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Dec 15, 2021 at 20:07 comment added whuber 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.
Dec 15, 2021 at 17:02 comment added Abel Gutiérrez @whuber althought I solved the problem with other method, I don't undertand yours. (Statitstics and probability has always been hard to me)
Dec 15, 2021 at 16:13 comment added BruceET 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.
Dec 15, 2021 at 11:33 comment added Abel Gutiérrez Ok, this is not the CDF, which is $1-P(T>t)$. Thanks!
Dec 15, 2021 at 10:54 comment added Abel Gutiérrez @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 $1-e^{-(\lambda_1+\lambda_2)t}$
Dec 14, 2021 at 17:59 comment added whuber 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.
S Dec 14, 2021 at 17:13 review First questions
Dec 14, 2021 at 17:18
S Dec 14, 2021 at 17:13 history asked Abel Gutiérrez CC BY-SA 4.0