I have a problem with the next fact:
Let $X_1,...,X_n$ be random variables.
Why $$P\left(\min(X_1,X_2,...,X_n)>r\right)=P(X_1>r,X_2>r,...,X_n>r)?$$
Could you clarify me please? I can´t see very clear it, is part of a proof.
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2$\begingroup$ If all the values are larger than $r$, then the smallest one is (and vice-versa -- if the smallest one is larger than $r$, so are all the others). $\endgroup$– Glen_bJan 26, 2018 at 14:14
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1 Answer
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For any $X_1,\dots,X_n$, whether or not exponential or independently distributed:
If $\min(X_1,X_2,...,X_n)>r$, then $X_1>r,X_2>r,...,X_n>r$.
If $X_1>r,X_2>r,...,X_n>r$, then $\min(X_1,X_2,...,X_n)>r$.
Therefore, the event $\min(X_1,X_2,...,X_n)>r$ occurs if and only if (i.e., is equivalent to) the event $X_1>r,X_2>r,...,X_n>r$ occurs. Therefore the probability of these events must be equal.
answered Jan 25, 2018 at 21:14