# Is there a tight lower bound for the expectation of the product of two positive valued random variables?

Let $$X,Y$$ be two (dependent) random variables with $$\mathbb{P}(X\ge 0)=\mathbb{P}(Y\ge 0)=1$$.

I want to find a tight lower bound of $$\mathbb{E}(XY)$$ when $$X,Y$$ are non-negative, almost surely.

Note that one trivial lower bound is $$0$$. But I want to know if there is an elegant method in literature for finding a tight lower bound, which has information of only the individual moments of the random variables $$X,Y$$.

I realize that the following lower bound can be obtained simply by using Cauchy-Scwartz inequality $$\mathbb{E}(XY)\ge \mu_X\mu_Y-\sigma_X\sigma_Y,$$ where $$\mu_X=\mathbb{E}X,\ \sigma_X^2=\mathrm{Var}(X)$$, and similarly $$\mu_Y,\sigma_Y$$ defined for $$Y$$. However, I don't know that if $$X,Y$$ are non-negative, almost surely, then whether this lower bound is non-negative. A little bit of algebra shows that this claim is equivalent to the following claim $$\frac{\mu_X^2}{\mu_{X^2}}+\frac{\mu_Y^2}{\mu_{Y^2}}\ge 1.$$ However, when $$X=Y$$, this claim is equivalent to proving that $$\mu_X^2\ge \frac{\mu_{X^2}}{2},$$ which, however, is false in general because Holder's inequality implies that $$|\mu_X|\le \frac{\sqrt{\mu_{X^2}}}{\sqrt{2}}$$. Any ideas how I can proceed? Thanks in advance.

• In what terms do you want the tight lower to be? Apr 7 '19 at 1:39
• I want the lower bound to be at least non-negative and to be equal to $\mu_{X^2}$ when $X=Y$. Apr 7 '19 at 4:51
• "I want the lower bound to be at least non-negative and to be equal to $\mu_{x^2}$ when X=Y" I interpret X=Y as meaning that $X=Y$ in distribution, because, you are assuming no knowledge of the dependency between $X$ and $Y$, so you can not also assume $X=Y$ a.s. Given that interpretation, why would the quoted statement be true? For instance, let $X = 1$ or $3$, w.p. $1/2, 1/2$. Then $EX^2 = 9/2$. Then $Y = 4 - X$ has the same distribution as $X$, but $E(XY) = 3 \ne E(X^2)$. Apr 7 '19 at 11:07

$$\newcommand{\de}{\delta} \newcommand{\si}{\sigma} \newcommand{\ep}{\varepsilon}$$ Let us present the exact lower bound on $$EXY$$ in terms of $$\mu_1:=\mu_X$$, $$\mu_2:=\mu_Y$$, $$\si_1:=\si_X$$, $$\si_2:=\si_Y$$, as follows:

The minimum of $$EXY$$ over all nonnegative random variables (r.v.'s) $$X$$ and $$Y$$ with prescribed positive values of $$\mu_1$$, $$\mu_2$$, $$\si_1$$, $$\si_2$$ is $$\begin{equation*} (\mu_1\mu_2-\si_1\si_2)_+, \tag{1} \end{equation*}$$ where $$u_+:=\max(0,u)$$.

Indeed, by rescaling, without loss of generality $$\begin{equation*} \mu_1=\mu_2=1. \end{equation*}$$ Let $$\begin{equation*} v_j:=1+\mu_j^2,\quad r_j:=1/v_j; \end{equation*}$$ everywhere here $$j\in\{1,2\}$$. Observe that $$\begin{equation*} \mu_1\mu_2-\si_1\si_2\le0\iff\si_1\si_2\ge1\iff r_1+r_2\le1. \tag{2} \end{equation*}$$ Consider the two possible cases, according to this observation.

Case 1: $$r_1+r_2\le1$$. Then the bound (1) is $$0$$. On the other hand, let $$S:=\{0,1,2\}$$ with the probability measure on $$S$$ assigning masses $$1-r_1-r_2,r_1,r_2$$ to the points $$0,1,2$$, respectively. Let r.v.'s $$X$$ and $$Y$$ be defined on $$S$$ as follows: $$\begin{equation*} X(0)=X(2)=0,\ X(1)=v_1,\quad Y(0)=Y(1)=0,\ Y(2)=v_2. \end{equation*}$$ Then $$\begin{equation*} EX=EY=1,\quad \operatorname{Var}X=\si_1^2,\quad \operatorname{Var}Y=\si_2^2,\quad EXY=E0=0. \end{equation*}$$ So, the bound $$(1)$$ is exact in Case 1.

Case 2: $$r_1+r_2\ge1$$. Then the bound (1) follows indeed by Cauchy--Schwarz: $$\begin{equation*} EXY-\mu_1\mu_2=E(X-\mu_1)(Y-\mu_2)\ge-\si_1\si_2. \end{equation*}$$ By (2), in Case 2 we have $$\si_1\si_2\le1$$. So, we can find $$p\in(0,1)$$ such that for $$q:=1-p$$ we have $$\begin{equation*} \si_2^2\le p/q\le1/\si_1^2. \end{equation*}$$ Let now $$S:=\{0,1\}$$ with the probability measure on $$S$$ assigning masses $$q,p$$ to the points $$0,1$$, respectively. Let r.v.'s $$X$$ and $$Y$$ be defined on $$S$$ as follows: $$\begin{equation*} X(0)=1-\si_1\sqrt{p/q},\quad X(1)=1+\si_1\sqrt{q/p}, \end{equation*}$$ $$\begin{equation*} Y(0)=1+\si_2\sqrt{p/q},\quad Y(1)=1-\si_2\sqrt{q/p}. \end{equation*}$$ Then $$X,Y\ge0$$, $$\begin{equation*} EX=EY=1,\quad \operatorname{Var}X=\si_1^2,\quad \operatorname{Var}Y=\si_2^2,\quad EXY=1-\si_1\si_2. \end{equation*}$$ So, the bound $$(1)$$ is exact in Case 2 as well. $$\Box$$

You were pretty close to this answer.

• Great! Thanks for the full answer. Apr 8 '19 at 9:12