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Suppose ABC=B for some column stochastic matrices A, B, and C. Can the following implication be made without further restrictions: There necessarily exists a column stochastic matrix D such that DB=BC?

I think this is implicated by lemma 1 of Rauh et al. - Coarse-graining and the Blackwell order and the theorem of Blackwell, Sherman and Stein. Unfortunately the paper contains just a very limited proof. It uses the statement that the capacity of a pre-garbling is bounded by the original experiment/channel and states this as a well-known fact.

My current efforts to prove this special step by myself failed. I would therefore appreciate any assistance.

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    $\begingroup$ I fixed the typo while changing the arXiv link to point to the PDF, but I was sorry to have to get rid of the phrase "I would … appreciate any assistants." $\endgroup$
    – LSpice
    Commented Jan 30, 2019 at 18:54
  • $\begingroup$ "The given example": example of what? "example is implicated": how can one imply/implicate an example? since I don't guess what is meant, I haven't edited this. $\endgroup$
    – YCor
    Commented Jan 30, 2019 at 19:34
  • $\begingroup$ Thanks. I have corrected that one. The term "example" was a remnant of an earlier version of this question. $\endgroup$
    – JayDoe
    Commented Jan 30, 2019 at 19:39

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Pre-garbling is equivalent to changing the input distribution to the channel defined by $B$. Since the capacity of $B$ is given by maximising the mutual information between the input and the output of the channel over all possible input distributions you cannot increase the capacity by pre-garbling.

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