Let :
- $(X_i)_{i \geq 1}$ uniform on $[0,1]$ independants
- $S_n=\sum_{k=1}^n X_k$
- $H_n$ is the cumulative fonction of $S_n$
- I prove that $H_{n+1}(x)= \int_{x-1}^{x} H_n(t) dt$ for all $x$
- I want to deduce from 1. $H_n(x)$ for $0\leq x \leq 1$
The result is given on several topics, like this one. How can I deduce it from 1. ?
My attempt :
For all $x$ :
$
\begin{align*}
H_{n+1} (x)
&= P(S_{n+1} \leq x) \\
&= P(S_{n} + X_{n+1} \leq x) \\
& = \int_{0}^{1} P( S_n +u \leq x) du \\
&= \int_{0}^{1} P( S_n \leq x-u) du \\
&= \int_{x-1}^{x} P( S_n \leq s) ds \\
&=\int_{x-1}^{x} H_n(s) ds \\
\end{align*}
$
$ \begin{align*} H_{n+1} (x)&=\int_{0}^{x} H_n(s) ds \\ H_2(x) &= \frac{x^2}{2} \\ H'_{n+1}&=H_n \\ d^k H_{n+1}&=H_{n+1-k} \\ \forall n \quad H(0)&=0 \\ \end{align*} $
$$P(S_n \leq x) = \frac{ x^n}{ n!} $$
