I understand that the probability of the sum of n uniform random variables less than 1 is 1/n!, but why the expected sum of n uniform random variables is the summation of 1/n!? How to derive this intuitively?
Thanks, L
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1Welcome to MSE. Please use MathJax. – José Carlos Santos Aug 05 '17 at 13:49
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1I have no idea what "the probability of the sum of n uniform random variables less than 1 is 1/n!" might mean. – lulu Aug 05 '17 at 13:54
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Anything to do with this problem? https://math.stackexchange.com/questions/2368661/avg-number-of-random-numbers-between-0-and-1-required-to-add-up-to-1/2368670#2368670 – Angina Seng Aug 05 '17 at 13:57
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Can you clarify your question? If not, I think it should be closed. – lulu Aug 05 '17 at 14:06
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What does the question have to do with coin tosses?? – bof Aug 05 '17 at 23:47
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Trying some extended mind-reading skills. I hope I understood the question correctly.
Given the sum of n independent and identically distributed U(0, 1) random variables: $ X=\sum _{k=1}^{n}U_{k}$.
Now $P(X<1)=\frac{1}{n!}$.
To show: the expected number of distributions, n, such that $P(X>1)$ is $\sum_{n=0}^{\infty}(\frac{1}{n!}) =e$
How to show this:
the probability that the sum of n variates is bigger than 1 while the sum of n-1 variates is smaller than one is the same as the chance that the sum of n variates is bigger than 1 minus the chance that n-1 variates are bigger than 1:
$(1-\frac{1}{n!})-(1-\frac{1}{(n-1)!})=\frac{1}{n(n-2)!}$
So the desired expected value is $\sum_{n=1}^{\infty} n\cdot \frac{1}{n(n-2)!}=\sum_{n=0}^\infty \frac{1}{n!}=e$
Erik Jurriën
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