Prove that two random variates must be negatively correlated

Question

Problem statement

Given the probability density $$ p(s,t|a,b,c) = \cases{Z(a,b,c)^{-1} \exp(-as-bt-cst) &if $s \geq 1$ and $t \geq 1$ \\ 0 &otherwise} \label{1} \tag{1}$$ where $$ \begin{align} Z(a,b,c) &\equiv \int \int \exp(-as-bt-cst) \ ds \ dt = \frac{1}{c} \exp\Big(\frac{ab}{c}\Big) \Gamma\Big(0,\frac{(a+c)(b+c)}{c}\Big) \\ \Gamma(0,x) &= \int_x^\infty \exp(-t) \frac{dt}{t} \end{align} $$ and $(a,b,c)$ are "admissible" such that $Z(a,b,c)$ exists, i.e., $$ c > 0, \quad a + c > 0, \quad b + c > 0.$$

Prove (or disprove) that for any such admissible $(a,b,c)$: $$ \langle st \rangle \leq \langle s \rangle \langle t \rangle. \label{2} \tag{2}$$ In other words, the random variates $(s,t)$ must be negatively correlated.

Comments

Lower bound on $\langle st \rangle$

For all $(s,t)$ in the support of \eqref{1} we have $1 \leq (s,t) \leq st$. So $1 \leq \max(\langle s \rangle, \langle t \rangle) \leq \langle st \rangle$.

The case $c=0$

If $c=0$ the random variates $(s,t)$ are independent and thus uncorrelated, such that \eqref{2} reduces to $\langle st \rangle = \langle s \rangle \langle t \rangle.$

Background

Given the constraints: $$ 1 \leq (s,t) < L, \quad \langle s \rangle = S, \quad \langle t \rangle = T, \quad \langle st \rangle = U $$ the density \eqref{1} is the maximum entropy distribution with respect to Lebesgue measure for $L \rightarrow \infty$ where $(a,b,c)$ are admissible Lagrange multipliers which solve the simultaneous equations: $$ -\nabla_{a,b,c} \log Z \equiv (\langle s \rangle, \langle t \rangle, \langle st \rangle) = (S,T,U). \label{3} \tag{3}$$ Thus by construction \eqref{1} is in the exponential family.

I implemented a numerical solver that finds admissible $(a,b,c)$ for given $(S,T,U)$ but it fails badly when $U$ is moderately larger than $ST$: it drives $c \rightarrow 0$ but cannot accurately satisfy \eqref{3}. Further numerical investigation led me to believe (95% sure) that \eqref{3} can only be solved when $U \leq ST$, but after much effort I could not prove this.

I tried several approaches to prove \eqref{2} (e.g. directly attacking \eqref{2}; bounding $\langle st \rangle$ with the moment generating function of \eqref{1}) but the only noteworthy result I could get is the following:

For $c \approx 0$, the solution to \eqref{3} is roughly: $$ a = \frac{1}{S-1}, \quad b = \frac{1}{T-1}, \quad c = -\frac{U - ST}{S^2(T-1)^2 + T^2(S-1)^2 + (S-1)^2 (T-1)^2}. $$ This can be obtained by a first-order expansion of \eqref{3} at $c = 0$ and solving for $(a,b,c)$. Since $c > 0$, we must have that $U < ST$, such that \eqref{2} holds.

I'm unsure what the significance of this result in relation to proving \eqref{2} might be.

Answer 1

Short answer

Your conjecture is true. Long answer is follows.

Long answer

I'll start by giving two simple inequalities, one is a special case of an important inequality, one is an inequality I crafted just for my work.

Preliminaries

Theorem 1 (Special case of FKG inequality)
Let $X$ be real random variable, $f$ and $g$ be two real increasing functions such that $f(X),g(X) \in L^2$. Then, $$ \mathbb{E}( f(X)g(X)) \ge \mathbb{E}(f(X))\mathbb{E}(g(X)) $$ Demonstration 2
We extend our probability space to have another random variable $X_1$ which independent of and identically distributed to $X$.
By the monotonicity of $f,g$, we have: $$ \mathbb{E}\left[ (f(X)-f(X_1))(g(X)-g(X_1) \right] \ge 0 $$

Our desired inequality is just the our expression after expanding.
Done.
Theorem 3 ( Logdensity and the morphing of distribution functions)
Let $X$ and $Y$ be two real random variables, $h$ and $g$ are their respective pseudo densities (definitions below). Suppose that : $$ \frac{h(t)}{h(s)} \ge \frac{g(t)}{g(s)} \forall t\ge s$$

Let $f$ be any increasing function such that $\mathbb{E}( f(X))$ and $\mathbb{E}( f(Y))$ are well defined, then
$$ \mathbb{E}(f(X)) \ge \mathbb{E}(f(Y))$$

Remark 4 the above inequality is taked in the sense that $\frac{a}{b} \ge \frac{c}{d} \leftrightarrow ad-bc \ge 0$ for all $b,d \ge 0$ ) .
Remark 5 all the conditions about $L^2$ about the well-definedness are just there to make all the statements valid, i.e, they are just technical, just forget it. The essence is the inequalities
Remark 6: (Definition of pseudo density) A measurable function $f$ is called a pseudo density if $f$ is nonnegative and $0< \int f < +\infty$. Clearly, by then $\tilde{f}:= \dfrac{f}{|f|_1}$ is a density function.
Demonstration 6

Under our assumption on the monotonicity and the inequality for logdensity, we have:

$$ \int_{\mathbb{R}^2} \left[ f(s)- f(t)\right] [ h(s)g(t)-h(t)g(s) ]dsdt \ge 0$$ Thus, $$ \Vert g \Vert_1 \int_{\mathbb{R}} f(s)h(s)ds \ge \Vert h \Vert_1 \int_{\mathbb{R}} f(s)g(s)ds $$ Done.
Remark 7 FKG's inequality has some interesting generalizations, but they are not within our scope.
Remark 8 The intuition behind those inequalities is that if we assign something big with other big things, something small with other small things, we should have some kind of increment.
Remark 9 If $\log h, \log g$ is well defined and have derivatives, the above inequality condition case be rephrased as $ \dfrac{\partial \log h }{\partial t} \ge \dfrac{\partial \log g }{\partial t}$

Back to our main question

Main question

Let $h_s := p( s , \cdot |a,b,c)$
If $(a,b,c)$ is an admissble choice, we have: $$ \dfrac{ \partial^2 \log h_s(t)}{\partial s \partial t} = -c < 0 $$ Thus for all $s_1< s_2$, we have: $$ \dfrac{\partial \log h_{s_1} }{\partial t} \ge \dfrac{\partial \log h_{s_2} }{\partial t}$$

Besides, we see that $h_s$ is the peusdo density of the conditional law of $T$, given $\{S=s\}$
. As $f(x):=x$ is an increasing function, we hence imply that : $$ g(s):= \mathbb{E}\left( T| S=s \right) \text{ is a decreasing function}$$

So, we see:
$$\langle ST \rangle = \mathbb{E}(ST) = \mathbb{E}\left( S \mathbb{E}(T|S) \right)= \mathbb{E}\left( f(S)g(S) \right) \underbrace{\le}_{FKG's} \mathbb{E}(S) \mathbb{E}(g(S)) = \langle S \rangle \langle T \rangle $$ (Note that $f$ is increasing, while $g$ is decreasing)
So the answer is Affirmative.

Comments

• So you see the sign of $\dfrac{ \partial^2 \log p}{\partial s \partial t}$ implies the sign of correlation.
• The theorem 3 gives some sense of the behaviour of $\mathcal{L}(T|S=s)$ when $s$ varies.