부러진 막대의 가장 큰 조각 분포 (간격)

길이 1의 스틱을 무작위 로 $k+1$ 조각으로 균일하게 끊습니다. 가장 긴 조각의 길이 분포는 무엇입니까?

더 공식적으로하자 BE IID U ( 0 , 1 ) 및하자 ( U ( 1 ) , ... , U는 ( k는 ) ) , 관련 주문 통계 수 즉, 우리는 단순히 같은에서 샘플 주문 그 방법 U ( 1 ) ≤ U ( 2 ) ≤ , ... , ≤ U ( K ) . 방해$(U_1, \ldots U_k)$$U(0,1)$$(U_{(1)}, \ldots, U_{(k)})$$U_{(1)} \leq U_{(2)} \leq, \ldots , \leq U_{(k)}$$Z_k = \max \left(U_{(1)}, U_{(2)}-U_{(1)}, \ldots, U_{(k)} - U_{(k-1)}, 1-U_{(k)}\right)$.

I am interested in the distribution of $Z_k$. Moments, asymptotic results, or approximations for $k \uparrow \infty$ are also interesting.

With the information given by @Glen_b I could find the answer. Using the same notations as the question

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

where $a_+ = a$ if $a > 0$ and $0$ otherwise. I also give the expectation and the asymptotic convergence to the Gumbel (NB: not Beta) distribution

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

The material of the proofs is taken from several publications linked in the references. They are somewhat lengthy, but straightforward.

1. Proof of the exact distribution

Let $(U_1, \ldots, U_k)$ be IID uniform random variables in the interval $(0,1)$. By ordering them, we obtain the $k$ order statistics denoted $(U_{(1)}, \ldots, U_{(k)})$. The uniform spacings are defined as $\Delta_i = U_{(i)} - U_{(i-1)}$, with $U_{(0)} = 0$ and $U_{(k+1)} = 1$. The ordered spacings are the corresponding ordered statistics $\Delta_{(1)} \leq \ldots \leq \Delta_{(k+1)}$. The variable of interest is $\Delta_{(k+1)}$.

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

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

We will now prove that $E \left( \prod_{i=1}^j \mathbb{1}_i \right) = (1-jx)_+^k$, 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

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

The last equality is a classic representation of harmonic numbers $H_i = 1+ \frac{1}{2}+ \ldots + \frac{1}{i}$, which we demonstrate below.

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

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. 그래픽 개요

아래 그림은 다른 값에 대한 가장 큰 조각의 분포를 보여줍니다. $k$. 에 대한$k=10, 20, 50$, 나는 점근선 Gumbel 분포 (가는 선)를 겹쳐 놓았습니다. Gumbel은 작은 값에 대한 매우 나쁜 근사치입니다.$k$그래서 나는 사진에 과부하가 걸리지 않도록 생략합니다. Gumbel 근사값은$k \approx 50$.

6. 참고 문헌

위의 증거는 참고 문헌 2와 3에서 가져온 것입니다. 인용 문헌에는 임의의 순위 정렬 간격 분포, 한계 분포 및 정렬 된 균일 간격의 일부 대체 구성과 같은 더 많은 결과가 포함되어 있습니다. 주요 참고 문헌에 쉽게 접근 할 수 없으므로 전체 텍스트에 대한 링크도 제공합니다.

1. Bairamov et al. (2010) 주문 된 균일 간격에 대한 결과 제한 , 통계 지, 51 : 1, pp 227-240
2. Holst (1980) 무작위로 끊어진 막대기 조각의 길이에서 , J. Appl. 잠언 17, pp 623-634
3. 파이크 (1965) 간격 , JRSS (B) 27 : 3, pp. 395-449
4. Renyi (1953) 주문 통계 이론 , 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!

2005 년 시에나 (이탈리아)에서 회의에 대한 답변을 작성했습니다.이 논문 (2006)은 내 웹 사이트 (pdf)에 나와 있습니다. 모든 간격 (최소에서 최대)의 정확한 분포는 75 및 76 페이지에 있습니다.

2016 년 9 월 맨체스터 (영국)에서 열린 RSS 컨퍼런스에서이 주제에 대한 프레젠테이션을하고 싶습니다.