Let a pair of random variables $(X,Y)$ be continuous random variables (i.e., they both have density with respect to Lebesgue measure) with joint distribution $P_{XY}$. The maximal correlation $\rho_m(X;Y)$ between $X$ and $Y$ is defined as $$\rho_m(X;Y):=\sup_{f,g} \rho(f(X); g(Y)),$$ where $\rho(\cdot; \cdot)$ denotes the correlation coefficient (aka Pearson correlation coefficient). R\'{e}nyi suggested the following simpler representation for maximal correlation: $$\rho_m(X;Y):=\sup_{f,g}\mathbb{E}[f(X)g(Y)],$$ 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 by Renyi as follows: $$\rho^2_m(X;Y)=\max_{f}\mathbb{E}[(\mathbb{E}[f(X)|Y])^2],$$ where $f$ satisfies the above conditions as well. One can easily see that for any arbitrary random variables $X$ and $Y$, $0\leq\rho_m(X;Y)\leq 1$ where 0 is achieved iff $X$ and $Y$ are independent and 1 is achieved iff there exists a pair of functions $f$ and $g$ such that $f(X)=g(Y)$ almost surely.

It is a well-known fact that for jointly Gaussian $(X,Y)$, we have $\rho_m^2(X; Y)=\rho^2(X; Y)$.

Suppose now that only $Y$ is Gaussian. I would like to know a necessary and sufficient condition on $P_{XY}$ such that $\rho_m^2(X; Y)=\rho^2(X; Y)$. In particular, I am interested to see if this statement is correct: if $\rho_m^2(X;Y)$ is close to $\rho^2(X;Y)$, then $X$ is close to a Gaussian random variable.

  • $\begingroup$ Douglas, you are right, the equality can not imply Gaussianity. However, I am looking for a necessary and sufficient condition on $P_{XY}$ with Gaussian $Y$ such that we have the equality. Can you also elaborate a bit on "then interpolate to get equality". $\endgroup$ – math-Student Jan 12 '16 at 20:58
  • $\begingroup$ I was mistaken about interpolation. Your statement seems reasonable to me now. $\endgroup$ – Douglas Zare Jan 13 '16 at 0:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.