Tại sao số lượng biến thống nhất liên tục trên (0,1) cần thiết cho tổng của chúng vượt quá một biến có nghĩa là

Chúng ta hãy tổng hợp một luồng các biến ngẫu nhiên, $X_i \overset{iid}\sim \mathcal{U}(0,1)$ ; Đặt $Y$ là số lượng số hạng chúng ta cần cho tổng số vượt quá một, nghĩa là $Y$ là số nhỏ nhất sao cho

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

Tại sao giá trị trung bình của $Y$ bằng hằng số Euler $e$ ?

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

Quan sát đầu tiên: $Y$ có CDF dễ chịu hơn PMF

Các hàm xác suất khối lượng $p_Y(n)$ là xác suất mà $n$ là "chỉ vừa đủ" cho tổng số vượt quá sự hiệp nhất, tức là $X_1 + X_2 + \dots X_n$ vượt quá một khi $X_1 + \dots + X_{n-1}$ không không phải.

Phân phối tích lũy $F_Y(n) = \Pr(Y \leq n)$ đơn giản yêu cầu $n$ là "đủ", tức là $\sum_{i=1}^{n}X_i > 1$ không hạn chế bao nhiêu. Điều này có vẻ như một sự kiện đơn giản hơn nhiều để đối phó với xác suất.

Quan sát thứ hai: $Y$ lấy các giá trị nguyên không âm để $\mathbb{E}(Y)$ có thể được viết theo CDF

Clearly $Y$ can only take values in $\{0, 1, 2, \dots\}$, so we can write its mean in terms of the complementary CDF, $\bar F_Y$.

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

In fact $\Pr(Y=0)$ and $\Pr(Y=1)$ are both zero, so the first two terms are $\mathbb{E}(Y) = 1 + 1 + \dots$.

As for the later terms, if $F_Y(n)$ is the probability that $\sum_{i=1}^{n}X_i > 1$, what event is $\bar F_Y(n)$ the probability of?

Third observation: the (hyper)volume of an $n$-simplex is $\frac{1}{n!}$

The $n$-simplex I have in mind occupies the volume under a standard unit $(n-1)$-simplex in the all-positive orthant of $\mathbb{R}^n$: it is the convex hull of $(n+1)$ vertices, in particular the origin plus the vertices of the unit $(n-1)$-simplex at $(1, 0, 0, \dots)$, $(0, 1, 0, \dots)$ etc.

For example, the 2-simplex above with $x_1 + x_2 \leq 1$ has area $\frac{1}{2}$ and the 3-simplex with $x_1 + x_2 + x_3 \leq 1$ has volume $\frac{1}{6}$.

For a proof that proceeds by directly evaluating an integral for the probability of the event described by $\bar F_Y(n)$, and links to two other arguments, see this Math SE thread. The related thread may also be of interest: Is there a relationship between $e$ and the sum of $n$-simplexes volumes?

Fix $n \ge 1$. Let

$$U_i = X_1 + X_2 + \cdots + X_i \mod 1$$ be the fractional parts of the partial sums for $i=1,2,\ldots, n$. The independent uniformity of $X_1$ and $X_{i+1}$ guarantee that $U_{i+1}$ is just as likely to exceed $U_i$ as it is to be less than it. This implies that all $n!$ orderings of the sequence $(U_i)$ are equally likely.

Given the sequence $U_1, U_2, \ldots, U_n$, we can recover the sequence $X_1, X_2, \ldots, X_n$. To see how, notice that

• $U_1 = X_1$ because both are between $0$ and $1$.

• If $U_{i+1} \ge U_i$, then $X_{i+1} = U_{i+1} - U_i$.

• 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 $f(1) = e.$