Capacity of 2 BSCs I have a channel with input/output = $\{1,2,3,4\}$ and my transition matrix looks like:
$\left[\begin{array}{cccc}
1-a & a & 0 & 0 \\
a & 1-a & 0 & 0 \\
0 & 0 & 1-b & b \\
0 & 0 & b & 1-b
\end{array}\right]$
I know that this is the combination / sum of two BSC, but my first problem is how to proof it.
The second one is that I want to calculate the capacity of the channel. I know that if I consider one BSC, the capacity would be $1 - H(a)$.
Let $X_1 = {0,1} = Y_1 $ and $X_2,Y_2$ analogously.
I already started like this:
$I(X;Y) = I(X_1, X_2;Y_1;Y_2) \\
 = H(Y_1,Y_2) + H(Y_1,Y_2|X_1,X_2) \\
 = H(Y_1) + H(Y_2) + H(Y_1,Y_2|X_1,X_2) \\
$
I can say that $H(Y_1,Y_2) = H(Y_1) + H(Y_2)$ because they are independent. However, I'm stuck here.
 A: This is a combination of two paralell separable BSC "subchannels". 
A standard trick for this kind of problem is to introduce an additional random variable that signals which of the two subchannels was used (say $Z=1$ if $X\in \{1,2\}$, $Z=2$ if $X\in \{3,4\}$). The key is that knowing $Y$ one knows $Z$. Hence  we write the chain rule for mutual information: $$I(X;Y,Z) = I(X;Y) + I(X;Z|Y) = I(X;Z) + I(X;Y|Z) $$
and we note that $I(X;Z|Y)=0$ (knowing $Y$, $Z$ is a constant). Further, letting $\alpha=P(Z=1)$: $$I(X;Z)=H(Z) - H(Z|X)=H(Z)=h(\alpha)$$ where $h(\cdot)$ is the binary entropy function, and 
$$I(X;Y|Z) = \alpha I(X_1;Y_1) + (1-\alpha)I(X_2;Y_2)$$
where $X_1,Y_1$ are the inputs and outputs of the first subchannel, and so on.
Then $$C=\max_{p(X)} I(X;Y)=\max_{\alpha,p{(X_1)},p{(X_2)}} h(\alpha)+\alpha I(X_1;Y_1) + (1-\alpha)I(X_2;Y_2)\\=\max_{\alpha} h(\alpha)+\alpha C_1 + (1-\alpha)C_2 $$
Maximizing over $\alpha$ we get the general formula for parallell (separable) channels: $2^C=2^C_1+2^C_2$
