# 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.

## 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.

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 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.
• I interpret your second inequality (Theorem 3) roughly as: "If one density ($h$) decays slower as another one ($g$), then the expectation of an increasing function $f$ under $h$ is larger than under $g$." Commented Nov 13, 2020 at 18:17