Let $X$ and $Y$ be two $\rho$ correlated Gaussian vectors, such that $X,Y\sim N(0,1)^n$ and $E[X\_iY\_i]=\rho$.
Let $M\_X = f(X)$ and $M\_Y = f(Y)$ be $k$-bit functions of $X$ and $Y$, that is $H(X)=H(Y)=k$.


We then have that $M\_X - X - Y - M\_Y$ is a Markov Chain.


By the standard Strong Data Processing Inequality (SDPI), we get $I(M\_X; Y) \le \rho^2 I(M\_X; X) \le \rho^2 k$ and likewise $I(M\_Y; X) \le \rho^2 k$.


I would like to show that $$I(M\_X ; M\_Y) \le \rho^2 k / 2.$$


This is motivated by the fact that when $M\_X$ and $M\_Y$ are seen as approximations to $X$ and $Y$, then one will need twice as good approximations when both $X$ and $Y$ are approximated, rather than if only one of them is (the case $I(X ; M\_Y)$.)


I'm aware of [the general result by Yury Polyanskiy and Yihong Wu](https://arxiv.org/abs/1508.06025), which lets one compute an end-to-end SDPI for a Markov Chain (or graph). However, in that case, it is assumed that each link loses some constant factor, wherein my case the two links $M\_X - X$ and $Y - M\_Y$ are rate capped rather than noisy.


Do you know any information-theoretical tricks, or known relations, that I'm unaware of which might be useful?



Strong Data Processing Inequality for capped channels
What happens if $n = k = 1$ and $X = Y$? In this case $\rho = 1$. Let $M\_X = 0$ if $X < 0$ and 1 otherwise. Let $M\_Y = 0$ if $Y < 0$ and 1 otherwise. Then $I(M\_X;M\_Y) = H(M\_X) - H(M\_X|M\_Y) = 1$. This seems to contradict your wanted inequality.


