I am trying to understand some probability theory, but I'm stuck on a very fascinating problem that I do not manage to solve. After the statement of the problem, I'll list some of my unsuccessful thoughts.
Let $X_1, \dots, X_n,\dots$ be a sequence of independent random variables, uniformly distributed on $[0,1]$. Define $$N_n = \min \{ k: \text{such that} \sum_1^k X_i > \frac{n}{2}+ \sqrt{\frac{n}{12}} \}$$
Compute: $\lim_n \mathbb{P}(\frac{N_n}{n} > 1).$
Unsuccessful thoughts:
- There might be an implicit hint to use the Irvin-Hall distribution formula.
- I tried to readjuct the definition of $N_n$ by $$N_n = \min \{ k: \text{such that } S_n - \sum_k^n X_i > \frac{n}{2}+ \sqrt{\frac{n}{12}} \}$$ in order to apply some variant of the central limit theorem, but without success.
- Numerical estimations show that the limit should be close to $1$.
- Intuition suggests that the expected value of $N_n$ should be $n$, but how to show it? And how to use it in order to solve the exercise?
Actually, CLT works. Let $S_n=\sum_{i=1}^n X_n$, then by CLT we have $$\frac{S_n-\frac n2}{\sqrt{\frac n{12}}}\xrightarrow{D} N,$$ where $N\sim N(0,1)$.Hence $$P\left(\frac{N_n}n>1\right)=P\left(S_n\leq \frac n2+\sqrt{\frac n{12}}\right)=P\left(\frac{S_n-\frac n2}{\sqrt{\frac n{12}}}\leq 1\right)\to P(N\leq 1).$$
Addendum： $A=\{\frac{N_n}n>1\}=\{S_n\leq \frac n2+\sqrt{\frac n{12}}\}=B$.
If $\omega\in A$, then $N_n(\omega)>n$, and then $S_n(\omega)\leq \frac n2+\sqrt{\frac n{12}}$, since otherwise $S_n(\omega)> \frac n2+\sqrt{\frac n{12}}$ implies that $N_n(\omega)\leq n$. Hence $A\subset B$.
If $\omega \in B$, then $S_n(\omega)\leq \frac n2+\sqrt{\frac n{12}}$, and then $N_n(\omega)>n$. Because otherwise $N_n(\omega)=k\leq n$, we have $S_k(\omega)> \frac n2+\sqrt{\frac n{12}}$, then $S_n(\omega)=S_k(\omega)+\sum_{i=k+1}^n X_i(\omega)\geq S_k(\omega)>\frac n2+\sqrt{\frac n{12}}$, contrading to "$\omega \in B$". Hence $B\subset A$.