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.'