Répartition du plus gros fragment d'un bâton cassé (espacements)

Soit un bâton de longueur 1 cassé en $k+1$ fragments uniformément au hasard. Quelle est la distribution de la longueur du plus long fragment?

Plus formellement, soit $(U_1, \ldots U_k)$ soit IID $U(0,1)$ , et soit $(U_{(1)}, \ldots, U_{(k)})$ les statistiques d'ordre associées, c'est-à - dire que nous commandons simplement l'échantillon dans d'une manière que $U_{(1)} \leq U_{(2)} \leq, \ldots , \leq U_{(k)}$ . Laisser$Z_k = \max \left(U_{(1)}, U_{(2)}-U_{(1)}, \ldots, U_{(k)} - U_{(k-1)}, 1-U_{(k)}\right)$ .

Je m'intéresse à la distribution de $Z_k$ . Les moments, les résultats asymptotiques ou les approximations de $k \uparrow \infty$ sont également intéressants.

Avec les informations fournies par @Glen_b, j'ai pu trouver la réponse. En utilisant les mêmes notations que la question

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

où si a > 0 et 0 sinon. Je donne également l'attente et la convergence asymptotique à la distribution de Gumbel ( NB : pas Beta)$a_+ = a$$a > 0$$0$

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

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

Soit des variables aléatoires uniformes IID dans l'intervalle ( 0 , 1 ) . En les commandant, nous obtenons les statistiques d'ordre k notées ( U ( 1 ) , … , U ( k ) ) . Les espacements uniformes sont définis comme Δ i = U ( i ) - U ( i - 1 ) , avec U $(U_1, \ldots, U_k)$$(0,1)$$k$$(U_{(1)}, \ldots, U_{(k)})$$\Delta_i = U_{(i)} - U_{(i-1)}$$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. 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

Ce n'est pas une réponse complète, mais j'ai fait quelques simulations rapides, et voici ce que j'ai obtenu:

Cela semble remarquablement beta-ish, et cela a un peu de sens, car les statistiques d'ordre des distributions uniformes iid sont beta wiki .

Cela pourrait donner un point de départ pour dériver le pdf résultant.

Je mettrai à jour si j'arrive à une solution finale fermée.

À votre santé!

I produced the answer for a conference in Siena (Italy) in 2005. The paper (2006) is presented on my web-site here (pdf). The exact distributions of all the spacings (smallest to largest) are found on pages 75 & 76.

I'm hoping to give a presentation on this topic at the RSS Conference in Manchester (England) in September 2016.