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Apr 13, 2017 at 12:58 history edited CommunityBot
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Jul 29, 2014 at 13:54 comment added vsoftco thanks. I am slightly unfamiliar with the notation, but my understanding is that basically $C_2$ has to behave like the identity on the support of $p(x) C_1$? Sorry for the long strings of comments.
Jul 29, 2014 at 13:51 vote accept vsoftco
Jul 29, 2014 at 3:19 comment added math-Student Suppose $p(x)$ and $p(y|x)$ are given. Then your problem is to characterize the sufficient statistics of $Y$ with respect to $X$. Lets call this $T(Y)$. Then $C_2$ is simply equal to $p(T(y)|y)=I_{\{(T(y)=y\}}$. It can be shown that $T(y)$ can be characterized as the following: $T:\mathcal{Y}\to \mathcal{P}(\mathcal{X})$ defined by $y\to p(x|y)$ where $\mathcal{P}(\mathcal{X})$ is the simplex of probability measures over alphabet $\mathcal{X}$. If this is not clear (which I think it is) please let me know,
Jul 29, 2014 at 3:15 history edited math-Student CC BY-SA 3.0
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Jul 29, 2014 at 0:49 comment added vsoftco PS: I edited the question to make it more clear.
Jul 29, 2014 at 0:48 comment added vsoftco Thanks @SAmath, was aware that this is the case, from the famous book. I was actually a bit unclear in my question. I would like to be able to say something about the transition matrices $C_1$ and $C_2$, i.e. given an input $p(x)$ and a channel $C_1$ (basically a transition matrix $p(y|x)$), what channels $C_2$ make give equality in the data processing. Intuitively, $C_2$ has to be somehow "correctable", that is, all information about the input should be present at the output of $C_2$. I find hard to formalize this and come up with an explicit form of $C_2$ as a function of $p(x)$ and $C_1$
Jul 29, 2014 at 0:26 history answered math-Student CC BY-SA 3.0