4 ms·
Thanks. I think I have a better understanding now, though still not having a complete grasp. >> f(x) = r_1 + x r_2* >> The right-hand plot shows the output
by alok-g 4mo ago
Thanks. I think I have a better understanding now, though still not having a complete grasp.
>> f(x) = r_1 + x r_2*
>> The right-hand plot shows the output of the kernel function for two indices x and x'.
The kernel function here would be k(x, x') = Cov[f(x), f(x')] = Cov[f(r_1 + x * r_2), f(r_1 + x' * r_2)].
In this case, I am guessing we should be able to figure what k(x, x') would be, but perhaps would not be x * x' for this case. x * x' sounds to be a very special case.
- magicalhippo 4mo ago> k(x, x') = Cov[f(x), f(x')] = Cov[f(r_1 + x * r_2), f(r_1 + x' * r_2)]. As I understand it, it would instead be k(x, x') = Cov[f(x), f(x')] = Cov[r_1 + x * r_2, r_1 + x' * r_2] I admit I haven't run through the full math. Given the definition of covariance I see how you get the x * x' term, but you're right in that it's not immediately obvious the other parts cancel fully.
- alok-g 4mo agoThanks for the correction; silly miss on my part.
- magicalhippo 4mo agoSo working the math a bit, it seems clear the author implicitly assumes the random variables follows a standard normal distribution, so zero mean (E[r] = 0) and unit variance (Var(r) = 1). In that case, you end up with a lot of E[...] = 0 and Var(...) = 1 terms and are left with the x*x' cross term.
- alok-g 4mo agoAnd then in the general case, the answer I guess would be a additive superposition of multiple functions including x.x' ... Hence, x.x' serves for purposes of explanation that the original author is aiming for.