The concatenation of two binary symmetric channels with parameter $p\in[0,1]$ (that is, the channel flips a bit with probability $p$) is a binary symmetric channel with parameter $2p(1-p)$.
In a paper I'm reading they say that the probability of a bit-flip of a concatenation of $m$ binary symmetric channels with parameter $p$ is $(1-(1-2p)^m)/2$. Can anyone help me understand why is this true? I think they may have meant that $(1-(1-2p)^m)/2$ is an upper bound on the error parameter of the concatenation channel, since all my calculations suggest that the parameter of this channel should be considerably more complex.
As I understand it, in simple terms, I need the probability of the set of binary sequences of length $m\in\mathbb{N}$ with odd number of ones, where the distribution of each bit is ${Bernoulli}(p)$.
A reference to a chapter in a book is provided but I wasn't able to find the relevant part in the chapter.
Thanks a lot to all the helpers!
We have, $$(b+a)^m-(b-a)^m = \sum_{k=0}^m {m\choose k}\ \left[1-(-1)^{k}\right]a^k b^{m-k} =\\=2\sum_{k \text{ odd}} {m\choose k} a^k b^{m-k}.$$ Now the channel flips if and only if there are an odd number of flips so the right hand side is twice the probability you want if you let $a=p,b=1-p,$ yielding half the left hand side as the answer $$ \frac{(b+a)^m-(b-a)^m}{2}=\frac{1-(1-2p)^m}{2}. $$