Let $\{X_i\}$ be a sequence of i.i.d continuous RV. Call $i$ record-setting if $$X_i > \max_{1 \leq j < i} X_j.$$
It is well–established on math.SE and elsewhere that the events "$X_n$ is record-setting" and "$X_{n+1}$ is record-setting" are independent. Below is a "proof" that they are indeed dependent. Where's the error?
Let $$M_{n} := \max\limits_{1 \leq j \leq n} X_j.$$
The event "$n$ and $n+1$ are both record-setting" occurs if and only if:
- $X_n > M_{n-1}$
- $X_{n+1} > M_{n-1}$
- $X_{n+1} > X_n$.
By symmetry, event #1 has probability $1/n$. Since the $X$ are i.i.d, event #2 has probability $1/n$ as well, and is independent of event #1. Finally, given events #1 and #2, the conditional probability of event #3 is $1/2$. Therefore the probability that both $n$ and $n+1$ are record-setters is $$P(A\cap B) = \frac 1 {2n^2} \neq \frac 1 {n(n+1)} = P(A)P(B)$$ and so the events are dependent.
Best Answer
Events #1 and #2 are not independent of each other. Indeed, they are conditionally independent of each other given $M_{n-1}$, but not unconditionally independent of each other. Intuitively, this should make sense, since the event $\{ X_n > M_{n-1}\}$, tells us $M_{n-1}$ is more likely to be 'smaller,' which makes the event $\{X_{n+1} > M_{n-1}\}$ more likely to occur.
For a concrete example, take $X_i \sim \text{Uniform}[0,1]$ and $n = 2$. Then $M_1 = X_1$, and it is a straightforward computation (by conditioning on $X_1$) to show that $\mathbb{P}(X_2 > X_1, X_3 > X_1) = 1/3$, not $1/4$.