Finding the cdf of a distribution Two people wish to explore Asia, where both tourists have a lifetime that is exponentially distributed with mean $\theta$. $T$ is the length of time that Asia is being explored by at least one of the tourists.
How do I find the cdf of the distribution given the following:
1) Both tourists leave the hotel at time $t = 0$, where their lifetimes are independently distributed.
2) The second tourist leaves the hotel on the event of the first tourist becoming sick. 
 A: The more difficult part of the problem is translating the "words" into technical language. 
For 1), let $X$ be the lifetime of Tourist A, and $Y$ the lifetime of Tourist B. Exploration continues until both die. If $U$ is the length of time that exploration continues, then $U=\max(X,Y)$. 
The probability that $U\le u$ is the probability both are dead by time $U$. The density function of $X$ is $\frac{1}{\theta}e^{-x/\theta}$ (for $\theta\gt 0$). It follows that the probability A is dead by time $u$ is $\int \frac{1}{\theta}e^{-x/\theta}\,dx$. This is $1-e^{-u/theta}$ (for $u\gt 0$). We get the same expression for $\Pr(Y\le u)$. Thus by independence
$$F_U(u)=(1-e^{-u/\theta})^2$$
for $u\gt 0$, and $F_U(u)=0$ elsewhere. 
2) suppose that Tourist A leaves first. Then exploration time $V$ is given by $V=X+Y$. To find $\Pr(V\le v)$ for $v\gt 0$, we integrate the joint density  function over the part of the first quadrant that is below the line $x+y=v$. Thus for positive $v$ we have
$$F_V(v)=\int_{x=0}^v \left(\int_{y=0}^{v-x}\frac{1}{\theta^2}e^{-x/\theta}e^{-y/\theta}\,dy\right)\,dx.$$
The integration is straightforward. For generalizations, please see the gamma distribution, or more specifically the Erlang distribution. 
