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In arXiv:2207.14734 the authors claim that it is "straightforward to show that" their equation 8 holds:

$$\mathrm{Pr}_{x\sim q}[x:f(x)\geq \mu] \geq \frac{1}{M}$$

where we have an objective function maps a bitstring to some value $f:( 0,1)^n\rightarrow[0,M]$. Moreover, $q(x)=|\langle x|\gamma\beta\rangle|^2$ with $|\gamma,\beta\rangle$ being the state after a run of the QAOA circuit. $\gamma$ is the expectation value of this objective function, i.e. $\mu = E_{x\sim q} f(x)$. Furthermore, the problem graph is $G=(V,E)$ and the values $n$ and $M$ are defined as $n=|V|$ and $M=|E|\leq n^2$.

From which theorem/lemma can the above equation be derived? Intuitively, I guess I understand what it means, but I do not know how it is derived even though it is supposed to be "straightforward to show".


Cross-posted on physics.SE

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  • $\begingroup$ In the paper, after Eq 8, the authors state that $M$ is the number of trials. So if we peform a single trial, the probability is $\geq$ 1, which is not true in general. So, to me, it is not clear if $M$ should be the number of trials or the size of the maxcut. $\endgroup$
    – MonteNero
    Commented Mar 1, 2023 at 1:24
  • $\begingroup$ @MonteNero I see your problem, however, $M=|E|$ so $M$ is the same as the number of edges in the problem graph. If we had $M=1$ a Max Cut Problem would be trivial. Thus I guess your case should not be the issue - nevertheless, I think they say that $M$ is the number of trials is only due to the fact that $M$ is in the denominator in equ. 8. If you check out Equ. 13, they arrive at the conclusion that one will need at most $5^k$. So they do not define $M$ to be the number of trials, it's just implied by the structure of the equation. $\endgroup$
    – Juri V
    Commented Mar 1, 2023 at 8:30
  • $\begingroup$ If $M=1$, maxcut is not trivial: there will be 4 candidate solutions (assuming there are 2 adjacent vertices) and half of these solutions give maxcut of size 0. Yet Eq 8 tells that the sample is higher than the average with probability 1. So this formula may already break at a very simple case where we have $n=2$. Additional assumptions on $q(x)$ are needed to guarantee Eq 8. So, now, why should it hold for $n>2$? $\endgroup$
    – MonteNero
    Commented Mar 1, 2023 at 13:54
  • $\begingroup$ @MonteNero Ok yes sorry, I withdraw this claim with the triviality. However, they seem to assume that $q(x)=|\langle{x}| \gamma,\beta \rangle|^2$ is based on the already optimized parameters $\gamma$ and $\beta$. Thus it is maybe more reasonable that the LHS of Eq. 8 will be fulfilled with certainty in the case of $M=1$? $\endgroup$
    – Juri V
    Commented Mar 1, 2023 at 17:28
  • $\begingroup$ It is clear from the get go that they mean gamma and beta are optimized. But what does it really mean mathematically for $q(x)$? I can say, gamma and beta are such that $q(x)=1$ for an optimal $x$ and $q(x)=0$ otherwise. Or I could also say optimal parameters are such that $q(x) = .00001$ for all optimal $x$. In the first case, equality obviously holds. But this is not a reasonable assumption in general. $\endgroup$
    – MonteNero
    Commented Mar 1, 2023 at 18:19

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