$P(U_2<U_3<...<U_n$ and $U_1 \in (U_2, U_2 + c))$, $U_i$ are i.i.d uniform(0,1) r.v's and $c>0$

Question

I'm trying to calculate the probability $P(U_2<U_3<...<U_n, ~\text{and}~~ U_1 \in (U_2, U_2 + c))$, $U_i$ are i.i.d uniform(0,1) r.v's and $a>0$ and $P(U_1<U_3<...<U_n, ~\text{and}~~ U_2 \in (U_3, U_3 + c))$.

I know that the probability of an increasing sequence for i.i.d random variables is $P(U_1<U_2<...<U_n)= \frac{1}{n}\frac{1}{n-1} \cdots\frac{1}{2} \cdot 1 = \frac{1}{n!}$. Because $U_1$ has 1 chance over n to be the smallest, then $U_2$ has 1 chance over $n-1$ options of being the smallest after $U_1$, and so on.

The problem is when one of those i.i.d uniform(0,1) random variables is not in the right order anymore and it belongs to one of the intervals $U_i \in (U_{i+1}, U_{i+1}+ c)$ for a positive constant $c$. I know this probability will depend on $c$ but I'm trying to find it exactly or approximate this probability as better as possible. Any ideas?

Answer 1

\begin{eqnarray*} && \Pr(U_2<U_3<\cdots<U_n, U_1\in(U_2,U_2+c)) \\ &=& \mathbb{E}(\Pr(U_2<U_3<\cdots<U_n, U_1\in(U_2,U_2+c)|U_2)) \\ &=& \mathbb{E}(\Pr(U_2<U_3<\cdots<U_n|U_2)\Pr(U_1\in(U_2,U_2+c)|U_2)) \ \ \ldots \text{Because Events $U_2<U_3<\cdots<U_n$ and $U_1\in(U_2,U_2+c)$ are conditionally independent given $U_2$} \\ &=& \mathbb{E}(\Pr(U_2<U_3<\cdots<U_n, \min(U_3,\ldots, U_n)>U_2|U_2)\Pr(U_1\in(U_2,\min(U_2+c,1))|U_2)) \\ &=& \mathbb{E}(\Pr(U_2<U_3<\cdots<U_n|U_2,\min(U_3,\ldots, U_n)>U_2)\Pr(\min(U_3,\ldots, U_n)>U_2|U_2)\Pr(U_1\in(U_2,\min(U_2+c,1))|U_2)) \\ &=& \mathbb{E}\left(\dfrac{1}{(n-2)!}(1-U_2)^{n-2}(\min(U_2+c,1)-U_2)\right) \\ &=& \dfrac{1}{(n-2)!}\mathbb{E}\left((1-U_2)^{n-2}(\min(U_2+c,1)-U_2)\right) \\ &=& \dfrac{1}{(n-2)!}\mathbb{E}\left(U_2^{n-2}(\min(1-U_2+c,1)-(1-U_2))\right) \ \ldots \text{ Since } U_2, 1-U_2 \text{ are identically distributed.} \\ &=& \dfrac{1}{(n-2)!}\mathbb{E}\left(U_2^{n-2}(\min(c,U_2))\right) \\ &= & \begin{cases} \dfrac{1}{(n-2)!}\mathbb{E}\left(U_2^{n-1}\right) = \dfrac{1}{n((n-2)!)} \ \ \ \ \ \ldots \text{ if } c\geq 1 \\ \dfrac{1}{(n-2)!}\displaystyle\left(\int_0^c u^{n-1} du+ \int_c^1cu^{n-2}du\right)= \dfrac{1}{(n-2)!}\displaystyle\left(\frac{c^n}{n}+ \frac{c}{n-1}-\frac{c^n}{n-1}\right)\ \ \ \ \ \ldots \text{ if } c< 1\end{cases} \end{eqnarray*}