Let a stick of length 1 be broken in $k+1$ fragments uniformly at random. What is the distribution of the length of the longest fragment?

More formally, let $(U_1, \ldots U_k)$ be IID $U(0,1)$, and let $(U_{(1)}, \ldots, U_{(k)})$ be the associated order statistics, i.e. we simply order the sample in such a way that $U_{(1)} \leq U_{(2)} \leq, \ldots , \leq U_{(k)}$. Let $Z_k = \max \left(U_{(1)}, U_{(2)}-U_{(1)}, \ldots, U_{(k)} - U_{(k-1)}, 1-U_{(k)}\right)$.

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

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    $\begingroup$ This is a well studied problem; see R. Pyke (1965), "Spacings," JRSS(B) 27:3, pp. 395-449. I'll try to come back to add some information later unless someone beats me to it. There's also a 1972 paper by the same author ("Spacings revisited") but I think what you're after is pretty much all in the first. There's some asymptotics in Devroye (1981), "Laws of the Iterated Logarithm for Order Statistics of Uniform Spacings" Ann. Probab., 9:5, 860-867. $\endgroup$
    – Glen_b
    Jul 22, 2015 at 4:57
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    $\begingroup$ Those should also give some good search terms to find later work if you need it. $\endgroup$
    – Glen_b
    Jul 22, 2015 at 5:10
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    $\begingroup$ This is awesome. The first reference is hard to find. For those interested, I put it on The Grand Locus. $\endgroup$
    – gui11aume
    Jul 22, 2015 at 15:59
  • $\begingroup$ Please correct the misprint: $Y_{(k)}$ instead of $U_{(k)}$. $\endgroup$
    – Viktor
    Sep 9, 2017 at 19:39
  • $\begingroup$ Thanks @Viktor! For such small things, don't hesitate to do the edit yourself (I think that it will be reviewed by other users for approval). $\endgroup$
    – gui11aume
    Sep 11, 2017 at 8:50

3 Answers 3


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. 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 $k \approx 50$.

Distribution of the largest fragment of a broken stick

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
  • $\begingroup$ Brilliant. By the way, is there a known asymptotics to $E(Z_k ^2)$? $\endgroup$
    – Amir Sagiv
    Feb 19, 2018 at 21:52
  • $\begingroup$ @AmirSagiv this is a good question. I had a quick look at the references and I could not find it. I could also not adapt the proof above. This made me realize that I don't know what the distribution of a square of a Gumbel is. Perhaps a good place to start? $\endgroup$
    – gui11aume
    Feb 21, 2018 at 14:30
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    $\begingroup$ $gui11aume Look here : mathoverflow.net/a/293381/42864 $\endgroup$
    – Amir Sagiv
    Feb 21, 2018 at 20:50
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    $\begingroup$ @AmirSagiv This is a very good post. For some reason, I misunderstood your question and thought you were interested in the asymptotic distribution of $Z_k^2$ (even though your comment was very clear), so my comment above is not so relevant. $\endgroup$
    – gui11aume
    Feb 22, 2018 at 23:16
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    $\begingroup$ @Masacroso, the figure is made with standard R. Every feature is added manually. The background is created with rect, the white lines with abline, the legend with legend and the gradient with colorRampPalette. The style was chosen to match my blog. You can see the code to produce relatively similar figures at this post. Click on the Penrose triangle to display the code. $\endgroup$
    – gui11aume
    Jan 25, 2020 at 22:02

This is not a complete answer, but I did some quick simulations, and this is what I obtained: Histogram of the longest fragment

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.


  • $\begingroup$ Just one more thing, the shape of histogram for increasing k doesn't change considerably, apart from getting "squished" close to 0. $\endgroup$
    – Lima
    Jul 22, 2015 at 0:39
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    $\begingroup$ Thank you for your thoughts @Lima (and welcome to Cross Validated). I think your answer can be improved. First, I would refrain from making statements without proof. If this is incorrect, you may put the people who see this thread on the wrong track. Second, I would document what you did. Without the value of $k$ that you used nor the code, the figure does not help anybody. Finally, I would copy-edit the answer and remove everything that is not directly answering the question. $\endgroup$
    – gui11aume
    Jul 23, 2015 at 7:26
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    $\begingroup$ Thanks for the suggestions. They're valid beyond stack exchange, and I'll remember to use them. $\endgroup$
    – Lima
    Jul 28, 2015 at 1:42

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.

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    $\begingroup$ Welcome to the site. We are trying to build a permanent repository of high-quality statistical information in the form of questions & answers. Thus, we're wary of link-only answers, due to linkrot. Can you post a full citation & a summary of the information at the link, in case it goes dead? Also, please don't sign your posts here. Every post has a link to your userpage where you can post that information. $\endgroup$ Mar 23, 2016 at 14:42

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