Распределение наибольшего фрагмента сломанной палки (промежутки)

Пусть палка длиной 1 разбита на $k+1$ фрагменты равномерно случайным образом. Каково распределение длины самого длинного фрагмента?

Более формально, пусть $(U_1, \ldots U_k)$ будет IID $U(0,1)$ , и $(U_{(1)}, \ldots, U_{(k)})$ будет ассоциированная статистика порядка, т.е. мы просто упорядочим выборку в таком виде таким образом, что $U_{(1)} \leq U_{(2)} \leq, \ldots , \leq U_{(k)}$ . Позволять$Z_k = \max \left(U_{(1)}, U_{(2)}-U_{(1)}, \ldots, U_{(k)} - U_{(k-1)}, 1-U_{(k)}\right)$ .

Я заинтересован в распределении $Z_k$ . Моменты, асимптотические результаты или приближения для $k \uparrow \infty$ также интересны.

С информацией, предоставленной @Glen_b, я смог найти ответ. Используя те же обозначения, что и вопрос

$$P(Z_k \leq x) = \sum_{j=0}^{k+1} { k+1 \choose j } (-1)^j (1-jx)_+^k,$$

где если a > 0 и 0 в противном случае. Я также даю ожидание и асимптотическую сходимость к распределению Гамбеля ( NB : не бета)$a_+ = a$$a > 0$$0$

$$E(Z_k)= \frac{1}{k+1}\sum_{i=1}^{k+1}\frac{1}{i} \sim \frac{\log(k+1)}{k+1}, \\ P(Z_k \leq x) \sim \exp\left(- e^{-(k+1)x + \log(k+1)} \right).$$

Материал доказательств взят из нескольких публикаций, ссылки на которые приведены в ссылках. Они несколько длинные, но прямые.

1. Доказательство точного распределения

Пусть - IID равномерных случайных величин в интервале ( 0 , 1 ) . Упорядочив их, мы получаем статистику k порядка, обозначенную ( U ( 1 ) , … , U ( k ) ) . Равномерное расстояние определяется как Δ i = U ( i ) - U ( i - 1 ) , где U $(U_1, \ldots, U_k)$$(0,1)$$k$$(U_{(1)}, \ldots, U_{(k)})$$\Delta_i = U_{(i)} - U_{(i-1)}$и$U_{(0)} = 0$ . Упорядоченные интервалы - это соответствующие упорядоченные статистические данные Δ ( 1 ) ≤ … ≤ Δ ( k + 1 ) . Интересующая переменная Δ ( k + 1 ) .$U_{(k+1)} = 1$$\Delta_{(1)} \leq \ldots \leq \Delta_{(k+1)}$$\Delta_{(k+1)}$

For fixed $x \in (0,1)$, we define the indicator variable $\mathbb{1}_i = \mathbb{1}_{\{\Delta_i > x\}}$. By symmetry, the random vector $(\mathbb{1}_1, \ldots, \mathbb{1}_{k+1})$ is exchangeable, so the joint distribution of a subset of size $j$ is the same as the joint distribution of the first $j$. By expanding the product, we thus obtain

$$P(\Delta_{(k+1)} \leq x) = E \left( \prod_{i=1}^{k+1} (1 - \mathbb{1}_i) \right) = 1 + \sum_{j=1}^{k+1} { k+1 \choose j } (-1)^j E \left( \prod_{i=1}^j \mathbb{1}_i \right).$$

$E \left( \prod_{i=1}^j \mathbb{1}_i \right) = (1-jx)_+^k$, which will establish the distribution given above. We prove this for $j=2$, as the general case is proved similarly.

$$E \left( \prod_{i=1}^2 \mathbb{1}_i \right) = P(\Delta_1 > x \cap \Delta_2 > x) = P(\Delta_1 > x) P(\Delta_2 > x | \Delta_1 > x).$$

If $\Delta_1 > x$, the $k$ breakpoints are in the interval $(x,1)$. Conditionally on this event, the breakpoints are still exchangeable, so the probability that the distance between the second and the first breakpoint is greater than $x$ is the same as the probability that the distance between the first breakpoint and the left barrier (at position $x$) is greater than $x$. So

$$P(\Delta_2 > x | \Delta_1 > x) = P\big(\text{all points are in } (2x,1) \big| \text{all points are in } (x,1)\big), \; \text{so} \\ P(\Delta_2 > x \cap \Delta_1 > x) = P\big(\text{all points are in } (2x,1)\big) = (1-2x)_+^k.$$

2. Expectation

For distributions with finite support, we have

$$E(X) = \int P(X > x)dx = 1 - \int P(X \leq x)dx.$$

Integrating the distribution of $\Delta_{(k+1)}$, we obtain

$$E\left(\Delta_{(k+1)}\right) = \frac{1}{k+1}\sum_{j=1}^{k+1}{k+1 \choose j}\frac{(-1)^{j+1}}{j} = \frac{1}{k+1}\sum_{j=1}^{k+1}\frac{1}{j}.$$

$H_i = 1+ \frac{1}{2}+ \ldots + \frac{1}{i}$

$$H_{k+1} = \int_0^1 1 + x + \ldots + x^k dx = \int_0^1 \frac{1-x^{k+1}}{1-x}dx.$$

With the change of variable $u = 1-x$ and expanding the product, we obtain

$$H_{k+1} = \int_0^1\sum_{j=1}^{k+1}{ k+1 \choose j }(-1)^{j+1}u^{j-1}du = \sum_{j=1}^{k+1}{k+1 \choose j}\frac{(-1)^{j+1}}{j}.$$

3. Alternative construction of uniform spacings

In order to obtain the asymptotic distribution of the largest fragment, we will need to exhibit a classical construction of uniform spacings as exponential variables divided by their sum. The probability density of the associated order statistics $(U_{(1)}, \ldots, U_{(k)})$ is

$$f_{U_{(1)}, \ldots U_{(k)}}(u_{(1)}, \ldots, u_{(k)}) = k!, \; 0 \leq u_{(1)} \leq \ldots \leq u_{(k+1)}.$$

If we denote the uniform spacings $\Delta_i = U_{(i)} - U_{(i-1)}$, with $U_{(0)} = 0$, we obtain

$$f_{\Delta_1, \ldots \Delta_k}(\delta_1, \ldots, \delta_k) = k!, \; 0 \leq \delta_i + \ldots + \delta_k \leq 1.$$

By defining $U_{(k+1)} = 1$, we thus obtain

$$f_{\Delta_1, \ldots \Delta_{k+1}}(\delta_1, \ldots, \delta_{k+1}) = k!, \; \delta_1 + \ldots + \delta_k = 1.$$

Now, let $(X_1, \ldots, X_{k+1})$ be IID exponential random variables with mean 1, and let $S = X_1 + \ldots + X_{k+1}$. With a simple change of variable, we can see that

$$f_{X_1, \ldots X_k, S}(x_1, \ldots, x_k, s) = e^{-s}.$$

Define $Y_i = X_i/S$, such that by a change of variable we obtain

$$f_{Y_1, \ldots Y_k, S}(y_1, \ldots, y_k, s) = s^k e^{-s}.$$

Integrating this density with respect to $s$, we thus obtain

$$f_{Y_1, \ldots Y_k,}(y_1, \ldots, y_k) = \int_0^{\infty}s^k e^{-s}ds = k!, \; 0 \leq y_i + \ldots + y_k \leq 1, \; \text{and thus} \\ f_{Y_1, \ldots Y_{k+1},}(y_1, \ldots, y_{k+1}) = k!, \; y_1 + \ldots + y_{k+1} = 1.$$

So the joint distribution of $k+1$ uniform spacings on the interval $(0,1)$ is the same as the joint distribution of $k+1$ exponential random variables divided by their sum. We come to the following equivalence of distribution

$$\Delta_{(k+1)} \equiv \frac{X_{(k+1)}}{X_1 + \ldots + X_{k+1}}.$$

4. Asymptotic distribution

Using the equivalence above, we obtain

$$\begin{align} P\big((k+1)\Delta_{(k+1)} - \log(k+1) \leq x\big) &= P\left(X_{(k+1)} \leq (x + \log(k+1))\frac{X_1 + \ldots + X_{k+1}}{k+1}\right) \\ &= P\left(X_{(k+1)} - \log(k+1) \leq x + (x + \log(k+1))T_{k+1}\right), \end{align}$$

where $T_{k+1} = \frac{X_1+\ldots+X_{k+1}}{k+1} -1$. This variable vanishes in probability because $E\left(T_{k+1}\right) = 0$ and $Var\big(\log(k+1)T_{k+1}\big) = \frac{(\log(k+1))^2}{k+1} \downarrow 0$. Asymptotically, the distribution is the same as that of $X_{(k+1)} - \log(k+1)$. Because the $X_i$ are IID, we have

$$\begin{align} P\left(X_{(k+1)} - \log(k+1) \leq x \right) &= P\left(X_1 \leq x + \log(k+1)\right)^{k+1} \\ &= \left(1-e^{-x - \log(k+1)}\right)^{k+1} = \left(1-\frac{e^{-x}}{k+1}\right)^{k+1} \sim \exp\left\{-e^{-x}\right\}. \end{align}$$

5. Graphical overview

The plot below shows the distribution of the largest fragment for different values of $k$. For $k=10, 20, 50$, I have also overlaid the asymptotic Gumbel distribution (thin line). The Gumbel is a very bad approximation for small values of $k$ so I omit them to not overload the picture. The Gumbel approximation is good from $k \approx 50$.

6. References

The proofs above are taken from references 2 and 3. The cited literature contains many more results, such as the distribution of the ordered spacings of any rank, their limit distribution and some alternative constructions of the ordered uniform spacings. The key references are not easily accessible, so I also provide links to the full text.

1. Bairamov et al. (2010) Limit results for ordered uniform spacings, Stat papers, 51:1, pp 227-240
2. Holst (1980) On the lengths of the pieces of a stick broken at random, J. Appl. Prob., 17, pp 623-634
3. Pyke (1965) Spacings, JRSS(B) 27:3, pp. 395-449
4. Renyi (1953) On the theory of order statistics, Acta math Hung, 4, pp 191-231

This is not a complete answer, but I did some quick simulations, and this is what I obtained:

This looks remarkably beta-ish, and this makes a bit of sense, since the order statistics of i.i.d. uniform distributions are beta wiki.

This might give some starting point to derive the resulting p.d.f..

I'll update if I get to a final closed solution.

Cheers!

Я подготовил ответ для конференции в Сиене (Италия) в 2005 году. Документ (2006 год) представлен на моем веб-сайте здесь (pdf) . Точные распределения всех расстояний (от самых маленьких до самых больших) приведены на страницах 75 и 76.

Я надеюсь выступить с докладом на эту тему на конференции RSS в Манчестере (Англия) в сентябре 2016 года.