Double Stochastic Gradient Descent

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1

In the Double Stochastic Gradient Descent method. The Hamiltonian was given as

H = 4 +2I \otimes X + 4 I ⊗ Z −X⊗X + 5 Y ⊗ Y + 2 Z ⊗ X

And so the idea of “Double” stochastic means when we iterate through our SGD, we only sample the Pauli terms from H and evaluate the expectation of those terms?

For instance, we can set to sample 3 out of 5 Pauli terms at each iteration. At the first iteration, we might sample, IX,XX, ZX then at the next step we might sample IX, XX, YY,… and so on. Am I thinking of this right? If so, how many terms were set to get the results presented below?

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Thanks!

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And so the idea of “Double” stochastic means when we iterate through our SGD, we only sample the Pauli terms from H and evaluate the expectation of those terms?

@KAJ226, exactly right! The stochasticity comes from two sources:

  1. The finite number of shots
  2. The sampling of a random subset of the Hamiltonian terms

Hence ‘doubly stochastic’.

In this particular example, the number of sampled Hamiltonian terms was set to 1, via n=1 in the below snippet:

def loss(params):
    return 4 + (5 / 1) * circuit(params, n=1)
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I just have a question that what is this loss exactly, from what I understand we sample terms from hamiltonian and look at the moving average to see if this converges, so what is this def loss(), and why is it returning 4+(5/1)* circuit()? Shouldn’t it just be circuit()?

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Hi @Kutubkhan_Bhatiya, this comes from the Hamiltonian, and the fact that we are only sampling 1 out of 5 terms:

H = 4 +2I \otimes X + 4 I ⊗ Z −X⊗X + 5 Y ⊗ Y + 2 Z ⊗ X

Here, the 4 + comes from the constant 4 term in the Hamiltonian. The 5/1 comes from the fact that we are only sampling 1 term from the remaining 5.

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Oh okay, that makes a lot of sense, so if we sampled let’s say 2 out of 5 terms, would it be 5/2? Thanks for replying, I didn’t think I would get a reply on such an old post.

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That’s right @Kutubkhan_Bhatiya, it would be 5/2

There’s no problem that it’s an old post, we’re here to help!