Binary Symmetric Channels: Crossover probability and reliability

Question

I am having difficulties understanding how it is possible to create, given one binary symmetric channel (BSC) with crossover probability $p<\frac{1}{2}$, an equivalent BSC with crossover probability $p'>\frac{1}{2}$, while maintaining reliability of the underlying transmission.

To allow discussion of reliability w.r.t BSC, I want to introduce transmission schemes. Let us suppose we are dealing with the transmission scheme described in the following figure:

The current question places the given BSC as the channel in the middle, and also ensures us the existence of an encoder-decoder pair, denoted $(E,D)$, that allows for reliable transmission. Let us denote this transmission scheme $T_1$.

I want to build a new transmission scheme, denoted $T_2$, with its own encoder-decoder pair, $(E',D')$, that builds on $T_1$, and for which the crossover probability is $p'>\frac{1}{2}$.

According to Wikipedia, the receiver can use bit-flip on top of the existing BSC. I am trying to translate this into a formal answer. I have:

1. The encoder stays the same. Thus, $E' = E$. .
2. The bit-flip is an preliminary operation performed by the old channel decoder. Let us denote the bit-flip operation as $BF$. Then, we have $D' = D\circ BF$.

But here as where I come across the following difficulty: Since I used bit-flip, and it is the only difference between the two transmission schemes, how is it possible that reliability is maintained?

Formally:

1. Let $T_1$ be the original transmission scheme with crossover probability $p$, and encoder-decoder pair $(E,D)$. It has output $\hat{W_1}$, with high probability.
2. Let $T_2$ be the new transmission scheme with crossover probability $p'$, and encoder-decoder pair $(E',D')$. It has output $\hat{W_2}$, with high probability.

But, since, the transmission schemes are different in and only in the bit-flip step, we have $\hat{W_1} \neq \hat{W_2}$, with high probability.

Then, let $c$ is the noise induced by the original channel, which is unchanged in both $T_1,T_2$. We know that $T_1$ is reliable, and so, with high probability: $$W = \hat{W_1} = D(E(W)+c)$$

But since $\hat{W_1} \neq \hat{W_2}$, we also have, with high probability: $$W\neq \hat{W_2} = D'(E'(W)+c)$$

So how is the new transmission still reliable?

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Edit:

I did not understand the following:

1. A binary symmetric channel is actually a set of channels, with a parameter that is changeable.
2. The function of a decoder.
3. If I know one transmission scheme satisfies what I want, I do not have to take the channel from it "as-is" to create a new transmission scheme. I might need to take the encoder of the decoder of that transmission scheme.

Thank you!

Answer 1

With respect to the question I had:

1. For $T_2$, I do not need to take the same BSC channel as in $T_1$. I need to put a BSC channel there with parameter $p'=1-p$. Then, the bit-flip is not the only difference between the two transmission schemes, but the parameter and the bit-flip operation.
2. The role of a decoder: Let us assume that we have a decoder with the following property: Given $k\in[0,1]$ and an input $L$, if $P(D(E(W)+c) \neq W) = P(L \neq W) < k$, the decoder knows how to reach the original $W$. Then, to create a successful new transmission that uses that same decoder, all I need to do to get a successful transmission in the new scheme is to ensure that there exists an input $L$ to the old decoder such that the decoder property holds for it. This allows the new transmission scheme to get $W$ with high probability. Notice that the relation $<k$ is not the only possible relation (or rather predicate) that has to occur. Any predicate on the probability can occur, as it depends on the original decoder.
3. Here I use $(E,D)$ from $T_1$, and them alone, for the creation of $T_2$. More precisely, I use $(E,D)$ to create $(E',D')$. The BSC channel from $T_1$ (with parameter $p$) is not the same BSC channel in $T_2$ (with parameter $p'=1-p$).