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I have no idea about how to prove the next:

Suppose we have two random variables, $t_1$ and $t_2$, that follow the distributions $\lambda_1e^{-\lambda_1 t_1}$ and $\lambda_2e^{-\lambda_2t_2}$, respectively.

Now we have the variable $t=min\{t_1, t_2\}$. Prove that $t$ follows an exponential distribution $\lambda e^{-\lambda t}$ and find the relation between $\lambda$, $\lambda_1$ and $\lambda_2$

All I know is how to generate $t_1$ and $t_2$. Suppose we have a variable $x$ following the distribution $f(x)$, then the acumulative function is $F(x_0)=\int_0^{x_0}f(x)dx$. Now if $\xi\in[0,1]$ uniformly,

$x'=F^{-1}(\xi)$ follows the distribution $f$

Therefore, for our case:

$t_1 = -\lambda_1 ln(1-\xi)\equiv-\lambda_1 ln(\xi)$

$t_2 = -\lambda_2 ln(1-\chi)\equiv-\lambda_2 ln(\chi)$

for $\xi,\chi\in[0,1]$ uniformly.

No idea how to prove the proposition above.

Improve this question
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    $\begingroup$ These random variables describe the first arrival times of a Poisson process. Assemble the two processes into one by taking the union of their events: you have a new Poisson process whose rate (obviously) is the sum of the two rates and its first arrival time (trivially) is the smaller of the two arrival times, QED. $\endgroup$
    – whuber ♦
    Commented Dec 14, 2021 at 17:59
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    $\begingroup$ @BruceET the link is broken. Nevertheless, I found this. But I'm still doing something wrong. If $T = min\{t_1,t_2\}$ and $t$ is the measured time, therefore: $P(T>t)=P(T_1>t,T_2>t)=(1-F_1(t))(1-F_2(t))=e^{-(\lambda_1-\lambda_2)t}$. It should be $1-e^{-(\lambda_1+\lambda_2)t}$ $\endgroup$
    – Abel Gutiérrez
    Commented Dec 15, 2021 at 10:54
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    $\begingroup$ Ok, this is not the CDF, which is $1-P(T>t)$. Thanks! $\endgroup$
    – Abel Gutiérrez
    Commented Dec 15, 2021 at 11:33
  • $\begingroup$ Trying link again. But it seems you've got the solution. // Please also consider method of @whuber's Comment. // Goal should be to learn about exponential distributions, not just to get past this particular problem. // Also, note that the dist'n of the maximum of two exponential distributions is not exponential. $\endgroup$
    – BruceET
    Commented Dec 15, 2021 at 16:13
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    $\begingroup$ My suggestions are elaborated and illustrated at stats.stackexchange.com/questions/215540, which is nearly the same question as yours. If you plan to have any more to do with exponential distributions, you will enjoy (and greatly benefit) from learning about their connection with Poisson processes. stats.stackexchange.com/questions/214421 contains some explanations. This connection often lets you sidestep having to compute various integrals and can make complicated questions solvable. $\endgroup$
    – whuber ♦
    Commented Dec 15, 2021 at 20:07

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