Is it true:$C = C_1 * C_2$ where $C_1$ and $C_2$ are the two cascaded BSC which makes $C$.

  • $\begingroup$ Adding "channel" to BSC is a bit redundant. Could you expand the acronym at least once, and provided more context on their transition probabilities? $\endgroup$ Dec 12, 2016 at 18:09
  • 1
    $\begingroup$ Perhaps the only cases when your conjectured result $C = C_1C_2$ is true is when both channels are noiseless ($C_1 = C_2 = 1 \implies C = 1$) or at least one channel has zero capacity in which case $C = 0 = C_1C_2$. $\endgroup$ Dec 14, 2016 at 2:50

1 Answer 1


No, this is not correct. Consider the chain of two BSCs with error probabilities $p_1$, $p_2$ as a single BSC with unknown error probability $p$.

Now, we know that in overall no error occurs, when:

  • neither BSC1 nor BSC2 create an error: $(1-p_1)(1-p_2)$
  • both channels make an error: $p_1p_2$.

Hence the overall probability of an error is

$$ p= 1-[(1-p_1)(1-p_2)+p_1p_2]=p_1+p_2-2p_1p_2 $$

which is also known as the binary convolution. Now, the capacity of the chain-BSC is given by $$C=1-H(p)$$ where $H(p)$ is the binary entropy function.

  • $\begingroup$ It should be $C=1-H(p)$. $\endgroup$
    – msm
    Dec 13, 2016 at 14:25
  • 1
    $\begingroup$ $p_1+ p_2 - 2p_1p_2$ is the probability that exactly one of the two channels make an error. $\endgroup$ Dec 14, 2016 at 2:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.