Let $m>n\ge1$ . Take different $x_{1},\ldots,x_{m}$ . By cyclically permuting the first $n$ numbers of any permutation of these $m$ numbers, we see that exactly $m!/n$ permutations are such that “a record occurs at time $n$ ”. Similarly, there are $m!/m=(m-1)!$ permutations such that “a record occurs at time $m$ ”. Further, among these permutations, by the first observation, exactly $(m-1)!/n$ are such that “records are set at both times $n$ and $m$ ”.

Now, split the whole probability space into $1+m!$ parts such that if there are equal numbers among the first $m$ values of random variables $X_{1},\ldots,X_{m}$ , we place $\omega$ into the first part, otherwise we place all $\omega$ corresponding to different permutations of the first $m$ values into different $m!$ parts, where in $m!/m$ parts we place only those permutations where the value of the $m$ th random variable is a record, in $m!/n$ parts where the value of the $n$ th random variable is a record, and in $(m-1)!/n$ parts where the value of both $n$ th and $m$ th random variables are records. Given that $\{X_{k}\}_{k\in\mathbb{N}}$ are independent identically distributed random variables, and they are continuous, the probability of the first part is $0$ , and the probabilities of all the other $m!$ parts are equal. We conclude that $\mathbb{P}(E_{m})=\tfrac{1}{m}$ (and, of course, $\mathbb{P}(E_{n})=\tfrac{1}{n}$ ), and $\mathbb{P}(E_{n}\cap E_{m})=\tfrac{1}{mn}=\mathbb{P}(E_{n})\mathbb{P}(E_{m})$ . Similarly, we can consider any finite number of times where records occur, concluding $\mathbb{P}(\cap_{k=1}^{m}E_{n_{k}})=\prod_{k=1}^{m}\mathbb{P}(E_{n_{k}})$ in general.

This certainly convinces me and “my tutor”.