$\mathbb{P}(\limsup_{n\rightarrow\infty}\frac{W(n)}{\sqrt{n}}>c)\geq \lim_{n\rightarrow{\infty}}\mathbb{P}(W(n)/\sqrt{n}>c)>0$

Let $W(t)$ be a Wienerprocess, why does the following inequality hold,

$$\mathbb{P}(\limsup_{n\rightarrow\infty}\frac{W(n)}{\sqrt{n}}>c)\geq \lim_{n\rightarrow{\infty}}\mathbb{P}(W(n)/\sqrt{n}>c)$$

Solution 1:

To make the Fatou's lemma idea suggested by @Dominik more detailed, we notice that by continuity of probability $$\mathbb{P}(\limsup A_n)=\mathbb{P}(\cap_{n\geq 1}\cup_{m\geq n}A_m)=\lim_{n\to\infty}\mathbb{P}(\cup_{m\geq n}A_m),$$

while $$\mathbb{P}(\cup_{m\geq n}A_m)\geq\mathbb{P}(A_m),$$

for any $m\geq n$, wihch yields $$\mathbb{P}(\cup_{m\geq n}A_m)\geq\sup_{m\geq n}\mathbb{P}(A_m).$$

By taking limits on both sides we get the formula you want.