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Let $X_i \stackrel{\mathcal L}{=} i \times U_i$ where $U_i$ are iid uniform $[0,1]$ time stamps $\sum$. (I don't quite get what time stamps means here, but I guess it means $U_i$ are uniformly distributed on $[0,1]$

The question is, for a certain $i$, would it be possible to calculate this probability:

$$ \Pr \{\cap_{j < i} (X_j < X_i) \} $$

In other words, what's the probability that $X_i$ is greater than any $X_j, j \in [1, i -1]$.

2 Answers 2

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Let $M_i = \max \{ X_j: j \le i-1\}$. If $k \le t \le k+1$ with $0 \le k \le i-1$, $P(M_i \le t) = \prod_{j=k+1}^{i-1} \frac{t}{j} = \frac{t^{i-1-k} k!}{(i-1)!}$. Thus $$P(M_i \le X_i) = \sum_{k=0}^{i-1} \frac{1}{i} \int_{k}^{k+1} \frac{t^{i-1-k} k!}{(i-1)!}\, dt = \sum_{k=0}^{i-1} \frac{k!}{i!} \frac{(k+1)^{i-k} - k^{i-k}}{i-k}$$ I don't think there's a closed form for this. The first few values (for $i$ from 1 to 10) are $$ 1,\frac{3}{4},{\frac {23}{36}},{\frac {163}{288}},{\frac {3697}{7200}},{ \frac {5113}{10800}},{\frac {38947}{88200}},{\frac {14070953}{33868800 }},{\frac {359861221}{914457600}},{\frac {1713552101}{4572288000}}$$

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Since $X_i$ and $X_j$ are independent for $i \not= j$:

$$ \mathbb{P}\left( \cap_{j

This reproduces Robert's result.

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