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Given U,V independent exponential R.Vs. Find the law of min(U,V).

I understand some of the basics, but I need exact formal proof at the beginning since I've seen very few of these questions. I will write the beginning which I hesitate of. Could you correct my mistakes? Or approve my 100% correctness.

My solution:

U and V are Random Variables. Thus we are looking for

$P\{\{t<U\} \bigcap \{t<V\}\}$

U, V are independent. Hence

$=P\{t<U\}\bigcap P\{t<V\}$

Which is in turn:

$=F_U(t)F_V(t)$

And then we calculate and differentiate.

Did I write it wrong somewhere? Is everything correct?

Vitali Pom
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    It's "almost" correct except the last equality. Note that the cumulative distribution function $F_U(t)$ is defined as the probability that $U\le t,$ not the other way round. That means, $P(U>t)=1-F_{U}(t).$ – Raghav Feb 01 '20 at 18:19
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    One more comment: ${t < U}$ and ${t < V}$ are events -- that is, sets -- so it's fine to write something like ${t < U } \cap {t < V}$, as you did. However, $P(t < U)$ is a number, so it doesn't make sense to write something like $P(t < U) \cap P(t < V)$ much like how it doesn't make sense to write $3 \cap 8$. Use multiplication there instead. – Aaron Montgomery Feb 01 '20 at 18:45
  • Wow. Many thanks! I'm surprised of myself, disapointed though that I had not understood it beforehand. :) Many thanks. – Vitali Pom Feb 01 '20 at 18:47
  • Thanks Aaron, many thanks I was making this mistake all the time. Will correct now. – Vitali Pom Feb 01 '20 at 18:50
  • Hi, sorry similar but no. As I stated I do know the answer, I just didn't know how to write it using my own words. I apologies for the dumb question, but I really had a hard time with these min variables. I needed someone to write it using theory of groups which the answers lack there. Will it help if I edit the title? I changed it I think 10 times while I was writing the question. Edit: I edited the title, let me know if it helps. – Vitali Pom Feb 01 '20 at 20:38

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