Irwin-Hall Laws from Uniform Sums

From Uniform Summands to an Irwin–Hall Law

Let \(X_1,\ldots,X_n\) be independent random variables, each uniformly distributed on the unit interval, and write \(S_n=X_1+\cdots+X_n\). The sum is supported on \([0,n]\), but its density changes form at the integers: as the sum grows, more of the unit cube contributes to the event that the total is below a given value. The resulting distribution is called the Irwin–Hall distribution.

There are two useful ways to identify this law. Independence gives a compact characteristic function, while geometric volume and inclusion–exclusion give an explicit cumulative distribution function and density. The latter formulas also make the polynomial pieces visible.

Characteristic Function

For a random variable \(X\), its characteristic function is \(\varphi_X(t)=E[e^{itX}]\), for real \(t\). For one unit-uniform variable, direct integration gives

The value at zero is also the continuous limit of the expression for \(t\ne 0\). Rewriting the numerator in terms of sine gives an equivalent form. Since the summands are independent, the characteristic function of their sum is the product of their characteristic functions.

The formula follows by integrating the uniform density and applying the product rule for characteristic functions of independent sums. The displayed sine form has a continuous value at \(t=0\), because \(\sin(t/2)/(t/2)\) tends to \(1\). Its phase factor \(e^{int/2}\) reflects that the sum is centered at \(n/2\).

Cumulative Distribution Function by Volume

The joint density of \((X_1,\ldots,X_n)\) is \(1\) on the unit cube \([0,1]^n\) and zero elsewhere. Thus \(P(S_n\le x)\) is the volume of the part of that cube satisfying \(u_1+\cdots+u_n\le x\). Without the upper constraints \(u_j\le 1\), the nonnegative simplex with coordinate sum at most \(x\) has volume \(x^n/n!\) for \(x\ge 0\). The unit-cube constraints can be imposed by inclusion–exclusion.

Proof. Start with the region \(u_j\ge 0\) and \(\sum_{j=1}^n u_j\le x\). For \(x\ge 0\), its volume is \(x^n/n!\); for \(x<0\), it is empty. One way to verify the volume is induction on \(n\): the one-dimensional length is \(x\), and integrating the \((n-1)\)-dimensional volume over the last coordinate gives

For each coordinate, the cube requires excluding the event \(u_j>1\). Apply inclusion–exclusion to these \(n\) excluded events. Choose a subset of \(k\) coordinates required to exceed \(1\), and subtract \(1\) from each of those coordinates. This translation changes the sum constraint from \(\sum_j u_j\le x\) to a nonnegative-coordinate constraint with total at most \(x-k\). Its volume is \((x-k)_+^n/n!\). There are \({n\choose k}\) choices of the subset, and inclusion–exclusion assigns the sign \((-1)^k\). Summing over \(k=0,\ldots,n\) gives the stated CDF. For \(x\le 0\) every positive-part term is zero. For \(x\ge n\), the event \(S_n\le x\) has probability one because \(S_n\le n\); the formula agrees with this since it was obtained by calculating the cube volume. \(\square\)

The Piecewise-Polynomial Density

For \(n\ge 2\), differentiating the CDF gives a continuous density (with the usual value at individual endpoints understood by continuity). The derivative of \((x-k)_+^n\) is \(n(x-k)_+^{n-1}\), including at \(x=k\). Hence:

For \(n=1\), the CDF has a corner at \(x=0\) and \(x=1\), so its density is instead the Uniform\((0,1)\) density almost everywhere. The formula above is stated for \(n\ge2\) to avoid interpreting its exponent-zero positive-part terms at those boundary points.

Checks, Symmetry, and a Common Pitfall

Each summand has mean \(1/2\) and variance \(1/12\). Independence therefore gives \(E[S_n]=n/2\) and \(\operatorname{Var}(S_n)=n/12\). These values provide simple checks on the scale and center of a derived law. There is also an exact symmetry: if \(X_j\) is Uniform\((0,1)\), then \(1-X_j\) has the same distribution. Independence is preserved by applying this transformation coordinatewise, so \(n-S_n=\sum_j(1-X_j)\) has the same distribution as \(S_n\). In particular, the density is symmetric about \(n/2\), and, because the distribution is continuous, \(F_{S_n}(n/2)=1/2\).

A common mistake is to use the simplex expression \(x^n/n!\) as the CDF numerator for every \(x\), then differentiate to claim the density is \(x^{n-1}/(n-1)!\) throughout the support. That expression counts the nonnegative simplex but ignores the restrictions \(u_j\le1\). For \(x>1\), parts of that simplex lie outside the unit cube and must be removed. Inclusion–exclusion supplies the correction terms \((-1)^k{n\choose k}(x-k)_+^n\); omitting them can even produce a density that increases beyond the center, contradicting the symmetry of the sum.