Capacity of sum of two channels

Question

Say channels $\Gamma_1$ and $\Gamma_2$ has capacity $C_1$ and $C_2$. Then what is the capacity of their sum $\Gamma_1 + \Gamma_2$? By searching, I found out that $C_{tot}=\log(2^{C_1}+2^{C_2})$, but I can't find the reason. (Sum of two channel is channel that has input channel of disjoint union of two input channels, and same forward probabilities.)

Answer 1

Let $X$ be the input of the composite channel, let the rv. $Z \in\{1,2\}$ indicate which channel was used. Let $P(Z=1)=p$.

Now, using the facts that $H(Z|X)=0$ and $H(Z|Y)=0$ and using the chain rule we get

$$I(X;Y)=H(X)-H(X|Y)=h(p) +p I(X_1;Y_1) + (1-p) I(X_2;Y_2) $$

where $h(p)$ is the binary entropy function. (Similar to this).

The maximum of each $I(X_i,Y_i)$ wrt to each input probability is the resp. capacity $C_i$. Then we are left to find the value of $p$ which maximizes $h(p) +p C_1 + (1-p) C_2$. That gives the desired result.