# Pointwise convergence of Random walk to the Brownian motion

Let $\{X_n\}_{n \in \mathbb{N}}$ be a sequence of i.i.d random variables taking value in the set $\{-1,1\}$ with $P(X_n = 1) = \frac{1}{2}$. Let $S_n = \sum\limits_{i=1}^n X_n$, and for $k \in \mathbb{N}$ define the random process $B^k$ as $$B^k(t) = \frac{S_{[kt]}}{\sqrt{k}} \qquad t \in \mathbb{R}$$

where $[x]$ is the integer part of $x$.

I know that $(B^k(t))_{t \in \mathbb{R}}$ converges to the standard Brownian motion in distribution, does it also converges pointwise? i.e. is there a random process $\{B(t)\}_{t\in \mathbb{R}}$ so that $$\forall t \in \mathbb{R} \quad \lim_{k \to \infty} B^k(t) = B(t) \quad \text{a.s}$$ and further that $B(t)$ is the Brownian motion.

• If the pointwise limit exists, then it must be BM. So the question really is if the limit exists. – Miheer May 5 '17 at 15:10
• A related question seems to be : is there pointwise convergence in the central limit theorem – Miheer May 5 '17 at 15:19

## 1 Answer

No, it doesn't. By the Kolmogorov 0-1 law, if the a.s. (or even i.p.) limit exists, it is a constant, which is absurd.

To see this, let's take $t=1$ for simplicity. Suppose that $S_k / \sqrt{k} \to Y$ a.s. Fix $j$ and write, for $k \ge j$,

$$\frac{S_k}{\sqrt{k}} = \frac{X_1 + \dots + X_{j-1}}{\sqrt{k}} + \frac{X_{j} + \dots + X_k}{\sqrt{k}}$$

As $k \to \infty$, the first term goes to zero a.s. (since the numerator does not depend on $k$). So we have

$$Y = \lim_{k \to \infty} \frac{X_{j} + \dots + X_k}{\sqrt{k}}$$

which shows that $Y \in \sigma(X_j, X_{j+1}, \dots)$. But $j$ was arbitrary, so this shows that $Y$ is a tail random variable. Kolmogorov says the tail $\sigma$-field is almost trivial, meaning that $Y$ is a.s. equal to a constant.

Alternatively, the law of the iterated logarithm for random walk will also show you that, almost surely, $\limsup S_k / \sqrt{k} = +\infty$ and $\liminf S_k / \sqrt{k} = -\infty$.