This is an extended comment.
To summarize, you have RVs $U\to X\to Y\to V$ where:
- $X\sim B(1/2)$,
- $Y=X+Z_Y$
- $U=X+(X\cdot Z_{U1} + (1-X)\cdot Z_{U0})$,
- $V=Y+(Y\cdot Z_{V1} + (1-Y)\cdot Z_{V0})$,
for RVs:
- $Z_Y\sim B(p_Y)$,
- $Z_{Ui}\sim B(p_{Ui})$, $i=0,1$
- $Z_{Vi}\sim B(p_{Vi})$, $i=0,1$
You are essentially wondering whether $U',V'$ with $U'\to X\to Y\to V'$ and:
- $U'=X+Z_{U'}$
- $V'=Y+Z_{V'}$
choosing:
- $Z_{U'}\sim B(p_{U'})$ so that $I(U';X)=I(U;X)$
- $Z_{V'}\sim B(p_{V'})$ so that $I(V';Y)=I(V;Y)$
gives $I(U';V')\geq I(U;V)$.
If all the $Z$'s turn out to be independent because of the Markov structure (not sure whether or not this is true), my inclination is that the answer to this question is always no for nontrivial cases, i.e. when (U,X) or (Y,V) are not BSCs
This is because there is a notion that uniform noise is the most degrading. Choosing $(X,U'),(Y,V')$ to be BSCs will make the noise distribution $(V'-U')$ not only uniform, but also uniform over each intermediate hop, i.e. 'maximally bad.'