# channel capacity of a general binary channel

I have found worked examples for special cases of binary channel such as the binary symmetric channel and the Z-channel. However, I am interested in a more general type of binary channel $X \to Y$ in which I only know that $P(Y=1|X=0)=a$ and $P(Y=1|X=1)=c$, hence one hopes that $0<a<c<1$.

The mutual information is defined to be $$I(X,Y):=\sum_{x,y \in \{0,1\}} P(X=x \wedge Y=y)\log_2 \left ( \frac{P(X=x \wedge Y=y)}{P(X=x)P(Y=y)} \right ).$$

Denoting $t=P(X=1)$, it follows that

$$I(X,Y)=(1-c)(1-t) \log_2 \left ( \frac{1-a}{1-c} \right ) + a \frac{1-c}{1-a}(1-t) \log_2 \left ( \frac{a(1-a)}{(c-a)t+a(1-c)} \right ) +ct \log_2 \left (\frac{c(1-a)}{(c-a)t + a(1-c)} \right ).$$ The channel capacity is defined as the supremum of the mutual information over all possible distributions of $X$, that is, over all $t \in [0,1]$.

Is there, in this case, a formula for the channel capacity as a function of $a$ and $c$ ?