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Ranks, Spacings, and Empirical Laws · Tutorial 483 of 994

Circular Spacings from Uniform Points

Learn how anchoring the circle at a sampled point gives a symmetric Dirichlet law for the gaps and makes exact gap probabilities accessible.

Advanced 9 min read

What You'll Learn

  • Define circular gaps using a sampled point as the starting anchor.
  • Derive the joint density of the gaps and identify its Dirichlet distribution.
  • Calculate the marginal law and tail probability of one gap.
  • Find the law of a sum of specified gaps.
  • Use simplex geometry and inclusion-exclusion to calculate probabilities involving large gaps.
  • Distinguish anchored circular spacings from gaps defined by starting at the smallest coordinate.

Why the Choice of Starting Point Matters

For points scattered around a circle, the gaps between consecutive points describe how evenly or unevenly they are distributed. Their laws are useful in problems about empty arcs, coverage, and the largest uncovered region. There is one important detail: a sequence of circular gaps needs a starting point. If that point is chosen as one of the sampled observations, the gaps have a particularly simple joint distribution.

Represent a circle of circumference one by \([0,1)\), with positions interpreted modulo one, and let \(X_1,\ldots,X_n\) be independent Uniform\((0,1)\) points. Anchor the traversal at \(X_1\) and proceed in the increasing direction around the circle. This convention—not starting at the smallest coordinate—will be used throughout.

Definition: For \(j=2,\ldots,n\), let \(Y_j=(X_j-X_1)\bmod 1\), the clockwise distance from \(X_1\) to \(X_j\). Write their increasing order as \(V_1<\cdots<V_{n-1}\), and put \(V_0=0\) and \(V_n=1\). The anchored circular spacings are \(D_i=V_i-V_{i-1}\), \(i=1,\ldots,n\). Thus \(D_1\) is the gap from \(X_1\) to the next point, and \(D_n\) is the final gap back to \(X_1\). Almost surely, \(D_i>0\) and \(\sum_{i=1}^n D_i=1\).

The Joint Dirichlet Distribution

Conditional on \(X_1\), the offsets \(Y_2,\ldots,Y_n\) are independent Uniform\((0,1)\) random variables: subtracting a fixed point modulo one preserves the uniform distribution. Their joint conditional law does not depend on \(X_1\). Consequently the ordered offsets have the usual order-statistic density, which lets us derive the joint law of the gaps directly.

Theorem: For \(n\geq 2\), the anchored circular spacings satisfy \((D_1,\ldots,D_n)\sim\operatorname{Dirichlet}(1,\ldots,1)\). In coordinates \(d_1,\ldots,d_{n-1}\), with \(d_n=1-\sum_{i=1}^{n-1}d_i\), their density is \((n-1)!\) on \(d_i>0\), \(\sum_{i=1}^{n-1}d_i<1\), and zero elsewhere.

Proof. Write \(m=n-1\). The ordered offsets \(V_1,\ldots,V_m\) have joint density \(m!\) on \(0<v_1<\cdots<v_m<1\). Indeed, the \(m!\) possible orderings of independent continuous uniform observations have equal density one, and the ordered region collects all of them. The transformation \(d_1=v_1,\ d_2=v_2-v_1,\ldots,d_m=v_m-v_{m-1}\) has triangular Jacobian determinant one. The final gap is \(d_{m+1}=1-v_m\), so the ordered region becomes \(d_i>0\) and \(\sum_{i=1}^m d_i<1\). The density in these first \(m\) coordinates is therefore \(m!=(n-1)!\). This is precisely the density of \(\operatorname{Dirichlet}(1,\ldots,1)\), whose constant on the unit simplex is \(\Gamma(n)=(n-1)!\). \(\square\)

The density is constant in simplex coordinates, but the gaps are not independent: they must add to one. The law is exchangeable, so any one gap has the same marginal distribution, and any specified collection of \(k\) gaps has the same sum distribution as any other collection of \(k\).

Worked Example: The Probability That One Gap Is Long

Suppose \(n=5\) points are sampled and the circle is anchored at \(X_1\). Find the probability that the gap immediately clockwise from \(X_1\) exceeds \(1/2\).

The event is \(D_1>1/2\). In the simplex of five gaps, shifting \(D_1\) down by \(1/2\) leaves nonnegative coordinates with total \(1/2\). The volume of this smaller simplex is \((1/2)^4\) times the volume of the original simplex: the simplex has dimension \(5-1=4\). Therefore \(P(D_1>1/2)=(1/2)^4=1/16\).

Equivalently, the marginal distribution is \(\operatorname{Beta}(1,4)\), whose density is \(4(1-d)^3\) for \(0<d<1\). Integrating its tail gives \(\int_{1/2}^1 4(1-d)^3\,\mathrm{d}d=(1/2)^4=1/16\), in agreement with the simplex calculation.

Marginal and Aggregated Gap Probabilities

The constant simplex density gives exact distributions for one gap and for sums of gaps. For a single gap, the marginal is \(\operatorname{Beta}(1,n-1)\); its density and upper-tail probability are

$$ f_{D_i}(d)=(n-1)(1-d)^{n-2},\quad 0<d<1, \qquad P(D_i>a)=(1-a)^{n-1},\quad 0\leq a\leq1. $$

More generally, let \(S_k\) be the sum of any specified \(k\) distinct gaps, where \(1\leq k\leq n-1\). Then \(S_k\sim\operatorname{Beta}(k,n-k)\), with density

$$ f_{S_k}(s)=\frac{(n-1)!}{(k-1)!(n-k-1)!}\, s^{k-1}(1-s)^{n-k-1},\quad 0<s<1. $$

To see this, group the gaps in the chosen set and let their total be \(s\). The \(k\) positive components within that group have a \((k-1)\)-dimensional simplex of scale \(s\), with volume factor \(s^{k-1}/(k-1)!\). The other \(n-k\) components have total \(1-s\), giving factor \((1-s)^{n-k-1}/(n-k-1)!\). Multiplying these factors by the joint density \((n-1)!\) yields the displayed density. This also verifies its normalization, since it is a beta density.

A related volume argument handles simultaneous lower bounds. For any fixed set of \(r\) distinct gaps and any \(a\geq0\),

$$ P(D_{i_1}>a,\ldots,D_{i_r}>a)= \begin{cases} (1-ra)^{n-1},&ra\leq1,\\ 0,&ra>1. \end{cases} $$

When \(ra\leq1\), subtract \(a\) from each of the \(r\) constrained coordinates. The remaining coordinates are nonnegative and sum to \(1-ra\); scaling a simplex of dimension \(n-1\) by \(1-ra\) multiplies its volume by \((1-ra)^{n-1}\). If \(ra>1\), the constraints require more than the total available length, so the event is impossible.

Worked Example: A Sum of Two Adjacent Gaps

For \(n=6\), find the probability that two specified adjacent gaps have combined length greater than \(1/3\). Adjacency does not change the law: the sum of any two specified gaps has a \(\operatorname{Beta}(2,4)\) distribution.

Its density is \(20s(1-s)^3\), since \(5!/(1!3!)=20\). Thus

$$ P(S_2>1/3)=\int_{1/3}^1 20s(1-s)^3\,\mathrm{d}s =1-\left[10s^2-20s^3+15s^4-4s^5\right]_{0}^{1/3}. $$

At \(s=1/3\), the expression in brackets is \(10/9-20/27+15/81-4/243=131/243\). Hence \(P(S_2>1/3)=1-131/243=112/243\), approximately \(0.461\).

Calculating Probabilities for the Largest Gap

The simultaneous-bound formula can be combined with inclusion-exclusion to study whether any gap exceeds a threshold. The method is especially transparent when two exceedances cannot occur together.

Worked Example: No Gap Exceeds Half the Circle

For \(n=4\), find the probability that every anchored circular gap is at most \(1/2\). Let \(A_i=\{D_i>1/2\}\). For each \(i\), \(P(A_i)=(1-1/2)^3=1/8\). Two distinct gaps cannot both exceed \(1/2\), because their sum would exceed the total length one. Thus all pairwise intersections have probability zero, and inclusion-exclusion gives

$$ P\left(\max_{1\leq i\leq4}D_i\leq1/2\right) =1-P\left(\bigcup_{i=1}^4 A_i\right) =1-\sum_{i=1}^4P(A_i) =1-4\left(\frac18\right)=\frac12. $$

The boundary case of a gap exactly \(1/2\) has probability zero because the gaps have a continuous distribution, so using “at most” rather than “less than” does not alter the result.

A Common Mistake: Starting at the Smallest Coordinate

A frequent error is to sort the original coordinates and assume that the cyclic gaps beginning at the minimum are also \(\operatorname{Dirichlet}(1,\ldots,1)\). The anchor convention matters. Starting at \(X_1\) chooses a sampled point without regard to its position relative to the coordinate cut at zero. Starting at the smallest coordinate selects a gap according to its relationship to that cut and generally gives a different law.

For example, with two points \(U_1,U_2\), let \(U_{(1)}<U_{(2)}\). The gap starting at the minimum is \(G=U_{(2)}-U_{(1)}\). Its density is \(2(1-g)\), \(0<g<1\), not the uniform density that a \(\operatorname{Dirichlet}(1,1)\) gap would have. In contrast, the gap from the anchored point \(X_1\) to \(X_2\), measured clockwise modulo one, is Uniform\((0,1)\). Therefore, before applying the Dirichlet law, specify where the circular sequence starts.

Key takeaway: With \(n\) independent uniform points on a circle, the \(n\) clockwise gaps starting at a specified sampled point have the \(\operatorname{Dirichlet}(1,\ldots,1)\) law. This yields Beta laws for individual gaps and their sums, while simplex scaling gives exact probabilities for simultaneous gap constraints.

Check Your Understanding

Try these questions using the anchored convention.

  1. For \(n=4\), what is the probability that a specified gap exceeds \(1/3\)?
  2. For \(n=7\), what distribution does the sum of three specified gaps have?
  3. For \(n=5\), can two specified gaps both exceed \(0.6\)? Explain using the total circumference.
  4. Why does anchoring at the smallest coordinate not, in general, give the Dirichlet law derived here?
  5. For \(n=3\), find the probability that two specified gaps are both greater than \(1/4\).