Htykuut

I am working on the following problem:

Let $X_1, X_2, ldots, X_n, ldots$ be iid. random variables, each of Uniform$(0,1)$ distribution. Denote by $f_n(x)$ the density of the random variable $S_n := sum_{k = 1}^n X_k$. Then
begin{align*}
f_n(x) = frac{1}{(n-1)!} sum_{k = 0}^{[x]} (-1)^k binom{n}{k} (x-k)^{n-1},
end{align*}
where $[x]$ denotes the floor function.

I tried this, but I think something went wrong:

Prove by induction. For $n = 1$
begin{align*}
f_1(x) = frac{1}{(1-1)!} sum_{k = 0}^{[x]} (-1)^k binom{1}{k} (x-k)^{1-1} &= sum_{k = 0}^{[x]} (-1)^k binom{1}{k}
=begin{cases}
1, & 0 le x < 1 \
0, & text{otherwise}
end{cases}
end{align*}
Furthermore
begin{align*}
f_{n+1}(x) &= mathbb P(S_{n+1} = x) = mathbb P(S_n + X_{n+1} = x) \
&= sum_{m = 0}^{infty} mathbb P(S_n + X_{n+1} = x mid X_{n+1} = m) mathbb P(X_{n+1} = m) \
&= sum_{m = 0}^{infty} mathbb P(S_n + m = x) cdot 1_{[0,1]}(m)
= mathbb P(S_n = x) + mathbb P(S_n = x-1) \
&= frac{1}{(n-1)!} sum_{k = 0}^{[x]} (-1)^kbinom{n}{k}(x-k)^{n-1} + frac{1}{(n-1)!} sum_{k = 0}^{[x-1]} (-1)^kbinom{n}{k}(x-1-k)^{n-1} \
&= frac{1}{(n-1)!}left(sum_{k = 0}^{[x]} (-1)^kbinom{n}{k}(x-k)^{n-1} + sum_{k = 0}^{[x]-1} (-1)^kbinom{n}{k}(x-1-k)^{n-1}right) \
&= frac{1}{(n-1)!}left(x^{n-1}+sum_{k = 1}^{[x]} (-1)^kbinom{n}{k}(x-k)^{n-1} + sum_{k = 1}^{[x]} (-1)^{k-1}binom{n}{k-1}(x-1-(k-1))^{n-1}right) \
&= frac{1}{(n-1)!}left(x^{n-1}+sum_{k = 1}^{[x]} (-1)^kbinom{n}{k}(x-k)^{n-1} - (-1)^{k}binom{n}{k-1}(x-k)^{n-1}right).
end{align*}

How can I solve this?

Edit:
begin{align*}
f_{n+1}(x) &= (f_n * f_{X_{n+1}})(x)
= int_{-infty}^infty f_{n}(y) f_{X_{n+1}}(x-y) , dy \
&= int_{-infty}^infty frac{1}{(n-1)!} sum_{k = 0}^{[y]} (-1)^k binom{n}{k}(y-k)^{n-1} cdot 1_{(0,1)}(x-y) , dy \
&= frac{1}{(n-1)!} int_{-infty}^{infty} sum_{k = 0}^{[y]} (-1)^k binom{n}{k}(y-k)^{n-1} cdot 1_{(x-1, x)}(y), dy \
&= frac{1}{(n-1)!} int_{x-1}^{x} sum_{k = 0}^{[y]} (-1)^k binom{n}{k}(y-k)^{n-1} , dy
end{align*}

I want to swap the sum and the integral, but the sum depends on $y$.

$int f_{n}left(yright)f_{1}left(x-yright)dy=intfrac{1}{left(n-1right)!}sum_{k=0}^{leftlfloor yrightrfloor }left(-1right)^{k}binom{n}{k}left(y-kright)^{n-1}1_{left(0,1right)}left(x-yright)dy$

$=frac{1}{left(n-1right)!}sum_{k=0}^{leftlfloor x-1rightrfloor }left(-1right)^{k}binom{n}{k}int_{x-1}^{leftlfloor xrightrfloor }left(y-kright)^{n-1}dy+frac{1}{left(n-1right)!}sum_{k=0}^{leftlfloor xrightrfloor }left(-1right)^{k}binom{n}{k}int_{leftlfloor xrightrfloor }^{x}left(y-kright)^{n-1}dy$

$=frac{1}{n!}sum_{k=0}^{leftlfloor x-1rightrfloor }left(-1right)^{k}binom{n}{k}left[left(leftlfloor xrightrfloor -kright)^{n}-left(x-1-kright)^{n}right]+frac{1}{n!}sum_{k=0}^{leftlfloor xrightrfloor }left(-1right)^{k}binom{n}{k}left[left(x-kright)^{n}-left(leftlfloor xrightrfloor -kright)^{n}right]$

$=frac{1}{n!}sum_{k=0}^{leftlfloor x-1rightrfloor }left(-1right)^{k}binom{n}{k}left[left(x-kright)^{n}-left(x-1-kright)^{n}right]+left(-1right)^{leftlfloor xrightrfloor }binom{n}{leftlfloor xrightrfloor }left(x-leftlfloor xrightrfloor right)^{n}$

$=frac{1}{n!}sum_{k=0}^{leftlfloor xrightrfloor }left(-1right)^{k}binom{n}{k}left(x-kright)^{n}+frac{1}{n!}sum_{k=0}^{leftlfloor xrightrfloor }left(-1right)^{k}binom{n}{k-1}left(x-kright)^{n}$
(here $binom{n}{-1}:=0$)

$=frac{1}{n!}sum_{k=0}^{leftlfloor xrightrfloor }left(-1right)^{k}left[binom{n}{k}+binom{n}{k-1}right]left(x-kright)^{n}$

$=frac{1}{n!}sum_{k=0}^{leftlfloor xrightrfloor }left(-1right)^{k}binom{n+1}{k}left(x-kright)^{n}$

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

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.