45
  • $\begingroup$ thanks a lot for this great answer Robin! I apologize it probably wasn't so clear from my post, this was mostly intuitive and clear to me already (it was really nice seeing it again so well written and clear though, thanks!). I was hoping to validate the intuition with math. The answer I posted is an attempt at what I was hoping to see in an answer. $\endgroup$
    – Oliver
    Sep 12, 2022 at 18:33
0
$\begingroup$

After playing around with a lot of algebra I've come to a conclusion though i don't feel confident enough to accept my answer I thought I may provide reasoning.

If we have two gaussian from the GP, $f(x_1) \sim N(\mu_1, \Sigma_{11})$ and $f(x_2) \sim N(\mu_2, \Sigma_{22})$ The conditional distribution $$ f(x_1) | f(x_2) = y \sim N(\mu^*, \Sigma^*)$$

has mean function $$\mu^* = \mu_1 + \Sigma_{12}\Sigma_{22}^{-1}(y-\mu_2)$$ and Covariance $$\Sigma^* = \Sigma_{11} - \Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}$$

Now $\Sigma^*$ turns out to be $0$ if $x1 = x2$

because $\Sigma_{12}$ should equal $\Sigma_{22}$ and by the same reason we get $$\mu^* = \mu_1 + (y-\mu_2)$$ but $\mu_1 = \mu_2$ therefore we have $\mu^* = y$ since the variance = 0, we get a perfectly interpolated point.

Feel free to correct me, this took a long think and I am not confident on it haha.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.