# Let $U_1,U_2,…$ be a sequence of independent uniform $(0, 1)$random variables, question about $\Pr(N > n)$

Let $$U_1,U_2,...$$ be a sequence of independent uniform $$(0, 1)$$random variables and let $$N:=\min\{n\geq 2: U_n>U_{n-1}\}$$ $$M:=\min\{n\geq 2: U_{1}+\cdots+U_n>1\}$$ Show that surprisingly, $$N$$ and $$M$$ have the same probability distribution, and their common mean is e!

This is Example 3.28 on page 124 in Ross's book (Introduction to Probability Models-11th edition)

My question is that: why $$\Pr(N > n)$$ is $$\Pr({U_1 > U_2 > ... > U_n})$$ but not $$\Pr({U_n > U_{n-1} > ... > U_1})$$ which is $$1/n!$$?

What does it mean that $$N>n$$? It means that for $$k=2,3,...,n$$ we can't have $$U_k>U_{k-1}$$, otherwise $$N$$ would be at most $$k$$. So $$N>n$$ if and only if $$U_1\geq U_2\geq...\geq U_n$$. Now, since the random variables are continuous we have $$P(U_1\geq U_2\geq...\geq U_n)=P(U_1>U_2>...>U_n)$$.
• why continuous mean $Pr(U_1 \geq U_2)=Pr(U_1 >U_2)$ – user469065 Sep 27 at 19:04