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by magicalhippo 2 days 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.
2 comments

So 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.
And 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.
Thanks for the correction; silly miss on my part.