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Let $𝑋$ and $π‘Œ$ be random variables. Then the maximal correlation $\rho_{m}(X;Y)$ is defined as:

$$\rho_{m}(X;Y):=\max_{f,g}\mathbb{E}[f(X)g(Y)],$$

where the maximization is taken over real-valued functions $f$ and $g$ such that $\mathbb{E}[f(X)]=\mathbb{E}[g(Y)]=0$ and $\mathbb{E}[f^2(X)]=\mathbb{E}[g^2(Y)]=1$.

A single-function characterization of maximal correlation correlation was given by RΓ©nyi as follows:

$$\rho_{m}^2(X;Y):=\max_{f}\mathbb{E}[(\mathbb{E}[f(X)|Y])^2],$$ where $f$ satisfies the above conditions as well.

I'm interested to show that for any X,Y and Z random variables: $$\rho_{m}((X,Z);Y) \ge \rho_{m}(X;Y)$$

How can we prove this basic definition ?

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    $\begingroup$ Not sure if I understand your notation right, but isn't it just the observation that any function of $X$ can be considered as a function of $X,Z$? $\endgroup$
    – fedja
    Commented Mar 6, 2021 at 13:39
  • $\begingroup$ I have re-edited the post by specifying that X, Y and Z are any random variables, thanks for your remark. I see your point by comparing f(X) and f(X,Z), but in this case we will want $f(X,Z) \ge f(X)$. Notice that we can also invert the problem which is exactly the same: $$\rho_{m}(Y;(X,Z)) \ge \rho_{m}(Y;X)$$ So here we want to compare now : $\max_{f}\mathbb{E}[(\mathbb{E}[f(Y)|X,Z])^2] \ge \max_{f}\mathbb{E}[(\mathbb{E}[f(Y)|X])^2] $ $\endgroup$
    – Vince_maths
    Commented Mar 6, 2021 at 14:10
  • $\begingroup$ I think this questions should be moved to MSE. (Hint: in your final formulation, just condition the integrand on the lhs with resp. to $X$ , and use Jensen's inequality for conditional expectations, before taking expectations.) $\endgroup$
    – esg
    Commented Mar 8, 2021 at 19:29
  • $\begingroup$ Thank you for your comment esg, I think this is exactly the good way to prove this. I just don’t understand exactly how to use the first condition that you mentioned with the $X$ on the lhs just before the Jensen inequality. You wanted to say: $max_{f}𝔼[(𝔼[𝑓(π‘Œ)|𝑋,𝑍])^2|𝑋]$ ? $\endgroup$
    – Vince_maths
    Commented Mar 9, 2021 at 9:22
  • $\begingroup$ No, I wanted to say: for any (admitted) $f$ we have (by Jensen's inequality) $\mathbb{E}\big[\big(\mathbb{E}(f(Y)|X,Z)\big)^2|X\big]\geq \big[\mathbb{E}(f(Y)|X,Z)|X\big]^2= \big[\mathbb{E}(f(Y)|X)\big]^2$ (a.s.), and thus $\mathbb{E}\big[\big(\mathbb{E}(f(Y)|X,Z)\big)^2]\geq \mathbb{E} \big[\big(\mathbb{E}(f(Y)|X)\big)^2\big]$. The rest is obvious. $\endgroup$
    – esg
    Commented Mar 9, 2021 at 17:12

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