Prove the inequality: $|Cov[f(X),g(Y)]|\leq|\rho|\sqrt{Var[f(X)]Var[g(Y)]}$, where $X,Y$ are jointly Gaussian with $\mathbb{E}[XY]=\rho$ and $X,Y\sim\mathcal{N}(0,1)$, $f,g\in L^2(\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx)$.

My attempt: Write out the chaotic expansion of $f, g$ as $f(x)=\sum_{k=0}^{\infty}a_kH_k(x)$ and $g(x)=\sum_{k=0}^{\infty}b_kH_k(x)$, where $H_k$ is the k-th Hermite polynomial.

It is easy to see that $a_0=\mathbb{E}[f(X)]$ and $b_0=\mathbb{E}[g(X)]$. By some calculation, I obtain $$Var[f(X)]=\sum_{k=1}^\infty k!a_k^2$$ and $$Var[g(Y)]=\sum_{k=1}^\infty k!b_k^2,$$ and $$Cov[f(X),g(Y)]=\sum_{k=1}^\infty k!a_kb_k\rho^k.$$

So we can write out the Left Hand Side of the claimed inequality as $$|Cov[f(X),g(Y)]|=|\sum_{k=1}^\infty k!a_kb_k\rho^k|$$ and the Right Hand Side $$|\rho|\sqrt{Var[f(X)]Var[g(Y)]}=|\rho|\sqrt{\sum_{k=1}^\infty k!a_k^2\sum_{l=1}^\infty l!b_k^2}=|\rho|\sqrt{\sum_{k,l=1}^\infty k!l!a_k^2 b_k^2}.$$ Then I have no clue how to proceed.


1 Answer 1


You are just one step away from the conclusion. Since $|\rho| \leq 1$, it follows from the Cauchy-Schwarz inequality that

\begin{align*} |\operatorname{Cov}(f(X), f(Y))| &\leq |\rho|\sum_{k=1}^{\infty} k! |a_k||b_k| \\ &\leq |\rho| \left( \sum_{k=1}^{\infty} k!|a_k|^2 \right)^{1/2}\left( \sum_{k=1}^{\infty} k!|b_k|^2 \right)^{1/2} \\ &= |\rho| \operatorname{Var}(f(X))^{1/2} \operatorname{Var}(g(X))^{1/2}. \end{align*}


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