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Let $T_x, T_y$ be two random variables that describe the remaining lifetime of two persons aged $x$ and $y$, respectively. The joint density of $T_x$ and $T_y$ is given by

$$f_{T_x,T_y}(s,t)=\begin{cases}\frac{2}{45}\cdot20^{-4}\cdot\Big(9 \cdot 20^2-(3s-t)^2\Big)&,s \in [0,20], t\in[0,60]\\0&, \mathrm{otherwise}\end{cases}$$

How can I use this to determine the probability that the person aged $y$ dies before the person aged $x$?

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You need to integrate the joint density over the region where $T_y$ is less than $T_x$. I suggest drawing a picture. Your integration region is the rectangle $(s,t) \in [0,20]\times[0,60]$ where $s$ corresponds to $T_x$ and $t$ corresponds to $T_y$. If you draw the line $s = t$ you should be able to work out the limits of integration from there.

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  • $\begingroup$ So what are my limits of integration? $\endgroup$ – user636610 Jan 26 at 17:30
  • $\begingroup$ I added some additional hints above. Short answer: draw a picture and it should be clear. $\endgroup$ – inhuretnakht Jan 26 at 18:24
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Draw a rectangle boundary for the joint PDF, where x axis in $[0,20]$ and y axis is $[0,60]$. We need $P(T_y<T_x)$. $y$ is smaller than $x$ below $y=x$ line. So, you need to integrate the joint in the region bounded by the lines $y=x,\ \ y=0, \ \ x=20$ as follows:

$$P(T_y<T_x)=\int_{0}^{20}\int_{0}^{x}f(x,y)dydx$$

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