为什么（0,1）上连续和变量的总和要超过1的数量均具有平均值

让我们总结随机变量，流$X_i \overset{iid}\sim \mathcal{U}(0,1)$ ; 令$Y$为总数需要超过1的项的数量，即$Y$是最小的项，使得

$$X_1 + X_2 + \dots + X_Y > 1.$$

为什么Y的均值$Y$等于欧拉常数$e$？

$$\mathbb{E}(Y) = e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$$

初步观察：$Y$具有比PMF更令人愉悦的CDF

概率密度函数$p_Y(n)$是概率$n$ “只是刚好够”总要超过1，即$X_1 + X_2 + \dots X_n$超过一个，而$X_1 + \dots + X_{n-1}$不不。

累积分布$F_Y(n) = \Pr(Y \leq n)$只需要$n$是“足够的”，即$\sum_{i=1}^{n}X_i > 1$与多少由没有限制。这似乎是处理概率的简单得多的事件。

第二个观察结果：$Y$取非负整数值，因此$\mathbb{E}(Y)$ 可以用 CDF表示

显然$Y$只能取值$\{0, 1, 2, \dots\}$，所以我们可以写它指的是在方面互补CDF，$\bar F_Y$。

$$\mathbb{E}(Y) = \sum_{n=0}^\infty \bar F_Y(n) = \sum_{n=0}^\infty \left(1 - F_Y(n) \right)$$

事实上$\Pr(Y=0)$和$\Pr(Y=1)$都是零，所以第一个两个术语是$\mathbb{E}(Y) = 1 + 1 + \dots$。

至于后来条款，如果$F_Y(n)$的概率是$\sum_{i=1}^{n}X_i > 1$，什么事件是$\bar F_Y(n)$的概率是多少？

第三次观察：$n$单形的（超）体积为$\frac{1}{n!}$

所述Ñ单纯形余心目中占据下的体积标准单元（ñ - 1 ）单纯形的所有阳性orthant的ř Ñ：它是的凸包（Ñ + 1 ）的顶点，尤其是起源加该单元的顶点（ñ - 1 ）在单纯形（1 ，0 ，0 ，... ），（0 ，1 ，0 ，$n$$(n-1)$$\mathbb{R}^n$$(n+1)$$(n-1)$$(1, 0, 0, \dots)$）等$(0, 1, 0, \dots)$

例如，2-单纯形与上面X 1 + X 2 ≤ 1具有区域$x_1 + x_2 \leq 1$2和3用单纯X1+X2+X3≤1具有体积$\frac{1}{2}$$x_1 + x_2 + x_3 \leq 1$6。$\frac{1}{6}$

用于证明通过直接评估为所描述的事件的概率的一体前进ˉ ˚F ý（Ñ ），并链接到其他两个参数，请参阅此数学SE线程。相关的线程可能也很有趣：e和n个单纯形体积之和之间是否存在关系？$\bar F_Y(n)$$e$$n$

修复Ñ ≥ 1。设U i = X 1 + X 2 + ⋯ + X $n \ge 1$模1是部分和的小数部分为我= 1 ，2 ，... ，Ñ。的独立的均匀性 X 1和 X 我+ 1个保证 Ü 我+ 1是一样可能超过 ü 我，因为它是小于它。这意味着所有 n ！序列（U i）的排序同样可能。

$$U_i = X_1 + X_2 + \cdots + X_i \mod 1$$ $i=1,2,\ldots, n$$X_1$$X_{i+1}$$U_{i+1}$$U_i$$n!$$(U_i)$

给定序列U 1，U 2，… ，U n，我们可以恢复序列X 1，X 2，… ，X n。要了解操作方法，请注意$U_1, U_2, \ldots, U_n$$X_1, X_2, \ldots, X_n$

• U 1 = X 1，因为两者都在 0到 1之间。$U_1 = X_1$$0$$1$

• 如果ü 我+ 1 ≥ ü 我，则X 我+ 1 = Ü 我+ 1 - ü 我。$U_{i+1} \ge U_i$$X_{i+1} = U_{i+1} - U_i$

• Otherwise, $U_i + X_{i+1} \gt 1$, whence $X_{i+1} = U_{i+1} - U_i + 1$.

There is exactly one sequence in which the $U_i$ are already in increasing order, in which case $1 \gt U_n = X_1 + X_2 + \cdots + X_n$. Being one of $n!$ equally likely sequences, this has a chance $1/n!$ of occurring. In all the other sequences at least one step from $U_i$ to $U_{i+1}$ is out of order. This implies the sum of the $X_i$ had to equal or exceed $1$. Thus we see that

$$\Pr(Y \gt n) = \Pr(X_1 + X_2 + \cdots + X_n \le 1) = \Pr(X_1 + X_2 + \cdots + X_n \lt 1) = \frac{1}{n!}.$$

This yields the probabilities for the entire distribution of $Y$, since for integral $n\ge 1$

$$\Pr(Y = n) = \Pr(Y \gt n-1) - \Pr(Y \gt n) = \frac{1}{(n-1)!} - \frac{1}{n!} = \frac{n-1}{n!}.$$

Moreover,

$$\mathbb{E}(Y) = \sum_{n=0}^\infty \Pr(Y \gt n) = \sum_{n=0}^\infty \frac{1}{n!} = e,$$

QED.

In Sheldon Ross' A First Course in Probability there is an easy to follow proof:

Modifying a bit the notation in the OP, $U_i \overset{iid}\sim \mathcal{U}(0,1)$ and $Y$ the minimum number of terms for $U_1 + U_2 + \dots + U_Y > 1$, or expressed differently:

$$Y = min\Big\{n: \sum_{i=1}^n U_i>1\Big\}$$

If instead we looked for:

$$Y(u) = min\Big\{n: \sum_{i=1}^n U_i>u\Big\}$$ for $u\in[0,1]$, we define the $f(u)=\mathbb E[Y(u)]$, expressing the expectation for the number of realizations of uniform draws that will exceed $u$ when added.

We can apply the following general properties for continuous variables:

$E[X] = E[E[X|Y]]=\displaystyle\int_{-\infty}^{\infty}E[X|Y=y]\,f_Y(y)\,dy$

to express $f(u)$ conditionally on the outcome of the first uniform, and getting a manageable equation thanks to the pdf of $X \sim U(0,1)$, $f_Y(y)=1.$ This would be it:

$$f(u)=\displaystyle\int_0^1 \mathbb E[Y(u)|U_1=x]\,dx \tag 1$$

If the $U_1=x$ we are conditioning on is greater than $u$, i.e. $x>u$, $\mathbb E[Y(u)|U_1=x] =1 .$ If, on the other hand, $x <u$, $\mathbb E[Y(u)|U_1=x] =1 + f(u - x)$, because we already have drawn $1$ uniform random, and we still have the difference between $x$ and $u$ to cover. Going back to equation (1):

$$f(u) = 1 + \displaystyle\int_0^x f(u - x) \,dx$$ , and with substituting $w = u - x$ we would have $f(u) = 1 + \displaystyle\int_0^x f(w) \,dw$.

If we differentiate both sides of this equation, we can see that:

$$f'(u) = f(u)\implies \frac{f'(u)}{f(u)}=1$$

with one last integration we get:

$$log[f(u)] = u + c \implies f(u) = k \,e^u$$

We know that the expectation that drawing a sample from the uniform distribution and surpassing $0$ is $1$, or $f(0) = 1$. Hence, $k = 1$, and $f(u)=e^u$. Therefore $f(1) = e.$