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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 $1/n!$?

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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 $P(U_1\geq U_2\geq...\geq U_n)=P(U_1>U_2>...>U_n)$.

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  • $\begingroup$ why continuous mean $Pr(U_1 \geq U_2)=Pr(U_1 >U_2)$ $\endgroup$ – user469065 Sep 27 at 19:04
  • $\begingroup$ Not only continuous, the fact that they are independent is also important. It follows from Fubini's theorem. Look at this question for example: math.stackexchange.com/questions/1422825/… $\endgroup$ – Mark Sep 29 at 5:51

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