Do binary symmetric channels maximize mutual information? Consider the following setup: $(X, Y)$ is a doubly symmetric binary source with parameter $0 < p < 1/2$, i.e., $X \sim \text{Bernoulli}(1/2)$, $Z \sim \text{Bernoulli}(p)$ and $Y = X \oplus Z$. Let $(U, V)$ be another pair of binary variables, where $U$ is derived from $X$ and $V$ is derived from $Y$, i.e., the Markov chain $U - X - Y - V$ holds. My question is the following:
Can you find binary random variables $U'$ and $V'$ such that $U' - X - Y - V'$ form a Markov chain and the channels $X \rightarrow U'$ and $Y \rightarrow V'$ are binary symmetric channels with ($I(\cdot;\cdot)$ denotes  mutual information.)
$I(U';X) \le I(U;X)$, $I(V';Y) \le I(V;Y)$, and $I(U'; V') \ge I(U;V)$ ?
The question can also be phrased in terms of the probabilities of the binary symmetric channels: Are there values $0 \le r,q \le 1/2$ such that
$1 - h(r) \le I(U;X)$, $1 - h(q) \le I(V;Y)$, and $1 - h(r * p * q) \ge I(U;V)$ ?
($h(x) = -x \cdot \log_2(x) - (1-x) \cdot \log_2(1-x)$ is the binary entropy function and $p * q = p*(1-q) + (1-p)*q$ the binary convolution.)
I did some numerical evaluation which suggests a positive answer to that question: Choosing $r,q$ such that $1 - h(r) = I(U;X)$ and $1 - h(q) = I(V;Y)$, it seemed that $1 - h(r * p * q) \ge I(U;V)$ is always satisfied. But I don't know how to formally prove this statement. I wasn't lucky trying to apply Mrs. Gerber's Lemma in this context. I also tried splitting the problem in two stages: First fix $V'=V$ and try to show that a BSC $X \rightarrow U'$ with $I(U';X) \le I(U;X)$ maximizes $I(U'; V')$. But I have a counterexample to this claim, so it appears necessary to consider both channels simultaneously.
 A: No. You can get a higher $I(U;V)$ using asymmetric channels. Below I construct a counterexample, but first a more succinct restatement of the question.

Restatement
To summarize, there is an input $U$ distorted by three independent binary hops, each described by $2\times 2$ stochastic matrices $C_L,\ C,\ C_R.$ Labeling all your RV's, $$U\overset{C_L}{\to}X\overset{C}{\to}Y\overset{C_R}{\to}V.$$ 
We are interested in maximizing $I(U;V)$ 
 subject to constraints: 


*

*$C$ is determined by nature.

*$C_L$ is such that $I(U;X) \leq r_L$, 

*$C_R$ is such that $I(Y;V) \leq r_R$,

*$U$'s distribution is such that the left channel's output (i.e. $X$) is $B(1/2)$.



Counterexample
You claim the $C_L,C_R$ which produce the maximum are binary symmetric. 


*

*If $C_L$ is a BSC with $B(1/2)$ output then its input must also be $B(1/2).$ For a given rate $r_L \leq 1$, then there is at most one 'positive' (i.e. can't be improved by relabeling the outputs) BSC $C_L$ whose output is $B(1/2).$  

*You have assumed $C$ is a BSC, so with a symmetric input its output is also symmetric. 

*For a rate $r_R\leq 1$ there is only one positive choice for $C_R.$


So to say they are binary symmetric is to determine all of $U,C_L$ and $C_R$.
Now take $C$ a perfect channel, 
$C= \left[\begin{smallmatrix}
    1 & 0 \\
    0 & 1
    \end{smallmatrix}\right]$ and $r_L=r_R=0.4.$ The associated positive BSC for this rate has crossover probability $\approx 0.15,$ and the end-to-end mutual information can be computed: 
$$I(U_{BSC}, V_{BSC})< 0.1895$$
However, trying randomly[1] you can find $U^\ast, C_L^\ast, C_R^\ast$ that satisfy all the mutual information constraints, but have greater $U$-to-$V$ mutual information. One I found happens to be quite close to a Z-channel:
\begin{equation}
   I(U^\ast; V^\ast) > 0.19,
\end{equation}
\begin{equation}
   C_L^\ast \approx \left[\begin{smallmatrix}0.2493 & 0.7507 \\ 0.9657 & 0.0343 \end{smallmatrix}\right], \qquad
    C_R^\ast \approx \left[\begin{smallmatrix} 0.9821 & 0.0179 \\ 0.3374 & 0.6626 \end{smallmatrix}\right], \qquad
   U^\ast \sim B(0.35)
\end{equation}

Discussion
This result is to be expected since there is a vague sense that uniform noise over a bounded space is the most degrading, even holding mutual information fixed. (by one heuristic "uniform noise means you can't precode to mitigate it")
A gentle introduction for a good, visualisable framework for studying binary symmetric channels is given in a short paper, Algebraic Information Theory for Binary Channels by Martin, Moskowitz and Allwein. Under this framework your maximization can be restated as a convex optimization problem for which I see no easy special cases. 
An easier-to-investigate (and arguably more interesting)  problem is one identical to yours that omits the fourth constraint that $X\sim B(1/2)$. But I could not find an easy path towards an answer for this either.  
For both of these there might be some magical connection to KL divergence which I am not seeing.

Code
[1]: Below is a crude counterexample finder. 
% Helper functions
    % Binary entropy
fn_h = @(p) -p.*log2(p) - (1-p).*log2(1-p); 
    % MI across mtx_bc when v_distn is input
fn_I = @(mtx_bc,v_distn) fn_h(v_distn(1)) + fn_h(v_distn*mtx_bc(:,1)) ...
    - nansum(nansum(-log2(diag(v_distn)*mtx_bc).*(diag(v_distn)*mtx_bc)));
    % Channel matrix when P(out=0|in=0)=pa, P(out=0|in=1)=pb
fn_mtxBC = @(pa,pb) [pa, 1-pa; pb, 1-pb];

% Set params
d_r_L = 0.4; 
d_r_R = 0.4;
d_xp = 0.146102; % solution to 1-H(p) = 0.4
mtxBSC = fn_mtxBC(d_xp, 1-d_xp);

% Search 
while true
    mtxL = fn_mtxBC(rand, rand);
    mtxR = fn_mtxBC(rand, rand);
    v_d = (mtxL'\[0.5, 0.5]')';
    if (abs(sum(v_d)-1) > 0.001 || ...
        min(v_d) < 0)
        continue
    end
    if(fn_I(mtxL, v_d)      > 0.4 || ...
       fn_I(mtxR, v_d*mtxL) > 0.4)
        continue;
    end
    fprintf('+\n');
    if fn_I(mtxL*mtxR, v_d) > fn_I(mtxBSC*mtxBSC, [0.5, 0.5])
        break;
    end
end

A: the channel does not maximize the mutual information, the source can select a distribution to maximize the mutual information and achieve a mutual information close to the capacity for a given channel. the best channel is an one to one mapping $f$ so $I(X,f(X))=I(X,X)=H(X)$. You can also use the data processing inequality for the Markov chain. 
$I(V;U)\leq min \{  I(V;X,Y),I(U;X,Y)\}=min \{  I(V;Y),I(U;X)\}$
$I(V',U')\leq min \{  I(V';X,Y),I(U';X,Y)\}=min \{  I(V';Y),I(U';X)\}\leq min \{  I(V;Y),I(U;X)\} $
Sorry I don't have enough points to comment.
