# Calculate channel capacity of general channel under constraint

Given a conditional distribution $P_{Y|X}$ I'd like to find the prior distribution $P_X$ that maximizes the mutual information $I(X;Y)$ with $P_Y(y)=\int P_{Y|X}(y|x)P_X(x)\text{dx}$ (this corresponds to finding the channel capacity $C(X;Y):=\max_{P_X}I(X;Y)$) subject to the constraint $E_{P_X}[-\log(X)]=a$.

In my particular case, $P_{Y|X}$ is a Bernoulli distribution with $X$ as its parameter and I'm looking for a distribution over the parameter.

My intuition would tell me this should be some Beta-distribution (something like $\text{Beta}(1/a,1)$?!) in my particular case, but I don't know how to approach such a problem, much less in the general case.

Could anyone point me in the right direction?

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What (and why) is the constraint $E[-\log(X)]=a$? – Ashok Feb 27 '13 at 13:09
Sorry, $E[-\log(X)]$ means the expected value of the random variable $X':=-\log(X)$. It's a little bit complicated to explain where this constraint comes from. In my particular case it's because the above stuff is a discrete approximization of a poisson process, where "heads" are thrown at a varying rate R. If time is discretized into little bins, this gives, for each of those bins, the probability $X$ to have no "heads" in that bin where $X = \exp(-R)$. The expected value of the rate $E[R]=E[-\log(X)]=a$ is fixed. If the constraint is the problem, $E[X]=a$ would also be ok. – user31757 Feb 27 '13 at 16:21
What is the range of $X$ (because you are saying $P_{Y|X}$ is Bernoulli with parameter $X$)? – Ashok Feb 28 '13 at 12:08
Sorry, I perhaps was a bit too unspecific there. $P_{Y|X}{Y=1|X=x}=x$ and $P_{Y|X}(Y=0|X=x)=(1-X)$ where $Y\in\{0,1\}$ and $X\in[0;1]$. – user31757 Mar 2 '13 at 3:48
wupps, should've been $P_{Y|X}(Y=1|X=x)=x$., ofc. – user31757 Mar 2 '13 at 3:50

Solution to the special case of $a=0.5$ under the less restrictive case $E[X]=a$. This case is trivial and is solved by $P(X=0)=P(X=1)=0.5$ in which the channel capacity is one which is the maximal value for your channel.