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

I've tried $\int_{0}^{\infty} P(X > z, Y > K/z)dz$, but this seems to overcount by quite a lot. Any advice on how I can express this value in the desired format?

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    $XY > K$ does not necessarily mean $X>z , Y>K/z$. As an example let K=100. From ($X>z , Y>K/z$), $X>10$ means $Y>10$. However, $X>10$ also contains $X=20$ and in this case $Y>5$ is enough (but it is assume that $Y>10$) – 2017-01-24

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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}(Xk)&=\iint_{xy>k}f_{X,Y}(x,y)\mathrm{d}x\mathrm{d}y \\ &=\iint_{xy>k}\frac{\partial}{\partial x}\frac {\partial}{\partial y}\mathbb{P}(X

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Since X and Y are independent events, $$ P(XY>K)=P(X>z \text{ and } Y>K/z)=P(X>z)\,P(Y>K/z). $$ So summing over all possible $z$ gives as a limit the integral $$P(XY>K) = \lim_{t\to0^+}\int_t^{1/t}{P(X>z)\,P(Y>K/z)\,dz}$$ (we need the limit with $t$ because we cannot divide by $0$ in $K/z$).

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    $X$ and $Y$ are random variables, not events, and nowhere does it say that they are independent. – 2017-01-24