I am reading through this paper which describes the use of a post-processing vector $\vec{c}$ with elements having values of $(-1,0,1)$. In equation 3 they give their solution as a linear combination of the measurement probabilities $\vec{p}$ with the post-processing vector $\vec{c}$. They justify this with the following:

"The measurement results give rise to an outcome probability vector $\vec{p} = (p_1,...,p_l,...)$. The desired output might be one of these probabilities $p_l$, or it might be some simple function of these probabilities. Hence, we allow for some simple classical post-processing of $\vec{p}$ in order to reveal the desired output."

It seems to me that they justify the use of $\vec{c}$ because it corrects the sign of the outcome probabilities. However, I am not satisfied with this explanation. I do believe that it is correct, but can someone provide a better justification for the use of this seemingly arbitrary post-processing method? Additionally, do you think there exists a derivation for how they arrive at the specific post-processing vector? Or, have they arrived at it empirically?

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    $\begingroup$ I don't know the exact context, but if you measure a probability vector $\vec p$, and the desired outcome (thinking in a supervised learning context) is some $\vec x_{\rm target}$, then doesn't this amount to solving the linear system $\vec x_{\rm target}=A\vec p$ for a matrix/vector $A$? If you want to learn such relation, you can set up a system of the form $\vec x_k=A\vec p_k$. This is also equivalent to learning an observable (which is nothing but linear post-processing done on observed probabilities) $\endgroup$
    – glS
    Aug 12, 2021 at 20:49
  • $\begingroup$ That makes sense in a supervised learning context where you know the desired solution. But what about the case when you don't know the exact solution, how do you determine the post-processing vector $\vec{c}$? $\endgroup$ Aug 12, 2021 at 20:59
  • $\begingroup$ well, you need to have a "desired outcome" specified somehow, no? How do they define it here? What's the precise task they are dealing with? $\endgroup$
    – glS
    Aug 12, 2021 at 21:18
  • $\begingroup$ I am not so concerned about their exact problem, rather, I have an outcome probability vector for an entirely different problem which has elements that are correct in magnitude but which has some elements with opposite sign. I can't find an issue with my algorithm so I am looking for a justification to flip some of the signs as was done in the cited paper. It is easy to setup the linear system as you have done, however, I need to be able to justify the fact that the outcome probability is correct but that it requires this post-processing step. $\endgroup$ Aug 12, 2021 at 21:34
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    $\begingroup$ I would say that in general, for the problem to be well-defined, and assuming it is a "learning problem" of some kind, the correct answer oughts to be known or codified in a known way. It would help if you could provide more details on the context/problem setting you are thinking about $\endgroup$
    – glS
    Aug 12, 2021 at 21:37


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