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Pick a point at random in the interval $[0,1]$, call it $P_1$.

Pick another point at random in the interval $[0,P_1]$, call it $P_2$.

Pick another point at random in the interval $[0,P2]$, call it $P_3$.


Let $S = P_1+P_2+P_3+\cdots$

What is the probability that $S$ is divergent?

Any thoughts?

P.S. random, in this particular case, means equidistributed. I.e. $P(a<P_1<b)=b-a$.

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Hint: Compute E(Pk) for every k and deduce E(S). If ever E(S) is finite, this tells you that... – Did May 19 '14 at 7:51
Well, its seems like E(P_k) = 1/2^k and E(S)=1. Is this legit reasoning? E(P_k)=0.5*E(P_k-1) =? 0.5^2*E(P_k-2)=... Also, if this is correct, it would imply that whatever the probability distribution, no matter how skewed towards 1 it is, the sum is always finite. Since 0<E(P1)<1 thus infinite sum: E +E^2 +E^3 +... converges. – Elie Bergman May 19 '14 at 8:01
Very much so. Well done. Let me suggest that you post your own solution as an answer (note that, after a while, you may even accept it). – Did May 19 '14 at 8:03
This is a tail event you are talking about and Kolmogorov's 0-1 law says that this probability is either 0 or 1, so which is it? – Georgy May 19 '14 at 8:21
@Georgy This seems offtopic to solve the present question. – Did May 19 '14 at 9:07

For every $k$, $\mathbb E(P_k)=1/2^k$ thus: $$\mathbb E(S) = \mathbb E(P_1)+\mathbb E(P_2)+\cdots = 1/2 + 1/4 + \cdots= 1$$ Since $\mathbb E(S)$ is finite it follows that $P(S=\infty) = 0$, otherwise the expectation $\mathbb E(S)$ would be infinite.

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Clearly it is possible for the sum to diverge. Do you mean that sum converges with probability $1$? This is not true though since you do need to know something about the PDF, otherwise, the atomic PDF that will always choose the harmonic series gives you a problem. – Ittay Weiss May 19 '14 at 8:21
Yes your right. I was assuming something crazy; that the PDF was in a sense "self similar" at all levels (as the uniform distribution is). – Elie Bergman May 19 '14 at 8:23
"Something crazy"? I fully disagree. Your remark that "No matter what the PDF for our selection, the sum will always be finite" is actually true, if suitably interpreted. To be precise, if $P_k=X_1X_2\cdots X_k$ for every $k$, for some i.i.d. $[0,1]$-valued sequence $(X_k)$, then $S$ is almost surely finite for every distribution of the random variables $X_k$ (the degenerate case when $X_k=1$ almost surely excepted, naturally). In other words, independence and equi-distribution are enough to conclude. (All these are (simple cases of) well-known models.) – Did May 19 '14 at 8:37
I'd say that proving $E(P_k)=2^{-k}$ requires some work... – 5xum May 19 '14 at 8:55
@Georgy I wonder what you are trying to achieve with this comment. Everybody even vaguely educated in probability theory is aware of the "modification of the notion of" convergence (rather than "limit") involved here, it is called almost sure convergence. There are quite rigorous ways to describe what you try to do with n1 and n2, and they lead to the conclusion that P(the series converges)=1 and P(the series diverges)=0. So, which mysteries remain? I see none. (Unrelated: what happened to your suggestion of a tail event/Kolmogorov's 0-1 law approach, should we forget it altogether?) – Did May 20 '14 at 8:28

protected by Alexander Gruber May 23 '14 at 20:09

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