Uniform distribution series first dropping point?

Question

Let $X_1, X_2,...$ be a sequence of independent uniform random variables on $(0,1)$. Define: $$N := \text{min} \{n\geq 2: X_n < X_{n-1}\}.$$ Calculate $E(N)$.

I think this problem asked about the expectations of the first dropping entry of the series, I also did a simulation and I think the answer is $e$? But I'm not sure how to compute it. I tried using the definition of expectations and I compute that $P(N=2)$ is $1/2$, but I stuck with computing $P(N=3)$. Can anyone tell me how to do this?

Answer 1

1. Check that $$N=2+\mathbf1_{\{X_1<X_2\}}+\mathbf1_{\{X_1<X_2<X_3\}}+\cdots$$
2. Check that $\Bbb P(X_1<\cdots<X_k)=\frac1{k!}$ for any $k\ge2$.
3. Take expectations & Fubinize.

Answer 2

Observe that: $$N>n\iff X_1\leq X_2\leq\cdots\leq X_n\text{ for }n=2,3,\dots$$so that by symmetry: $$\Pr(N>n)=\frac1{n!}\text{ for }n=2,3,\dots$$ Now apply:$$\mathbb EN=\sum_{n=0}^{\infty}\Pr(N>n)$$