Expressing P(XY > K) in terms of P(X > a, Y > b)? I'm trying to express the value $P(XY &gt; K)$ in terms of $P(X &gt; a, Y &gt; b)$ for some values of $a$ and $b$ that are probably related to $K$.  I'm generally looking for a formula where $P(X &gt; a, Y &gt; b)$ is present in some form or other - it's fine if it's inside an integral or whatnot.

I've tried $\int_{0}^{\infty} P(X &gt; z, Y &gt; K/z)dz$, but this seems to overcount by quite a lot.  Any advice on how I can express this value in the desired format?
Remember that $$f_{X,Y}(x,y)=\frac{\partial}{\partial x}\frac {\partial}{\partial y} F_{X,Y}(x,y)=\frac{\partial}{\partial x}\frac {\partial}{\partial y}\mathbb{P}(X&lt;x,Y&lt;y)$$
Then $$\begin{align} \mathbb{P}(XY&gt;k)&amp;=\iint_{xy&gt;k}f_{X,Y}(x,y)\mathrm{d}x\mathrm{d}y \\ &amp;=\iint_{xy&gt;k}\frac{\partial}{\partial x}\frac {\partial}{\partial y}\mathbb{P}(X&lt;x,Y&lt;y)\mathrm{d}x\mathrm{d}y\end{align}$$
The integral bounds will depend on the supports for both $x$ and $y$.
Since X and Y are independent events, $$ P(XY&gt;K)=P(X&gt;z \text{ and } Y&gt;K/z)=P(X&gt;z)\,P(Y&gt;K/z). $$
So summing over all possible $z$ gives as a limit the integral
$$P(XY&gt;K) = \lim_{t\to0^+}\int_t^{1/t}{P(X&gt;z)\,P(Y&gt;K/z)\,dz}$$
(we need the limit with $t$ because we cannot divide by $0$ in $K/z$).