The Exponential Representation of Uniform Spacings

Why Exponential Variables Describe Uniform Gaps

The gaps between ordered uniform observations are dependent: they are positive, but their sum must equal one. There is nevertheless a useful way to generate their joint distribution from independent random variables. Start with independent unit-exponential variables and divide each by their total. The resulting proportions have exactly the same distribution as the uniform spacings. This representation converts questions about gaps into calculations involving exponential sums and their ratios.

The normalization matters. The exponential variables before division are independent; the normalized proportions are not. Their shared denominator enforces the sum-to-one constraint.

The Spacings and the Representation

Definition: Let $U_1,\ldots,U_n$ be independent $U(0,1)$ random variables, and write their order statistics as $U_{(1)}<\cdots<U_{(n)}$, an event of probability one. Set $U_{(0)}=0$ and $U_{(n+1)}=1$. The $n+1$ uniform spacings are $$ D_i=U_{(i)}-U_{(i-1)},\qquad i=1,\ldots,n+1. $$ They are positive and satisfy $\sum_{i=1}^{n+1}D_i=1$.

For the density of these spacings, use coordinates $d_1,\ldots,d_n$, with $d_{n+1}=1-\sum_{i=1}^n d_i$. Their possible values form the open simplex $$ \mathcal S_n=\left\{(d_1,\ldots,d_n):d_i>0,\ \sum_{i=1}^n d_i<1\right\}. $$ The ordered uniform observations have joint density $n!$ on $0<u_1<\cdots<u_n<1$: the original joint density is one, and each strictly ordered vector can arise from $n!$ permutations of the observations. The transformation from ordered observations to the first $n$ spacings is triangular, with determinant one. Thus the joint density of $D_1,\ldots,D_n$ is $n!$ on $\mathcal S_n$, and zero elsewhere.

Theorem: Let $E_1,\ldots,E_{n+1}$ be independent $ \operatorname{Exp}(1)$ random variables, each with density $e^{-x}$ for $x>0$. Define $$ T=\sum_{j=1}^{n+1}E_j,\qquad P_i=\frac{E_i}{T},\quad i=1,\ldots,n+1. $$ Then $(P_1,\ldots,P_{n+1})$ has the same joint distribution as $(D_1,\ldots,D_{n+1})$, the spacings from $n$ independent $U(0,1)$ observations.

Proof by Change of Variables

Write $m=n+1$. Transform from $e_1,\ldots,e_m$ to $d_1,\ldots,d_n,t$ by setting $$ e_i=td_i\quad (i=1,\ldots,n),\qquad e_m=t\left(1-\sum_{i=1}^n d_i\right)=td_m. $$ The domain is $d\in\mathcal S_n$ and $t>0$. The inverse transformation is $t=\sum_{i=1}^m e_i$, $d_i=e_i/t$ for $i\le n$.

Here is the Jacobian calculation. In the derivative matrix, the first $n$ columns (derivatives with respect to the $d_i$) are $t(e_i-e_m)$, where $e_i$ in this expression denotes the $i$th coordinate vector in $\mathbb R^{n+1}$. The last column, for the derivative with respect to $t$, is $(d_1,\ldots,d_m)^\mathsf T$. Factor $t$ out of each of the first $n$ columns. The remaining determinant has absolute value one: adding the first $n$ rows to the last row makes its last-column entry $\sum_{i=1}^m d_i=1$, while the first $n$ columns have an identity block in the first $n$ rows and zeros in the last row. Therefore the absolute Jacobian is $t^n$.

The joint density of the independent exponentials is $\exp(-\sum_i e_i)=e^{-t}$. Including the Jacobian, the joint density in $(d_1,\ldots,d_n,t)$ is $t^n e^{-t}$ on $\mathcal S_n$ and $t>0$. Integrating out $t$ gives $$ f_{P_1,\ldots,P_n}(d_1,\ldots,d_n) =\int_0^\infty t^n e^{-t}\,dt =\Gamma(n+1) =n!. $$ This is exactly the joint density of the first $n$ uniform spacings. Hence the two spacing vectors have the same distribution. $\square$

The number of exponentials is $n+1$, not $n$: $n$ interior order statistics cut the interval into $n+1$ gaps. The proof also shows why the common total $T$ disappears from the proportions: integrating it out leaves the constant simplex density.

Aggregated Spacings and Exponential Sums

The representation is especially convenient when an event involves a sum of selected gaps. If $k$ of the $m=n+1$ proportions are added together, their sum is the ratio $A/(A+B)$, where $A$ is the sum of the corresponding $k$ exponentials and $B$ is the sum of the other $m-k$. These two sums are independent, since they use disjoint sets of independent exponentials.

A sum of $q$ independent unit-exponential variables has gamma density $x^{q-1}e^{-x}/(q-1)!$ for $x>0$. For $1\le k<m$, take $A$ and $B$ with shapes $k$ and $m-k$, and set $r=A/(A+B)$, $t=A+B$. The inverse is $A=rt$, $B=(1-r)t$; its Jacobian is $t$. Thus the joint density of $(r,t)$ is $$ \frac{r^{k-1}(1-r)^{m-k-1}t^{m-1}e^{-t}}{(k-1)!(m-k-1)!},\qquad 0<r<1,\ t>0. $$ Integrating over $t$ shows that the sum of any $k$ distinct spacings has density $$ \frac{(m-1)!}{(k-1)!(m-k-1)!}\,r^{k-1}(1-r)^{m-k-1},\qquad 0<r<1. $$ In standard terminology, this is a $\operatorname{Beta}(k,m-k)$ distribution. This result applies to a sum of selected gaps, not just to a consecutive block.

Worked Example: The Probability That One Gap Is Large

There are $n=3$ uniform observations and therefore $m=4$ spacings. Find $P(D_1>1/2)$.

The single spacing is $E_1/T$, which has a $\operatorname{Beta}(1,3)$ density. The formula above gives $3(1-r)^2$ for $0<r<1$. Consequently, $$ P(D_1>1/2)=\int_{1/2}^1 3(1-r)^2\,dr =\left[-(1-r)^3\right]_{1/2}^1 =\frac18. $$ As a direct check using the exponential representation, the event is $E_1>(E_2+E_3+E_4)$. Conditional on $S=E_2+E_3+E_4$, the exponential tail probability is $P(E_1>S\mid S)=e^{-S}$. Independence then gives $E[e^{-S}]=\prod_{j=2}^4 E[e^{-E_j}]=(1/2)^3=1/8$, in agreement with the density calculation.

Worked Example: A Sum of Two Nonconsecutive Gaps

For four uniform observations there are five spacings. Find $P(D_2+D_4>1/2)$.

The two selected exponentials sum to a gamma variable with shape 2, and the other three sum to an independent gamma variable with shape 3. Therefore $D_2+D_4$ has a $\operatorname{Beta}(2,3)$ density, $12r(1-r)^2$. Its upper-tail probability is $$ \begin{aligned} P(D_2+D_4>1/2) &=\int_{1/2}^1 12r(1-r)^2\,dr\\ &=\left[6r^2-8r^3+3r^4\right]_{1/2}^1\\ &=1-\left(\frac{6}{4}-1+\frac{3}{16}\right) =\frac{5}{16}. \end{aligned} $$ The calculation depends on how many gaps are selected, not on whether those gaps are adjacent.

Worked Example: Two Gaps Exceeding Thresholds

For $n=3$, find $P(D_1>1/4,\ D_2>1/4)$. The joint density of the first three spacings is $3!=6$ on the simplex. On the event in question, put $x_1=d_1-1/4$, $x_2=d_2-1/4$, and $x_3=d_3$. These are positive and satisfy $x_1+x_2+x_3<1/2$. This is the original three-dimensional simplex scaled by $1/2$ in each coordinate, so its volume is $(1/2)^3$ times the original simplex volume. The density is constant, and hence $$ P(D_1>1/4,\ D_2>1/4)=\left(\frac12\right)^3=\frac18. $$ More generally, for two specified distinct spacings and thresholds $a,b\ge 0$ with $a+b<1$, the same translation and scaling give $P(D_i>a,\ D_j>b)=(1-a-b)^n$. If $a+b\ge1$, the probability is zero.

Worked Example: A Ratio of Two Gamma Sums

With four independent unit-exponential variables, let $A=E_1+E_2$, $B=E_3+E_4$, and $R=A/(A+B)$. Find the density of $R$ and $P(R>1/2)$.

Each sum has gamma density $x e^{-x}$ for $x>0$, and $A,B$ are independent. Set $A=rt$, $B=(1-r)t$; the Jacobian is $t$. Substitution gives the joint density $$ (rt)((1-r)t)e^{-t}\,t=r(1-r)t^3e^{-t},\qquad 0<r<1,\ t>0. $$ Integrating over $t$ yields $$ f_R(r)=r(1-r)\int_0^\infty t^3e^{-t}\,dt = r(1-r)\Gamma(4) =6r(1-r). $$ The factor $t$ from the Jacobian is essential. The density integrates to one because $\int_0^1 6r(1-r)\,dr=1$. Finally, $$ P(R>1/2)=\int_{1/2}^1 6r(1-r)\,dr =\left[3r^2-2r^3\right]_{1/2}^1 =1-\frac12=\frac12. $$ In the spacing representation with three uniform observations, $R=(D_1+D_2)/(D_1+D_2+D_3+D_4)=D_1+D_2$.

Common Pitfalls and Why the Representation Helps

A frequent error is to treat the normalized variables $P_i=E_i/T$ as independent because the $E_i$ are independent. They are not: their sum is identically one. For example, with at least two gaps, knowing that one proportion is unusually large leaves less total mass for the others. Independence applies before normalization, not after it.

Another common error is to omit the Jacobian in a ratio transformation. In the last example, substituting $A=rt$ and $B=(1-r)t$ into the joint density gives $r(1-r)t^2e^{-t}$ before the Jacobian is included. The actual transformed density has the extra factor $t$, so it is $r(1-r)t^3e^{-t}$. Omitting that factor changes the integral over $t$ from $\Gamma(4)=6$ to $\Gamma(3)=2$, and produces a density that does not integrate to one.

The exponential representation is useful because it turns a constrained vector of gaps into proportions of independent positive variables. It gives a practical simulation method: generate $n+1$ independent unit exponentials, add them, and divide each by the total. It also makes sums of gaps accessible through gamma ratios. The resulting exact probabilities are valuable when a spacing event is awkward to express directly in terms of ordered observations.

Key takeaway: For $n$ independent uniform observations, generate all $n+1$ spacings in distribution by normalizing $n+1$ independent $\operatorname{Exp}(1)$ variables. Keep the shared denominator—and the Jacobian in any ratio calculation—explicit.

Check Your Understanding

Try these questions using the representation and calculations developed above.

How many independent unit-exponential variables are needed to represent the spacings from $n$ uniform observations, and why?
For four uniform observations, what is the distribution of the sum of three specified spacings?
For $n=2$, calculate the probability that a specified spacing exceeds $1/3$.
In the change of variables $A=rt,\ B=(1-r)t$, what is the absolute Jacobian, and which factor would be lost if it were omitted?
Explain why the normalized exponential proportions cannot be independent.