$X_1,X_2,$.....are i.i.d standard uniform U(0,1),let $$N:=\min\{n \ge2 | X_n>X_{n-1}\}$$ $$T=\min\{n \ge 1 | X_1+...+X_n>1\}$$
We have conclusions that:
- $\mathbb{E}[N]=e$
- $\mathbb{E}[T]=e$
Conclusion 1 is easy to prove, but how about 2?
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What is the relationship between $T$ and $e$? – gt6989b May 09 '19 at 11:43
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Sorry， E[T]=e，I made a clerical error – 张心以 May 09 '19 at 11:50
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Is there something missing in your definition of $N$ that would make it a number rather than a set of numbers? – David K May 09 '19 at 12:55
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I re-edit the problem, clearer now – 张心以 May 09 '19 at 13:01
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N is not a set, it is a random variable. – 张心以 May 09 '19 at 13:03
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also, the T is a random variable – 张心以 May 09 '19 at 13:03
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Is $N$ the least $n$ for which $X_n>X_{n-1}$? – paw88789 May 09 '19 at 13:04
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Yes, and T is the smallest n for which $\sum_{n=1}^{n}X_n>1$ – 张心以 May 09 '19 at 13:10
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https://math.stackexchange.com/q/214399/321264, https://math.stackexchange.com/q/111314/321264 – StubbornAtom May 02 '20 at 07:51
1 Answers
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Here is the proof of conclusion 1: $$\mathbb{P}(N>n)=\mathbb{P}(X_1>X_2>...>X_n)=\frac{1}{n!}$$ $$\mathbb{E}[N]=\sum_{n=1}^{\infty}\mathbb{P}(N>n)=e$$
张心以
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I use python to get conclusion 2, but I don't have specific proof, so I'm looking for the proof of conclusion 2. – 张心以 May 09 '19 at 13:30