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.



Is there a tight lower bound for the expectation of the product of two positive valued random variables?
$\newcommand{\de}{\delta}
\newcommand{\si}{\sigma}
\newcommand{\ep}{\varepsilon}$
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\*}
>  \max(0,\mu\_1\mu\_2-\si\_1\si\_2).
> \tag{1}
> \end{equation\*}
> 
> 
> 


Indeed, by rescaling, without loss of generality
\begin{equation\*}
 \mu\_1=\mu\_2=1.
\end{equation\*}
Let
\begin{equation\*}
r\_j:=\frac1{1+\mu\_j^2}
\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 in (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 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.


