I was reading a paper concerning probability theory.

We have that $X_i$, $i = 1,2,...$ i.i.d random variables, and $S_n = \sum_{i=1}^nX_i$ is defined as partial sum as usual. If

$$\frac{S_n}{n} \to 0 \quad \text{ in probability}$$

then the author says that we can easily get

$$\lim_{n\to \infty} \min_{1\leq k \leq n}\mathbb{P}\left(\frac{|S_n - S_k|}{n} < \epsilon\right) = 1 $$

for any $\epsilon > 0$. I tried to prove this but fail to give a rigorous proof. Really appreciate if any one could give some hints. Or if you don't agree with this, could you give a counter example to show that it does not hold.



$$\left\{ \left| \frac{S_n-S_k}{n} \right|< \epsilon \right\} \supseteq \left\{ \left| \frac{S_n}{n} \right|< \frac{\epsilon}{2} \right\} \cap \left\{ \left| \frac{S_k}{n} \right| < \frac{\epsilon}{2} \right\}$$

we have

$$\mathbb{P} \left( \left| \frac{S_n-S_k}{n} \right| < \epsilon \right) \geq \mathbb{P} \left( \left| \frac{S_n}{n} \right| < \frac{\epsilon}{2} \right) - \mathbb{P} \left( \left| \frac{S_k}{n} \right| \geq \frac{\epsilon}{2} \right)$$

which implies

$$\begin{align*} \min_{1 \leq k \leq n} \mathbb{P} \left( \left| \frac{S_n-S_k}{n} \right| < \epsilon \right) &\geq \mathbb{P} \left( \left| \frac{S_n}{n} \right| < \frac{\epsilon}{2} \right) - \max_{1 \leq k \leq n} \mathbb{P} \left( \left| \frac{S_k}{n} \right| \geq \frac{\epsilon}{2} \right) \\ &\geq 1- 2 \max_{1 \leq k \leq n} \mathbb{P} \left( \left| \frac{S_k}{n} \right| \geq \frac{\epsilon}{2} \right) \end{align*}$$

Therefore, the assertion follows if we can show

$$\lim_{n \to \infty} \max_{1 \leq k \leq n} \mathbb{P} \left( \left| \frac{S_k}{n} \right| \geq \frac{\epsilon}{2} \right) = 0. \tag{1}$$

Fix $\delta>0$. Since $S_n/n \to 0$ in probability there exists $N_0 \in \mathbb{N}$ such that

$$\mathbb{P} \left( \left| \frac{S_n}{n} \right| \geq \frac{\epsilon}{2} \right) \leq \delta \quad \text{for all $n \geq N_0$.} \tag{2}$$

Choosing $N_1 \geq N_0$ sufficiently large we can achive

$$\mathbb{P} \left( \left| \frac{S_k}{n} \right| \geq \frac{\epsilon}{2} \right) \leq \delta \quad \text{for all $k \in \{1,\ldots,N_0\}$ and $n \geq N_1$}. \tag{3} $$

On the other hand,

$$\mathbb{P} \left( \left| \frac{S_k}{n} \right| \geq \frac{\epsilon}{2} \right) \leq \mathbb{P} \left( \left| \frac{S_k}{k} \right| \geq \frac{\epsilon}{2} \right) \stackrel{(2)}{\leq} \delta \tag{4}$$

for all $k \in \{N_0+1,\ldots,n\}$ and $n \geq N_1 \geq N_0$. Combining (3) and (4) shows

$$\max_{1 \leq k \leq n} \mathbb{P} \left( \left| \frac{S_k}{n} \right| \geq \frac{\epsilon}{2} \right) \leq \delta$$

for all $n \geq N_1$. Since $\delta>0$ is arbitrary, this proves (1).

  • $\begingroup$ thank you much, so actually you divide k into two parts, $1\leq k \leq N_0$ and $N_0+1\leq k\leq n$ $\endgroup$ Jun 22 '16 at 1:51
  • $\begingroup$ nice proof. I understand it totally. :-) $\endgroup$ Jun 22 '16 at 1:52
  • 1
    $\begingroup$ @Schrödinger'sCat You are welcome. If you find the answer helpful, you can accept it by clicking on the tick next to it. $\endgroup$
    – saz
    Jun 22 '16 at 7:06

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