# On the stochastic definition of $e$

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.

• This was answered in another question, with $x=1$ plugged in: math.stackexchange.com/questions/8508/… Feb 6, 2015 at 17:59
• 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. Feb 6, 2015 at 19:32

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: $$\mathbb{P}[X_1+\ldots+X_m\leq 1]=\mu\left(\left\{(x_1,\ldots,x_m)\in[0,1]^m:x_1+\ldots+x_m\leq 1\right\}\right)=\frac{1}{m!}.$$
• It seems that for $m=2$ and support $(0,2)$ the proposed pdf, $\frac{t^{m-1}}{(m-1)!}$, does not integrate to 1.
• @ki3i: It is the pdf of $S_m$ only on $[0,1]$. Feb 6, 2015 at 20:52
• In (1) index $m$ should start at $1$ (not at $0$). Feb 9, 2015 at 10:20
• 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. Feb 9, 2015 at 14:24