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I've read on Wikipedia that one can give a stochastic representation of $e$:

In addition to exact analytical expressions for representation of $e$, there are stochastic techniques for estimating $e$. One such approach begins with an infinite sequence of independent random variables $X_1, X_2,\dots$, drawn from the uniform distribution on $[0, 1]$. Let $V$ be the least number $n$ such that the sum of the first $n$ samples exceeds $1$: $$V = \min \left \{ n \mid X_1+X_2+\cdots+X_n > 1 \right \}.$$ Then the expected value of $V$ is $e$: $\mathbb{E}(V) = e$.

I was wondering how to show (analytically) that $\mathbb{E}(V) = e$. I looked at the references but they seems to deal just with numerical aspects.

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    $\begingroup$ This was answered in another question, with $x=1$ plugged in: math.stackexchange.com/questions/8508/… $\endgroup$
    – Alex R.
    Feb 6, 2015 at 17:59
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    $\begingroup$ Or look at the discussion page of the wiki article. Or its archive. There was a long thread about the correctness and appropriateness of this example. $\endgroup$ Feb 6, 2015 at 19:32

1 Answer 1

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We have: $$\mathbb{E}[V]=\sum_{m=0}^{+\infty}\mathbb{P}[V> m]\tag{1}$$ and: $$ \mathbb{P}[V> m] = \mathbb{P}[X_1+\ldots+X_m\leq 1]\triangleq A_m.\tag{2} $$ The pdf of $S_m=X_1+\ldots+X_m$ can be computed by multiple convolution1: over the interval $[0,1]$ it is given by $\frac{t^{m-1}}{(m-1)!}$, hence $A_m=\frac{1}{m!}$ and: $$ \mathbb{E}[V]=\sum_{m\geq 0}\frac{1}{m!}=e $$ as wanted.

1) As an alternative approach, notice that: $$\mathbb{P}[X_1+\ldots+X_m\leq 1]=\mu\left(\left\{(x_1,\ldots,x_m)\in[0,1]^m:x_1+\ldots+x_m\leq 1\right\}\right)=\frac{1}{m!}.$$

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  • $\begingroup$ It seems that for $m=2$ and support $(0,2)$ the proposed pdf, $\frac{t^{m-1}}{(m-1)!}$, does not integrate to 1. $\endgroup$
    – ki3i
    Feb 6, 2015 at 20:46
  • $\begingroup$ @ki3i: It is the pdf of $S_m$ only on $[0,1]$. $\endgroup$ Feb 6, 2015 at 20:52
  • $\begingroup$ If you have Mathematica, just have a look at the output of PDF[UniformSumDistribution[4], x]. $\endgroup$ Feb 6, 2015 at 20:56
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    $\begingroup$ In (1) index $m$ should start at $1$ (not at $0$). $\endgroup$
    – drhab
    Feb 9, 2015 at 10:20
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    $\begingroup$ Again: things are wrong here. To repair it is enough to change "$\mathbb P[V\geq m]$" into "$\mathbb P[V>m]$" on the two spots where this expression occurs. $\endgroup$
    – drhab
    Feb 9, 2015 at 14:24

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