Let :
- $\forall n \geq 1, \quad S_n= \sum_{k=1}^{n} X_k \quad$ where $(X_k)_k$ are independent $U[0,1]$
- $\alpha \in \mathbb{R}$
- $T = \inf\{k \geq n, S_n > 1\}$
- We want to show that : $ \lim\limits_{n \to \infty} P(S_n \geq 1) \geq \frac{1}{\sqrt{2 \pi}} \int_ {\alpha}^{+\infty}e^{- \frac{t^2}{2}}dt$ and deduce from this inequality the distribution of $T$ . The distribution of $T$ can be found with other methods like this.
My attempt :
- For $ 0 \leq x \leq 1$, we compute $P(S_n \leq x)=\frac{x^n}{ n!}$
$ \begin{align*} \sigma_n &:= \int_{ [0,1]^n} \mathbb{1}_{ \sum_{k=1}^n x_i <1 } \quad dx_1 \dots d{x_n} \\ &=\int_{ [0,1]} \int_{ [0,1]^{n-1}} \mathbb{1}_{ \sum_{k=1}^{n-1} x_i <1-x_n } \quad dx_1 \cdots d{x_{n-1}} dx_n \\ &= \int_{ [0,1]} \int_{ [0,1]^{n-1}} \mathbb{1}_{ \sum_{k=1}^{n-1} x_i <1-x_n } \quad dx_1 \cdots d{x_{n-1}} dx_n \\ &= \int_{ [0,1]} (1-x_n)^{n-1} \int_{ [0,1]^{n-1}} \mathbb{1}_{ \sum_{k=1}^{n-1} x_i <1 } \quad dx_1 \cdots d{x_{n-1}} dx_n \\ &= \sigma_{n-1} \int_{ [0,1]} (1-x_n)^{n-1} dx_n \sigma_{n-1} \\ &= \frac{1}{n} \sigma_{n-1} \\ &= \frac{1}{n!} \\ \end{align*} $
$ \begin{align*} \mathbb{E}( \mathbb{1}_{ \sum_{k=1}^n X_i} <x) &= \int_{ [0,1]^n} \mathbb{1}_{ \sum_{k=1}^n x_i <x } \quad dx_1 \cdots d{x_n} \\ &= x^n \int_{ [0,1]^n} \mathbb{1}_{ \sum_{k=1}^n x_i <1 } \quad dx_1 \cdots d{x_n} \\ &= \frac{x^n}{ n!} (\star) \\ \end{align*} $
- Distribution of $T$
$ \begin{align*} \{S_n \leq 1 \} &= \{ T \geq n+1 \} \\ P(T=n) &= P(T\geq n) - P(T \geq n+1 ) \\ P(T=n) &= \frac{1}{n!} - \frac{1}{(n-1)!} \\ \end{align*} $
- There is an equality quite close that can be found by integration by parts : $$ \int_{x}^{+\infty} e^{-t} \frac{t^n}{n!} dt = \sum_{k=0}^{n} e^{-x} \frac{x^k}{k!} $$
