Density of the sum of $n$ uniform(0,1) distributed random variables

Solution 1:

Here is a totally different method. So you want to compute $I*I*\cdots*I = I^{*n}$, where $I$ is the indicator function of $[0,1]$, and the $*$ denotes convolution, and the superscript $I^{*n}$ denotes the $n$th convolution with itself.

Differentiate $n$ times. You will get $(\delta_1 - \delta_0)^{*n}$, where $\delta_x$ denotes the Dirac delta function centered at $x$. This is easily expanded using the Binomial Theorem, remembering that $\delta_x * \delta_y = \delta_{x+y}$.

Now take the indefinite integral $n$ times. The tricky bit is how to consider the constants of integration. But the easiest boundary condition is to suppose that all the derivatives disappear at $-\infty$. In this way $\int^n \delta_y (dx)^n = \frac1{(n-1)!} (x-y)_+^{n-1}$, where $x_+$ denotes $\max\{x,0\}$. (Please excuse the notation for $n$th indefinite integral.)

I got this method from an exercise in the book by Nakhle Asmar on Partial Differential Equations.

Solution 2:

$\int f_{n}\left(y\right)f_{1}\left(x-y\right)dy=\int\frac{1}{\left(n-1\right)!}\sum_{k=0}^{\left\lfloor y\right\rfloor }\left(-1\right)^{k}\binom{n}{k}\left(y-k\right)^{n-1}1_{\left(0,1\right)}\left(x-y\right)dy$

$=\frac{1}{\left(n-1\right)!}\sum_{k=0}^{\left\lfloor x-1\right\rfloor }\left(-1\right)^{k}\binom{n}{k}\int_{x-1}^{\left\lfloor x\right\rfloor }\left(y-k\right)^{n-1}dy+\frac{1}{\left(n-1\right)!}\sum_{k=0}^{\left\lfloor x\right\rfloor }\left(-1\right)^{k}\binom{n}{k}\int_{\left\lfloor x\right\rfloor }^{x}\left(y-k\right)^{n-1}dy$

$=\frac{1}{n!}\sum_{k=0}^{\left\lfloor x-1\right\rfloor }\left(-1\right)^{k}\binom{n}{k}\left[\left(\left\lfloor x\right\rfloor -k\right)^{n}-\left(x-1-k\right)^{n}\right]+\frac{1}{n!}\sum_{k=0}^{\left\lfloor x\right\rfloor }\left(-1\right)^{k}\binom{n}{k}\left[\left(x-k\right)^{n}-\left(\left\lfloor x\right\rfloor -k\right)^{n}\right]$

$=\frac{1}{n!}\sum_{k=0}^{\left\lfloor x-1\right\rfloor }\left(-1\right)^{k}\binom{n}{k}\left[\left(x-k\right)^{n}-\left(x-1-k\right)^{n}\right]+\left(-1\right)^{\left\lfloor x\right\rfloor }\binom{n}{\left\lfloor x\right\rfloor }\left(x-\left\lfloor x\right\rfloor \right)^{n}$

$=\frac{1}{n!}\sum_{k=0}^{\left\lfloor x\right\rfloor }\left(-1\right)^{k}\binom{n}{k}\left(x-k\right)^{n}+\frac{1}{n!}\sum_{k=0}^{\left\lfloor x\right\rfloor }\left(-1\right)^{k}\binom{n}{k-1}\left(x-k\right)^{n}$ (here $\binom{n}{-1}:=0$)

$=\frac{1}{n!}\sum_{k=0}^{\left\lfloor x\right\rfloor }\left(-1\right)^{k}\left[\binom{n}{k}+\binom{n}{k-1}\right]\left(x-k\right)^{n}$

$=\frac{1}{n!}\sum_{k=0}^{\left\lfloor x\right\rfloor }\left(-1\right)^{k}\binom{n+1}{k}\left(x-k\right)^{n}$

pfffffff.....In my language: een echte 'rotsom'.