3 ms·
> 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
by magicalhippo 5mo 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 5mo agoThanks for the correction; silly miss on my part.
- magicalhippo 5mo 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 5mo 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.