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

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  • $\begingroup$ Adding "channel" to BSC is a bit redundant. Could you expand the acronym at least once, and provided more context on their transition probabilities? $\endgroup$ – Laurent Duval Dec 12 '16 at 18:09
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    $\begingroup$ 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 $C = 0 = C_1C_2$. $\endgroup$ – Dilip Sarwate Dec 14 '16 at 2:50
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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.

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  • $\begingroup$ It should be $C=1-H(p)$. $\endgroup$ – msm Dec 13 '16 at 14:25
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    $\begingroup$ $p_1+ p_2 - 2p_1p_2$ is the probability that exactly one of the two channels make an error. $\endgroup$ – Dilip Sarwate Dec 14 '16 at 2:53

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