Prove directly from the definition of the Wiener process: $\mathbb P \left( \{W_{n^2}\}_{n=1}^{\infty} \text{ is a bounded sequence } \right) = 0$

Question

Let $W_t$ - standard Wiener process. Prove directly from the definition of the Wiener process: $$\mathbb P \left( \{W_{n^2}\}_{n=1}^{\infty} \text{ is a bounded sequence } \right) = 0$$ Tip: If $(W_{n^2})_n$ is bounded then also $(W_{n^4})_n$ is bounded.

When the sequence $\{W_{n^2}\}_{n=1}^{\infty}$ was limited then exist $M>0$ such that $|W_{n^2}|\le M$ for all $n\ge 1$.

From the definition of Wiener process: $$\forall_{0 \le s <t} W_t-W_s \sim \mathcal N(0,t-s)$$ so also $$W_{(n+1)^2}-W_{n^2} \sim \mathcal N(0,2n-1)$$

However, these observations don't seem to help me and I don't know what to use to prove my thesis.

Answer 1

Define $$ A_n:=\left\{\left\lvert W_{(n+1)^2}-W_{n^2}\right\rvert>\sqrt{n}\right\}. $$ By definition of the Brownian motion, the sequence $\left(A_n\right)_{n\geqslant 1}$ is independent and $\mathbb P(A_n)=\mathbb P\left( N\sqrt{2n+ 1}>\sqrt n\right)$, where $N$ has a standard normal distribution. Since $\sum_{n=1}^\infty \mathbb P(A_n)=\infty$,we conclude by the second Borel-Cantelli lemma that $\mathbb P\left(\limsup_n A_n\right)=1$ hence the probability that $\left(\left\lvert W_{(n+1)^2}-W_{n^2}\right\rvert\right)$ is bounded is $0$.