Question: Let $X_1, X_2 ... $ be RVs that are continuous, independent and equally distributed. We define $A_n= \lbrace X_n>max_{k<n} X_k\rbrace$. Prove that $P(\limsup A_n)=1$.

Thoughts: We studied the Borell Cantelli lemma, but I'm having difficulties proving that $\sum_{n=1}^\infty P(A_n)$ doesn't converge. Would love some help with that. Thanks

Answer(according to the hint): There are $n!$ ways to order the n RVs, so each permutation has a probability of $\frac 1{n!}$. Assuming that $X_n$'s is set to be largest, we can now permutate the next n-1 RVs so we have a probability $\frac {(n-1)!}{n!}=\frac 1n$ for such situation (=$A_n$). Since the Harmonic series diverges, we're through, using Borell-Cantelli.

  • $\begingroup$ Borel-Cantelli is offtopic. Just think about what the complement of $\limsup A_n$ represents. $\endgroup$ – Did Feb 23 '15 at 17:42

Hint: If $n$ randomly chosen people are standing in line, how often does it occur that the tallest person is at the end of the line?

To apply the second Borel Cantelli lemma, you also have to prove that the events $\{A_n\}$ are mutually independent.


We must show two things:

  1. $P(A_n) = 1/n$.

  2. $(A_n)_{n \ge 1}$ is an independent collection of events.

From these two, we can use BCL2.

  1. The short version is symmetry. $X_n$ has the same chance of being the max of $X_1, X_2, ..., X_n$ as any of the others

Continuity, independence and equal distribution are essential. See the proof here.

  1. $A_n$'s are independent because the $X_n$'s are independent.


$A_2 = \{X_2 = \max\{X_1, X_2\}\}$

$A_3 = \{X_3 = \max\{X_1, X_2, X_3\}\}$

Since the $X_n$'s are independent, the probability that $X_3$ is the maximum among $X_1, X_2, X_3$ is independent of which is the maximum among $X_1, X_2$


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