The Dirichlet Law of Uniform Spacings

Why the Joint Law of the Spacings Matters

The ordered observations from a uniform sample do more than divide the interval into consecutive pieces: their entire vector of gaps has a simple joint distribution. Knowing that law gives a precise description of how the sample partitions $[0,1]$, while also making clear why the gaps cannot vary independently: their total is fixed.

Let $U_1,\ldots,U_n$ be independent $\operatorname{Uniform}(0,1)$ random variables, and write their order statistics as $U_{(1)}<\cdots<U_{(n)}$. These inequalities hold with probability one because the sampling distribution is continuous. Include the two boundary gaps as well as the gaps between observations.

Definition: The $n+1$ uniform spacings are $S_1=U_{(1)}$, $S_i=U_{(i)}-U_{(i-1)}$ for $2\leq i\leq n$, and $S_{n+1}=1-U_{(n)}$. They are positive with probability one and satisfy $\sum_{i=1}^{n+1}S_i=1$.

The simplex constraint means that the joint law is not a density on an unrestricted $(n+1)$-dimensional region. We will describe it using $S_1,\ldots,S_n$ as coordinates, with $S_{n+1}=1-\sum_{i=1}^{n}S_i$. This is the same constraint emphasized in the earlier material on uniform spacings; the purpose here is to derive the full joint law on that constrained space.

From Order Statistics to Spacings

For independent uniform observations, the joint density of the order-statistic vector $(U_{(1)},\ldots,U_{(n)})$ is $n!$ on the region $0<u_1<\cdots<u_n<1$, and zero elsewhere. Apply the change of variables $s_1=u_1$, $s_i=u_i-u_{i-1}$ for $2\leq i\leq n$. Its inverse is $u_i=\sum_{j=1}^{i}s_j$.

The Jacobian matrix of the inverse transformation is lower triangular: its entry in row $i$, column $j$ is $1$ if $j\leq i$, and $0$ otherwise. Every diagonal entry is $1$, so the absolute value of its determinant is $1$. The order-statistic support becomes $s_i>0$ for $1\leq i\leq n$ and $\sum_{i=1}^{n}s_i<1$; the remaining spacing is then positive.

Theorem: With respect to ordinary $n$-dimensional Lebesgue measure $ds_1\cdots ds_n$, the density of $(S_1,\ldots,S_n)$ is $$ f(s_1,\ldots,s_n)= \begin{cases} n!, & s_i>0\ (1\leq i\leq n),\quad \sum_{i=1}^{n}s_i<1,\\ 0, & \text{otherwise}. \end{cases} $$ Equivalently, the full spacing vector has constant density $n!$ in the first $n$ simplex coordinates, with $S_{n+1}=1-\sum_{i=1}^{n}S_i$.

This density is a direct consequence of the change-of-variables formula: the order-statistic density contributes $n!$, and the transformation contributes a Jacobian factor of $1$. The support condition also captures the original ordering exactly. Indeed, positive spacings give $u_1=s_1>0$ and $u_i-u_{i-1}=s_i>0$, while $\sum_{i=1}^n s_i<1$ gives $u_n<1$.

The Dirichlet Distribution and the Simplex Density

For clarity, specify the coordinate convention in the definition of the Dirichlet law. For $k\geq2$, the open $(k-1)$-simplex consists of $s_1,\ldots,s_{k-1}>0$ with $\sum_{i=1}^{k-1}s_i<1$, and $s_k=1-\sum_{i=1}^{k-1}s_i$. A density on this simplex is understood with respect to $ds_1\cdots ds_{k-1}$.

Definition: For parameters $\alpha_1,\ldots,\alpha_k>0$, a random vector $S=(S_1,\ldots,S_k)$ has a $\operatorname{Dirichlet}(\alpha_1,\ldots,\alpha_k)$ distribution if its density in the first $k-1$ simplex coordinates is $$ f(s_1,\ldots,s_{k-1}) = \frac{\Gamma(\alpha_1+\cdots+\alpha_k)} {\prod_{i=1}^{k}\Gamma(\alpha_i)} \prod_{i=1}^{k}s_i^{\alpha_i-1}, $$ on the open simplex, and is zero elsewhere. Here $\Gamma(a)=\int_0^\infty t^{a-1}e^{-t}\,dt$ for $a>0$.

The normalizing constant in this definition follows from the simplex integral $$ \int_{\substack{s_i>0,\ 1\leq i\leq k-1\\ \sum_{i=1}^{k-1}s_i<1}} \prod_{i=1}^{k}s_i^{\alpha_i-1}\,ds_1\cdots ds_{k-1} = \frac{\prod_{i=1}^{k}\Gamma(\alpha_i)} {\Gamma(\alpha_1+\cdots+\alpha_k)}. $$ To see why, integrate $\prod_{i=1}^k y_i^{\alpha_i-1}e^{-y_i}$ over $y_i>0$. Integrating each coordinate separately gives $\prod_i\Gamma(\alpha_i)$. Alternatively, set $t=\sum_i y_i$, $s_i=y_i/t$ for $i<k$, and $s_k=1-\sum_{i<k}s_i$. The inverse is $y_i=ts_i$, and its Jacobian has absolute value $t^{k-1}$. The integral becomes the simplex integral above multiplied by $\int_0^\infty t^{\sum_i\alpha_i-1}e^{-t}\,dt=\Gamma(\sum_i\alpha_i)$. All these integrals are finite because every $\alpha_i$ is positive, so the change of variables and separation of the nonnegative integrals are valid.

Theorem: For $n$ independent $\operatorname{Uniform}(0,1)$ observations, their $n+1$ spacings have distribution $\operatorname{Dirichlet}(1,\ldots,1)$, with $n+1$ parameters.

To verify the identification, substitute $k=n+1$ and $\alpha_i=1$ into the Dirichlet density. Since $\Gamma(1)=1$ and $\Gamma(n+1)=n!$, the product of powers is $1$ and the density is $n!$. Its support and coordinate convention are exactly those in the spacing density derived above.

Worked Examples

Worked Example: Two Spacings from One Observation

Let $n=1$. Then $S_1=U_{(1)}=U_1$ and $S_2=1-U_1$. Using $s_1$ as the single simplex coordinate, the support is $0<s_1<1$, with $s_2=1-s_1$.

The joint density in that coordinate is $1!=1$. The Dirichlet formula with parameters $(1,1)$ gives $\Gamma(2)/(\Gamma(1)\Gamma(1))=1$, with both powers equal to zero. Thus $(S_1,S_2)\sim\operatorname{Dirichlet}(1,1)$. As a check, its density integrates to $\int_0^1 1\,ds_1=1$.

Worked Example: A Joint Spacing Probability for Two Observations

For $n=2$, the first two spacings have density $2$ on $s_1>0,\ s_2>0,\ s_1+s_2<1$, and $S_3=1-s_1-s_2$. Find $P(S_1>1/4,\ S_2>1/4)$.

The constraints restrict $s_1$ to $1/4<s_1<3/4$. For each such $s_1$, the allowed values of $s_2$ are $1/4<s_2<1-s_1$. Integrating the joint density gives $$ \begin{aligned} P(S_1>1/4,\ S_2>1/4) &=\int_{1/4}^{3/4}\int_{1/4}^{1-s_1}2\,ds_2\,ds_1\\ &=2\int_{1/4}^{3/4}(3/4-s_1)\,ds_1\\ &=2\left[\frac{3s_1}{4}-\frac{s_1^2}{2}\right]_{1/4}^{3/4}\\ &=2\left(\frac{9}{32}-\frac{5}{32}\right)=\frac14. \end{aligned} $$ The result is a probability between zero and one. Geometrically, requiring the first two gaps to exceed $1/4$ leaves a simplex of the same shape scaled by $1/2$ in two dimensions, whose area is $1/4$ of the original.

Worked Example: Four Gaps Each Exceed a Threshold

Take $n=3$, so the four spacings have density $3!=6$ in the first three coordinates. Find the probability that all four spacings exceed $1/8$.

Set $T_i=S_i-1/8$ for $i=1,\ldots,4$. The event is $T_i>0$ for every $i$, and the sum constraint becomes $\sum_{i=1}^4T_i=1-4/8=1/2$. In the first three coordinates, the allowed region therefore has volume $(1/2)^3/3!$. This volume formula follows by scaling the unit three-simplex, whose volume is $1/3!$, by $1/2$ in each coordinate. Hence $$ P(S_i>1/8\ \text{for all }i) =6\frac{(1/2)^3}{3!} =6\frac{1}{48} =\frac18. $$ The factor $6$ is the density; the volume alone is not the probability.

Normalization, Geometry, and a Common Pitfall

The density $n!$ is constant because all ordered configurations have the same density and the spacing transformation preserves volume. It is also correctly normalized: the region $s_i>0,\ \sum_{i=1}^n s_i<1$ has volume $1/n!$. One way to establish this is induction on $n$. For $n=1$, the region has length $1$. If the $(n-1)$-dimensional simplex with total at most $a$ has volume $a^{n-1}/(n-1)!$, then slicing the $n$-dimensional simplex by $s_n=t$ gives volume $$ \int_0^1\frac{(1-t)^{n-1}}{(n-1)!}\,dt =\frac{1}{n!}. $$ Multiplying this volume by $n!$ gives total probability one.

Common mistake: It is tempting to say that the $n+1$ spacings are independent because the sample observations are independent. That conclusion is false: the spacings obey $\sum_i S_i=1$, so fixing any $n$ of them determines the last one. The independence of the original observations supports the order-statistic density, but it does not make the transformed spacings independent.

A related mistake is to claim that the full vector has an ordinary $(n+1)$-dimensional density equal to $n!$. It lies on the hyperplane $\sum_i s_i=1$, which has zero $(n+1)$-dimensional Lebesgue measure. The stated density is instead with respect to the first $n$ coordinates on the simplex. This coordinate convention is essential both for the Jacobian calculation and for matching the Dirichlet density.

Key takeaway: The spacings from $n$ independent uniform observations are uniform over the $n$-dimensional simplex in the first $n$ coordinates. Their joint density is $n!$, and the full vector has the $\operatorname{Dirichlet}(1,\ldots,1)$ law.

Check Your Understanding

Use the simplex-coordinate convention from this tutorial.

Write the first $n$ spacings in terms of $U_{(1)},\ldots,U_{(n)}$, and state how the final spacing is determined.
Why does the transformation from the order statistics to the first $n$ spacings have Jacobian determinant $1$?
For $n=2$, state the support and density of $(S_1,S_2)$, and identify $S_3$.
Explain why the Dirichlet parameters are all equal to one and why the density constant is $n!$.
Why is the full $n+1$-component spacing vector not described by an ordinary $(n+1)$-dimensional Lebesgue density?