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Suppose $X$ and $Y$ are independent random variables with $X ~ \sim Pareto(\theta_x, c),\ Y \sim Pareto(\theta_y, c)$, where the p.d.f. of $Pareto(\theta, c)$ is

\begin{align} f(t) = \begin{cases} \frac{\theta c^\theta}{t^{\theta+1}} & t \geq c \\ 0 & \text{otherwise} \end{cases} \end{align}

and c.d.f. is

\begin{align} F(t) = \begin{cases} 1 - \frac{c^\theta}{t^\theta} & t \geq c \\ 0 & \text{otherwise} \end{cases} \end{align}

In this settings, I'd like to derive $P(XY > z)$ for some $z > c$.

Using independence of $X$ and $Y$,

\begin{align} P(XY > z) &= \int_{c}^\infty f_X(x) [1-F_Y(\frac{z}{x})] \ dx \\ &= \cfrac{\theta_x \theta_y c^{\theta_x+\theta_y}}{z^{\theta_y}} \int_{c}^\infty x^{\theta_y - \theta_x - 1} \ dx \end{align}

It looks like the integral part will diverge in some parameter settings.

My question is: why $P(XY > z)$ cannot be appropriately defined in such cases? My calculation is not correct at some point?

myuuuuun
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    Your mistake is to use the identity $$1-F_Y(z/x)=(cx/z)^{\theta_Y}$$ for every $x>C$ while this holds only if $$x\leqslant z/c$$ – Did Jun 23 '17 at 15:59
  • I understand. Thank you for your clear answer. – myuuuuun Jun 23 '17 at 16:16
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    Your use of $C$ and $c$ for the Pareto lower bounds is about the worst visual combination one could think of. How about $b$ and $c$? – wolfies Jun 23 '17 at 16:36
  • I'm sorry for making you confusing. $C$ and $c$ indicate a same constant ($C$ is a typo, I modified). Of course I can use different lower bounds for $X$ and $Y$ but I thought it's not essential to my problem. – myuuuuun Jun 23 '17 at 17:55

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