Moments and Covariances of Uniform Spacings

Why Spacing Moments Matter

A sample of \(n\) independent uniform observations divides the interval from 0 to 1 into \(n+1\) random gaps. The gaps cannot vary independently: if one is unusually large, the others must together be smaller because all \(n+1\) gaps have a fixed sum. Their means and variances describe the typical size of a gap, while their covariances quantify this dependence. These calculations also make it possible to assess the variability of a whole interval formed by adding several adjacent gaps.

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. We include both endpoint gaps in the spacing vector:

There are \(n+1\) gaps, and \(\sum_{i=1}^{n+1}D_i=1\). By the Dirichlet law of uniform spacings, this vector has distribution \(\operatorname{Dirichlet}(1,\ldots,1)\): its density is constant, equal to \(n!\), on the \(n\)-dimensional simplex of positive coordinates summing to 1.

Moments from the Dirichlet Density

For a Dirichlet vector with positive parameters \(\alpha_1,\ldots,\alpha_m\), let \(\alpha_0=\sum_{i=1}^m\alpha_i\). Its density on the simplex is proportional to \(\prod_i x_i^{\alpha_i-1}\), with normalizing constant \(\Gamma(\alpha_0)/\prod_i\Gamma(\alpha_i)\). Consequently, for nonnegative integers \(r_1,\ldots,r_m\), and \(r_0=\sum_i r_i\), integrating the monomial against this density gives

To see why the ratio has this form, the simplex integral of \(\prod_i x_i^{\beta_i-1}\) is \(\prod_i\Gamma(\beta_i)/\Gamma(\sum_i\beta_i)\) for positive \(\beta_i\). This identity follows by writing the product of gamma integrals in positive coordinates \(y_i\), then setting \(t=\sum_i y_i\) and \(x_i=y_i/t\). The Jacobian is \(t^{m-1}\), so the integral factors into a gamma integral in \(t\) and the simplex integral in \(x\). Applying the identity first with \(\beta_i=\alpha_i+r_i\) and then using the Dirichlet normalizing constant gives the displayed moment formula.

For uniform spacings, \(m=n+1\) and every \(\alpha_i=1\). The integer-argument identity \(\Gamma(q+1)=q!\) then gives the convenient formula

Means, Variances, and Covariances

The same calculation applies to every spacing, including the endpoint gaps. For the mean, set one exponent equal to 1. For a second moment, set one exponent equal to 2. For the product of two distinct gaps, set two exponents equal to 1. These choices give

Subtracting products of means yields the variance and covariance. This proves the following formulas for every \(1\leq i,j\leq n+1\):

Indeed, the variance calculation is

For distinct \(i\) and \(j\), the covariance calculation is

The negative sign is a mathematical expression of the fixed-total constraint. It is not merely a qualitative intuition: every pair of distinct gaps has the same strictly negative covariance.

Variance of a Sum of Gaps

Let \(A\) be any fixed set of \(k\) distinct gap indices, where \(1\leq k\leq n+1\), and define \(S_A=\sum_{i\in A}D_i\). Linearity of expectation gives \(E[S_A]=k/(n+1)\). To find its variance, expand the variance of the sum into individual variances and pairwise covariances. There are \(k\) variance terms and \(k(k-1)\) ordered cross terms:

The count \(k(k-1)\) matters: when a variance is expanded, each unordered pair appears twice, as both \(\operatorname{Cov}(D_i,D_j)\) and \(\operatorname{Cov}(D_j,D_i)\). Equivalently, one may count \(\binom{k}{2}\) pairs and multiply their sum by 2.

Intervals Other Than the Unit Interval

If \(X_1,\ldots,X_n\) are independent uniform observations on an interval \((a,b)\) of length \(L=b-a>0\), their ordered gaps, including both endpoint gaps, are \(L\) times the corresponding \(U(0,1)\) spacings. Multiplication by \(L\) multiplies means by \(L\), and variances and covariances by \(L^2\). In particular, a sum of \(k\) gaps has mean \(Lk/(n+1)\) and variance

Common Mistake and Practical Use

A frequent error is to treat the gaps as independent because each has the same marginal distribution. Identical distributions do not imply independence. For distinct spacings here, the covariance is negative, and independence would incorrectly force it to be zero. Another error is to calculate the variance of a sum using only the individual variances. The covariance terms are essential; in fact, they reduce the variance of a sum of several gaps.

The fixed-total identity provides a useful diagnostic. Since \(\sum_iD_i=1\), the variance of the sum must be zero. Substituting the common variance and covariance confirms this:

For a partial sum, the formula also has sensible boundary checks. With \(k=1\), it gives the variance of one gap. With \(k=n+1\), it gives zero because the sum is the constant 1. A sum of \(k\) gaps can also be viewed as a grouped spacing; its variability depends on the group size, while the negative covariances ensure that its complement moves in the opposite direction.