Consider two discrete memoryless channels with capacities $C_1$ and $C_2$. We have them cascaded in series i.e., the output of the Channel-1 (capacity $C_1$) is fed as input to Channel-2 (capacity $C_2$). Let $C$ be the overall capacity of this serial setup.
Is it true that $C \leq \min(C_1,C_2)$? Intuitively it appears true. I am looking for a formal proof.
As a follow-up can it be shown that $C = \min(C_1,C_2)$?
Thank you!
If the channels are combined in series, the random variables they transition form a markov chain, $X\stackrel{C_1}{-}Y\stackrel{C_2}{-}Z$. The first result can then be obtained using the data processing inequality.
For every input distribution $p_X$, we would have $I(X; Z) \leq I(X; Y)$ and $I(X; Z) \leq I(Y; Z)$ \begin{align*} C &= \max_{p_X} I(X; Z) \leq \max_{p_X} I(X; Y) = C_1\\ C&= \max_{p_X} I(X; Z) \leq \max_{p_X} I(Y; Z) \leq \max_{p_Y} I(Y; Z) = C_2 \end{align*} Where the last inequality occurs because the set of $p_Y$ which can be obtained through $p_Y(y) = \sum_x p(y|x) p_X(x)$ is a subset of the set of all probability distributions $p_Y$ on $Y$.
The second result is not true in general. As a counterexample, consider two $BSC(p)$ channels in series. They form a $BSC(2p(1-p))$ channel when combined.