Does stochastic domination of $X$ and $Y$ imply stochastic domination of $X \cdot Y$?

Question

Suppose the random variables $X \geq 0$ and $Y \geq 0$ are both stochastically dominated by $Z \geq 0$, i.e. \begin{align*} & P(X \leq x), P(Y \leq x) \geq P(Z \leq x) \ , \ \forall x \geq 0 \ . \end{align*} Write $X \overset{p}{\leq} Z$ and $Y \overset{p}{\leq} Z$ for stochastic domination.

Question: Can give a random variable $F(Z)$, such that $XY \overset{p}{\leq} F(Z)$? An obvious candidate would be $F(Z) = Z^2$, but I don't know how to go about proving this. Note that we make no assumptions on the dependence structure between $X$ and $Y$.

Any help is much appreciated!

Answer 1

The answer is yes.

Indeed, for any real $t\ge0$, \begin{equation*} P(XY\ge t)\le P(X\ge\sqrt t)+P(Y\ge\sqrt t) \le 2P(Z\ge\sqrt t)=:G(t); \tag{10}\label{10} \end{equation*} note that $G$ is a left-continuous (l.c.) function decreasing (non-strictly) from $2$ to $0$ on $[0,\infty)$. So, there exists \begin{equation*} t_*:=\max\{t\ge0\colon G(t)\ge1\}\in[0,\infty); \end{equation*} In particular, \begin{equation*} G(t)<1\text{ for }t>t_*. \end{equation*} Next, since $P(Z>z+1)$ is right-continuous (r.c.) in real $z$, for each real $t>t_*$ there exists \begin{equation*} f(t):=\min\{z\ge0\colon P(Z>z+1)\le G(t)\}\in[0,\infty), \tag{15}\label{15} \end{equation*} and the function $f$ is l.c. on $(t_*,\infty)$ and increasing (non-strictly) from $f(t_*+)$ to $\infty$ on $(t_*,\infty)$. So, for each real $z>f(t_*+)$ there exists \begin{equation*} h(z):=\max\{t>t_*\colon f(t)\le z\}\in(t_*,\infty), \end{equation*} and then for any real $t>t_*$ and any real $z>f(t_*+)$ \begin{equation*} h(z)\ge t\iff f(t)\le z. \tag{20}\label{20} \end{equation*} For real $z\ge0$, let finally \begin{equation*} F(z):=1(z\le f(t_*+))\,t_*+1(z>f(t_*+))h(z). \tag{30}\label{30} \end{equation*} Note that for any real $z>f(t_*+)$ there is some real $t>t_*$ such that $z>f(t)$ and hence, by \eqref{20}, $h(z)\ge t>t_*$. So, \begin{equation*} F(z)\ge t_* \text{ for all real }z\ge0. \end{equation*}

So, \begin{equation*} 0\le t\le t_*\implies P(F(Z)\ge t)=1. \tag{40}\label{40} \end{equation*}

Finally, by \eqref{30}, \eqref{20}, and \eqref{15}, \begin{align*} t>t_*\implies &P(F(Z)\ge t) \\ &\ge P(Z>f(t_*+),h(Z)\ge t) \\ &=P(Z>f(t_*+),f(t)\le Z) \\ &\ge P(Z>f(t)) \\ &>G(t). \tag{50}\label{50} \end{align*}

Thus, by \eqref{10}, \eqref{40}, and \eqref{50}, for all real $t\ge0$, \begin{equation} P(XY\ge t)\le P(F(Z)\ge t).\quad\Box \end{equation}

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Illustrating example: If $P(Z>z)=e^{-z}$ for real $z>0$, then $t_*=\ln^2 2$, $f(t)=\max(0,\sqrt t-\ln2-1)$ for $t>t_*$, $f(t_*+)=0$, and $h(z)=(z+\ln2+1)^2$ for real $z>0$, so that here \begin{equation*} F(z)=1(z=0)\,\ln^2 2+1(z>0)(z+\ln2+1)^2 \end{equation*} for real $z\ge0$.