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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.

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    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.* – whuber Dec 14 '21 at 17:59
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    @BruceET the link is broken. Nevertheless, I found [this](https://stats.stackexchange.com/questions/378724/the-probability-distribution-of-waiting-time-until-two-exponentially-distributed?rq=1). 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 $1-e^{-(\lambda_1+\lambda_2)t}$ – Abel Gutiérrez Dec 15 '21 at 10:54
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    Ok, this is not the CDF, which is $1-P(T>t)$. Thanks! – Abel Gutiérrez Dec 15 '21 at 11:33
  • Trying [link](https://math.stackexchange.com/questions/3655552/distribution-of-minimum-of-two-independent-exponential-distributions) 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. – BruceET Dec 15 '21 at 16:13
  • @whuber althought I solved the problem with other method, I don't undertand yours. (Statitstics and probability has always been hard to me) – Abel Gutiérrez Dec 15 '21 at 17:02
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    My suggestions are elaborated and illustrated at https://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. https://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. – whuber Dec 15 '21 at 20:07

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