Do product distributions (or graph products) eventually cluster as more products are taken? Say we have a joint distribution on a finite alphabet $\mathcal{X}\times \mathcal{Y}$. It could be a communication link where we want to send a random message $X$ over a channel, but it gets garbled into $Y$.
The joint distribution of $(X,Y)$ can be visualized as an undirected, weighted bipartite graph on the vertex set $\mathcal{X}\times \mathcal{Y}$. 
If we fix a small $\varepsilon > 0$ and repeat the experiment enough times, then by the joint-AEP, for $n$ large enough, the outcome will be nearly uniformly distributed in some small subset $S \subset \mathcal{X}^n \times \mathcal{Y}^n$. Then with up to negligible probability, the outcome corresponds to an edge picked (nearly) uniformly at random from the unweighted bipartite graph $G=(\mathcal{X}^n,\mathcal{Y}^n,E)$, where there is only an edge between $x$ and $y$ if $(x,y)\in S$.
If we repeat the experiment $n\cdot i$ times, then we will be choosing edges from some bigger graph $G_i$ on vertex set $\mathcal{X}^{n\cdot i}\times\mathcal{Y}^{n\cdot i} $ where there is an edge between $(x,y)$ if $(x,y)\in S\underset{i\ \text{times}}{\underbrace{\times \dots \times}} S$. 
This can be described succinctly by defining a shorthand operation $\tilde\ast$ on general undirected, unweighted bipartite graphs $G_A=(\mathcal{X}_A,\mathcal{Y}_A,E_A)$ and $G_B=(\mathcal{X}_B,\mathcal{Y}_B,E_B)$:

$G_A\tilde\ast G_B$ is the bipartite graph $(\mathcal{X}_A\times \mathcal{X}_B,\mathcal{Y}_A\times \mathcal{Y}_B,E_A\times E_B).$

Then $G_{i+1}= G_i \tilde\ast G$. Here is my question:

As $i$ gets large, then does the graph $G_i$ eventually cluster, and become close to a disjoint union of a bunch of complete bipartite graphs? (In particular, $2^{nI(X;Y)}$ complete bipartite graphs $K_{2^{nH(X|Y)},2^{nH(Y|X)}}$)

This would be a very handy property from the perspective of information theory and communications. It implies that if you have enough noisy observations $Y$, then on average you can extract $I(X;Y)$ bits of pure, noiseless information about $X$ from them.
We know from unrelated analysis that this information extraction can be done when $(X,Y)$ is multivariate Gaussian.
Update: This is false, but a similar statement about encoding a Gaussian source is true.

Things I have tried


*

*The planar separator theorem for arbitrary genus cannot help us. If it could, a contradiction would arise where $G_i$ eventually clusters arbitrarily well, which we know can't be true.

*Spectral clustering ideas tells us that if $G_i$ does eventually cluster, then the eigenvectors which correspond to the smallest eigenvalues of $G_i'$s Laplacian should eventually start looking like indicator functions for clusters as $i$ gets large. Unfortunately, even if the alphabet $\mathcal{X}\times \mathcal{Y}$ is small, the product distribution's alphabet gets large very fast, even after taking few products so I can't directly test this. 


Nonetheless, if $(X,Y) \sim \frac{1}{8}\left[\begin{smallmatrix}1 & 1 & 0 & 0 \\ 1& 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0& 0& 1 & 1\end{smallmatrix}\right],$ $G_5$'s Laplacian is computationally easy. Ignoring the trivial 0 eigenvalue, there are 10 small eigenvectors of $G_5$'s Laplacian. They look like:

While each eigenvector obviously preferences some nodes and discards others, no immediate indicator-function structure is apparent to me, at least without more intelligent sorting.
Any pertinent intuitions are appreciated. Thanks for reading
 A: This conjecture is false in general. Unfortunately the only way I can think to prove this is through information theory.
Consider a binary source $X\sim 
B(0.5)$ sent through an independent binary symmetric channel with crossover probability $p_c$ so that output $Y=1-X$ with probability $p_c$ and $X$ otherwise. Then $I(X;Y)=1-H(p_c)$ (here $H$ is the entropy of a Bernoulli RV at some probability).
The rate-distortion function for the binary-symmetric channel with Hamming distortion is well-known to be:
$$R(D)=H(p_c)-H(D);\quad 0 \leq D \leq \min\{1-p_c,p_c\}$$
Say $p_c=0.2.$ Then when encoding $Y$ at rate $I(X;Y)=1-H(p_c)\approx 0.278 \text{ bits}$, the lowest possible amount of Hamming distortion is: $R^{-1}(I(X;Y))\approx 0.0922>0.$ This can be attained in limit as the sequence of source observations grows long. Denote the bound-attaining sequence of encodings as $U^n_H$. Note that $\ell_{n\to\infty} \frac{1}{n}I(X^n, U^n_H) < I(X;Y),$ because of the positive amount of Hamming distortion.
The conjecture guarantees (for a long enough sequence of source observations) the existence of encoder for $Y^n$ which produces an output $U^n$ satisfying the following, for any $\varepsilon>0$: 


*

*$X^n \to Y^n \to U^n$

*$H(U^n)=I(X^n;Y^n)$

*$I(X^n;U^n)=I(X^n;Y^n)-\varepsilon-\mathcal{O}(1/n).$


See this post for the connection.
However, for small enough $\varepsilon$ this implies: $$\ell_{n\to\infty} \frac{1}{n} I(X^n;U^n)=I(X;Y)-\varepsilon >\ell_{n\rightarrow \infty}\frac{1}{n}I(X^n;U_H^n)$$
Then $U^n$ is encoded at the same rate as $U_H^n$ but, by the inequality, has less Hamming distortion. This violates the converse of the rate-distortion theorem.
