The main empirical finding which led to the NTK/GP/Mingard et al picture of neural nets is that, in practice, that linear approximation works quite well. As neural networks get large, their parameters change by only a very small amount during training, so the overall Δθ found during training is actually nearly a solution to the linearly-approximated equations.
Trying to check if I’m understanding correctly: does that mean that despite SGD doing a lot of successive changes that use the gradient at the successive parameter values, these “even out” such that they end up equivalent to a single update from the initial parameters?
Sort of. They end up equivalent to a single Newton step, not a single gradient step (or at least that’s what this model says). In general, a set of linear equations is not solved by one gradient step, but is solved by one Newton step. It generally takes many gradient steps to solve a set of linear equations.
(Caveat to this: if you directly attempt a Newton step on this sort of system, you’ll probably get an error, because the system is underdetermined. Actually making Newton steps work for NN training would probably be a huge pain in the ass, since the underdetermination would cause numerical issues.)
Trying to check if I’m understanding correctly: does that mean that despite SGD doing a lot of successive changes that use the gradient at the successive parameter values, these “even out” such that they end up equivalent to a single update from the initial parameters?
Sort of. They end up equivalent to a single Newton step, not a single gradient step (or at least that’s what this model says). In general, a set of linear equations is not solved by one gradient step, but is solved by one Newton step. It generally takes many gradient steps to solve a set of linear equations.
(Caveat to this: if you directly attempt a Newton step on this sort of system, you’ll probably get an error, because the system is underdetermined. Actually making Newton steps work for NN training would probably be a huge pain in the ass, since the underdetermination would cause numerical issues.)
By Newton step, do you mean one step of Newton’s method?
Yes.