Intro and Problem Statement

The Irwin–Hall distribution is a probability distribution of the sum of $$n$$ independent, uniformly-distributed, continuous random variables in the interval $$[0, 1]$$. The distribution is a piecewise polynomial function composed of $$n$$ sections of degree $$(n − 1)$$ which, combined, cover the interval $$[0, n]$$.

The zeroth piecewise segment is always of the form $$\frac{x^{n − 1}}{(n − 1)!}$$. The last segment is always of the form $$\frac{(−x + n)^{n − 1}}{(n − 1)!}$$. The segments in between are more complicated, and are where I'm running into trouble.

I want to find a pattern that allows me to generate the piecewise equation for an arbitrary segment in the distribution of an arbitrary number of variables.

The Basics

The probability distribution of a single variable looks like a horizontal line (degree zero). The equation of the probability distribution is:

$$f_X(x; 1) = \begin{cases} 1 & : 0 ≤ x ≤ 1 \\ 0 & : \text{otherwise} \end{cases}$$

If we add a second identical variable to the first, their combined probability distribution runs from 0 to 2, composed of two diagonal lines (degree one). The equation of the combined probability distribution of two variables is:

$$f_X(x; 2) = \begin{cases} x & : 0 ≤ x ≤ 1 \\ −x + 2 & : 1 < x ≤ 2 \\ 0 & : \text{otherwise} \end{cases}$$

If we add a third variable, the combined probability distribution runs from 0 to 3, composed of three parabolic arcs (degree two). The equation of the combined probability distribution of three variables is:

$$f_X(x; 3) = \begin{cases} \frac{x²}{2} & : 0 ≤ x ≤ 1 \\ \frac{−2 x² + 6x − 3}{2} & : 1 < x ≤ 2 \\ \frac{(−x + 3)²}{2} & : 2 < x ≤ 3 \\ 0 & : \text{otherwise} \end{cases}$$

As more and more variables are added (as $$n$$ gets large), their combined Irwin-Hall distribution approaches a normal distribution with mean $$μ = \frac{n}{2}$$ and variance $$σ² = \frac{n}{12}$$.

How I'm Building Up the Equations

The $$(n − 1)$$th derivative of the $$k$$th segment in the distribution, skipping the zeroth segment whose $$(n − 1)$$th derivative is always 1, is of the form $$\frac{(−1)^k}{k!} (n − 1) (n − 2) (n − 3) \, ... (n − k)$$, or equivalently $$\frac{(−1)^k (n − 1)!}{k! (n − 1 − k)!}$$. For example, in the case of four variables, the 3rd derivatives of the four segments are:

• $$\frac{ 1}{0!} = 1$$,
• $$\frac{−1}{1!} (4 − 1) = −3$$,
• $$\frac{ 1}{2!} (4 − 1) (4 − 2) = 3$$, and
• $$\frac{−1}{3!} (4 − 1) (4 − 2) (4 − 3) = −1$$.

In the case of five variables, the 4th derivatives of the five segments are:

• $$\frac{ 1}{0!} = 1$$,
• $$\frac{−1}{1!} (5 − 1) = −4$$,
• $$\frac{ 1}{2!} (5 − 1) (5 − 2) = 6$$,
• $$\frac{−1}{3!} (5 − 1) (5 − 2) (5 − 3) = −4$$, and
• $$\frac{ 1}{4!} (5 − 1) (5 − 2) (5 − 3) (5 − 4) = 1$$.

These numbers correspond to the $$(n − 1)$$th row of Pascal's triangle, with every other number negative. That is, for $$n = 4$$ variables, the 3rd derivatives of the four segments are the four terms in the 3rd row of the triangle, with odd positions negated.

To build up the full $$n$$-segment distribution curve, we take the integrals of these base numbers $$(n − 1)$$ times, adding a new constant term each time. The constant term for a given integration step depends on the number $$m$$ of integrations, including the current step (starting at one).

• The constant we add to the zeroth segment is always zero.
• The constant we add to the first segment is $$\frac{(−1)^{m − 1}}{0! × m!} (n)$$.
• The constant we add to the second segment is $$\frac{(−1)^{m}}{1! × m!} (2^{m − 1} n² − (1 + 2^{m − 1}) n)$$.
• The constant we add to the third segment is $$\frac{(−1)^{m − 1}}{2! × m!} (3^{m − 1} n³ − (2 × 2^{m − 1} + 3 × 3^{m − 1}) n² + 2! (1 + 2^{m − 1} + 3^{m − 1}) n)$$.
• The constant we add to the fourth segment is $$\frac{(−1)^{m}}{3! × m!} (4^{m − 1}n⁴ - (3 × 3^{m − 1} + 6 × 4^{m − 1}) n³ + (6 × 2^{m − 1} + 9 × 3^{m − 1} + 11 × 4^{m − 1}) n² − 3! (1^{m − 1} + 2^{m − 1} + 3^{m − 1} + 4^{m − 1}) n)$$

These are the expressions I want to figure out how to create more of.

As an example, the equation for the third segment ($$k = 3$$) after three integrations is:

$$\int \left( \int \left( \int \left( \frac{−1}{3!} (n − 1) (n − 2) (n − 3) \right) + \frac{n³ − 5n² + 6n}{2! × 1!} \right) − \frac{3n³ − 13n² + 12n}{2! × 2!} \right) + \frac{9n³ − 35n² + 28n}{2! × 3!}$$

In the case of four variables, that simplifies to:

$$\int \left( \int \left( \int \left( −1 \right) + 4 \right) − 8 \right) + \frac{32}{3}$$

Which expands into:

$$\frac{−x³}{6} + 2x² − 8x + \frac{32}{3}$$

Which can be re-written as:

$$\frac{−x³ + 12x² − 48x + 64}{3!}$$

Which factors into:

$$\frac{(−x + 4)³}{3!}$$

The whole process is complicated, but not difficult.

The Stumbling Block

The problem is, I can't figure out the patterns for creating constants beyond the fourth segment. For the first four segments, I have the patterns for their constant terms and can generate equations for the probability distributions of any number of variables. Using WolframAlpha, I've worked out the equations for generating the constants out to eight segments for values of $$n$$ up to 6, but by the seventh or eighth segment and variable it starts running out of computation time and won't give an answer. And using only the eight equations I am able to get, I have been unable to find the patterns for all the coefficients of those equations.

To recap, here are the equations for the terms that I have figured out:

1. $$0$$
2. $$\frac{(−1)^{m − 1}}{0! × m!} (n)$$
3. $$\frac{(−1)^{m}}{1! × m!} (2^{m − 1} n² − (1 + 2^{m − 1}) n)$$
4. $$\frac{(−1)^{m − 1}}{2! × m!} (3^{m − 1} n³ − (2 × 2^{m − 1} + 3 × 3^{m − 1}) n² + (1 + 2^{m − 1} + 3^{m − 1}) n)$$
5. $$\frac{(−1)^{m}}{3! × m!} (4^{m − 1}n⁴ - (3 × 3^{m − 1} + 6 × 4^{m − 1}) n³ + (6 × 2^{m − 1} + 9 × 3^{m − 1} + 11 × 4^{m − 1}) n² − 3! (1^{m − 1} + 2^{m − 1} + 3^{m − 1} + 4^{m − 1}) n)$$

Beyond four, though, I'm stuck. I feel like there should be a pattern to these equations that I can use to generate the equations for the constants for an arbitrary segment, but if there is one, I can't find it.

Some partial patterns do stand out to me:

• The expressions are polynomials in terms of $$n$$ with coefficients composed of sums of numbers raised to the power of $$(m − 1)$$
• The $$k$$th expression is $$k$$ terms long, with highest degree $$k$$ and containing terms of all lower degrees down to and including 1, with zero constant term
• The first term in the $$k$$th expression is $$k^{m − 1} × n^k$$
• The last term in the $$k$$th expression is of the form $$(−1)^{k − 1} × (k − 1)! × (1^{m − 1} + 2^{m − 1} + 3^{m − 1} + \, ... + k^{m − 1}) × n$$
• The terms alternate positive and negative — the first is always positive, the second is always negative, and so on

Here are the equations for the fifth segment's constants (numbered by $$m$$ value):

1. $$\frac{ 1}{4! × 1!} ( n⁵ − 14n⁴ + 71n³ − 154n² + 120n)$$
2. $$\frac{−1}{4! × 2!} ( 5n⁵ − 66n⁴ + 307n³ − 582n² + 360n)$$
3. $$\frac{ 1}{4! × 3!} ( 25n⁵ − 314n⁴ + 1367n³ − 2374n² + 1320n)$$
4. $$\frac{−1}{4! × 4!} ( 125n⁵ − 1506n⁴ + 6235n³ − 10230n² + 5400n)$$
5. $$\frac{ 1}{4! × 5!} ( 625n⁵ − 7274n⁴ + 28991n³ − 45814n² + 23496n)$$
6. $$\frac{−1}{4! × 6!} ( 3125n⁵ − 35346n⁴ + 136867n³ − 210822n² + 106200n)$$
7. $$\frac{ 1}{4! × 7!} (15625n⁵ − 172634n⁴ + 653927n³ − 989254n² + 492360n)$$
8. $$\frac{−1}{4! × 8!} (78125n⁵ − 846786n⁴ + 3153835n³ − 4708950n² + 2323800n)$$

Here are the equations for the sixth segment's constants (numbered by $$m$$ value):

1. $$\frac{−1}{5! × 1!} ( n⁶ − 20n⁵ + 155n⁴ − 580n³ + 1044n² − 720n)$$
2. $$\frac{ 1}{5! × 2!} ( 6n⁶ − 115n⁵ + 840n⁴ − 2885n³ + 4554n² − 2520n)$$
3. $$\frac{−1}{5! × 3!} ( 36n⁶ − 665n⁵ + 4630n⁴ − 14935n³ + 21734n² − 10920n)$$
4. $$\frac{ 1}{5! × 4!} ( 216n⁶ − 3865n⁵ + 25890n⁴ − 79775n³ + 110334n² − 52920n)$$
5. $$\frac{−1}{5! × 5!} ( 1296n⁶ − 22565n⁵ + 146530n⁴ − 436555n³ + 584174n² − 273000n)$$
6. $$\frac{ 1}{5! × 6!} ( 7776n⁶ − 132265n⁵ + 837690n⁴ − 2433935n³ + 3184734n² − 1464120n)$$
7. $$\frac{−1}{5! × 7!} ( 46656n⁶ − 777965n⁵ + 4828930n⁴ − 13767235n³ + 17730014n² − 8060520n)$$
8. $$\frac{ 1}{5! × 8!} (279936n⁶ − 4589665n⁵ + 28028490n⁴ − 78754775n³ + 100247214n² − 45211320n)$$

And here are the only equations I've been able to figure out for the seventh segment's constants (numbered by $$m$$ value):

1. $$\frac{ 1}{6! × 1!} ( n⁷ − 27n⁶ + 295n⁵ − 1665n⁴ + 5104n³ − 8028n² + 5040n)$$
2. $$\frac{−1}{6! × 2!} (7n⁷ − 183n⁶ + 1915n⁵ − 10185n⁴ + 28678n³ − 39672n² + 20160n)$$

I've tried different ways of factoring the equations, but haven't been able to find a pattern there either.

Is there a pattern for figuring out the rest of the coefficients for these equations to generate the constant terms for the integrals that lead to the probability distribution?