If $X$ is a random variable then the covariance of two nondecreasing transformations of $X$ is nonnegative (see here). This fact is very intuitive but it turns out that it does not immediately generalize to random vectors. I am wondering what is known about the class of random vectors for which this condition holds.
To formalize this question, assume from now on the following:
- A function $g: \mathbb R^n \to \mathbb R$ is called nondecreasing if it is nondecreasing in each component.
- Let $$X=(X_1,\dots,X_n): (\Omega, \mathcal A) \to (\mathbb R^n,\mathcal B^n)$$ be a random vector defined on the probability space $(\Omega, \mathcal A, P)$.
- Let $g, h: \mathbb R^n \to \mathbb R$ be measurable functions with $E[g(X)^2]<\infty$, $E[h(X)^2]<\infty$.
Theorem 1: If $g$ and $h$ are nondecreasing and $n=1$ then $$\mathrm{Cov}[g(X),h(X)] \ge 0.$$
Theorem 2: If $g$ and $h$ are nondecreasing and $n\ge 2$ then $\mathrm{Cov}[g(X),h(X)] \ge 0$ does not necessarily hold.
Question: For $n\ge 2$, what is known about the class of random vectors $X$ for which $$\mathrm{Cov}[g(X),h(X)] \ge 0\tag{*}$$ holds whenever $g$ and $h$ are nondecreasing?
Some observations:
- Note that $(*)$ is the same as $$E[g(X)h(X)] \ge E[g(X)]E[h(X)].$$
- If $X$ is such that $(*)$ holds for any nondecreasing $g$ and $h$ and $X$ possesses finite second moments then all components of $X$ are nonnegatively correlated. To see this, simply choose $g(x) = x_i$ and $h(x) = x_j$ for $i, j = 1,\dots, n$.
- Theorem 3: $(*)$ holds for any nondecreasing $g$ and $h$ if the components of $X$ are stochastically independent.
Proof of Theorem 1: See here.
Proof of Theorem 2: Let $n = 2$ and suppose $X=(X_1, X_2)$ takes on the values $(1,0)$ and $(0,1)$ with probability $0.5$ each. Define measurable, nondecreasing functions by $g(x_1, x_2) = x_1$ and $h(x_1, x_2) = x_2$. Then \begin{align} \mathrm{Cov}[g(X), h(X)] &= E[g(X)h(X)]-E[g(X)]E[h(X)]\\ &= E[X_1 X_2]-E[X_1]E[X_2]\\ &= 0-0.5\times0.5\\ &= -0.25. \end{align}
Proof of Theorem 3: We proceed by induction on the number of components $n$. If $n=1$ we can simply apply Theorem 1. Now suppose $X$ has $n+1$ components. Write $X=(X_1, Y)$ where $Y$ consists of the last $n$ components. Then $$E[g(X_1,Y)h(X_1,Y)|Y=y] \ge E[g(X_1,Y)|Y=y]\,E[h(X_1,Y)|Y=y]$$ since $g(\cdot,y)$ and $h(\cdot,y)$ are nondecreasing functions for any $y$. Therefore \begin{align} E[g(X)h(X)] &= E[\,E[g(X_1,Y)h(X_1,Y)|Y]\,] \\ &\ge E[\,E[g(X_1,Y)|Y]\,E[h(X_1,Y)|Y]\,]. \end{align} further, since $Y$ and $X_1$ are independent, $$\tilde g(y) := E[g(X_1,Y)|Y=y] = E[g(X_1,y)]$$ is a nondecreasing function of the $n$-dimensional vector $y$. Hence, by applying the induction assumption, \begin{align} E[\,E[g(X_1,Y)|Y]\,E[h(X_1,Y)|Y]\,] &= E[\,\tilde g(Y)\,\tilde h(Y)\,]\\ &\ge E[\,\tilde g(Y)]\,E[\tilde h(Y)\,] \\ &= E[g(X)]\,E[h(X)]. \end{align}
It follows that $E[g(X)h(X)] \ge E[g(X)]E[h(X)].$