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Let $X_1,X_2,\dots,X_n$ be $n$ independent uniform random variables on the intervals $[a_1, b_1],[a_2,b_2],\dots,[a_n,b_n]$ respectively. And let $Y=X_1+X_2+\dots+X_n$.

How do I calculate the CDF (cumulative distribution function) of $Y$?

I believe can be done using convolution consecutively as this question implies for the case $n=3$ and $a_1=\dots=a_n=0$ and $b_1=\dots=b_n=1$. This case is as I understand it the equvalent of the Irwin-Hall-distribution with $n=3$. This leads me to think that there might be another way of calculating the CDF of Y using the Irwin-Hall-distribution?

As I see it, my problem is equivalent to having be $n$ independent uniform random variables $X_1',X_2',\dots,X_n'$ on the intervals $[0, c_1],[0,c_2],\dots,[0,c_n]$ respectively, where $c_i=b_i-a_i$ for $i\in\{1,2,\dots,n\}$, and then shifting the CDF of the corresponding sum $Y'=X_1'+X_2'+\dots+X_n'$ by $a_1+a_2+\dots+a_n$. This makes my problem resembling the Irwin-Hall-distribution more than my original problem, except for the $c_i$'s not necessarily being 1. The Wikipedia page mentions an extension of the Irwin-Hall-distribution, where it is possible to multiply the summands by a factor (maybe I can use the $c_i$'s?)

Can anyone verify if this is a possible approach (and if so how to procced, since the Wiki page is limited), or if I need to do the comprehensive calculations using convolutions, or maybe a third method using a smart trick?

afd
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  • I have not much hope. The fact that the $c_i=b_i-a_i$ can be distinct seems to spoil everything. – drhab Feb 07 '23 at 12:03

1 Answers1

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Your idea of rescaling the random variables works, but the final result (regardless of how you choose to obtain it) is messy. In general the answer is a piece-wise polynomial function of degree $n$ that changes character at $2^n$ points. (That messiness probably explains why it is hard to track down in standard references.)

Adopting your notation for the rescaled variables, note that $U_i= X'_i/c_i$ lies in the unit interval $[0,1]$.

Method 1. The CDF you seek for $Y'= \sum_i X_i$ is defined by $Prob( Y'>\alpha)$. Equivalently $Prob( \sum_i c_i U_i> \alpha )$.

Geometrically, this probability can be regarded as the volume of the region inside the n-dimensional unit cube $[0,1]^n$ contained within the half-space carved out by the linear inequality $\sum_i c_i U_i>\alpha$. This volume turns out to be a piece-wise polynomial function of $\alpha$, also known as a spline. The formula changes structure every time the hyperplane level set $\alpha$ cuts across one of the $2^n$ vertices of this cube. One explicit formula for it is described in the following expository article:

"Spline Notation Applied to a Volume Problem", D.L. Barrow and P.W. Smith The American Mathematical Monthly, Vol. 86, No. 1 (Jan., 1979), pp. 50-51.

Method 2. For small value of n, and for explicit values of the $c_i$, you can also work out the detailed formula for the CDF using Laplace transform methods. This is feasible if you have access to something like Mathematica. (The Laplace transform is handy whenever you need to convolve two or more functions.)

MathFont
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  • Thanks! This turned out to be quite helpful. Using Laplace transform methods I was able to derive a formula for the CDF of $Y'$ which was (not surprisingly) identical to the formula from the article you mentioned: $$F_{Y'}(y)=\frac{1}{n!\cdot\prod_{i=1}^n c_i}\sum_{C'\in \mathcal{P}(C)}(-1)^{|C'|}(y-\sum_{c\in C'}c)^n\cdot u(y-\sum_{c\in C'}c)$$ where $C={c_i;\mathrm{for};i=1,\dots,n}$ and $\mathcal{P}(C)$ is the power set of $C$ and $u$ is the unit step function. Since I want to calculate $P(Y'\in[\alpha; \beta))$ and probably only for $n\leq 4$, this now seems like a manageable task. – afd Feb 13 '23 at 09:00