discretized Brownian motion

Question

These are the definitions I'm working with:

• A (standard) Brownian motion in $\mathbb{R}$ is a stochastic process $W(t)$ $(t \geq 0)$ such that the following properties hold:

  1. $W(0) = 0$ almost certain
  2. $W(t) - W(s) \sim\mathcal{N}(0, t - s)$ (for 0 $\leq s \leq t$)
  3. For every $0 = t_0 < t_1 < ... < t_N$, $W(t_j) - W(t_{j-1})$ $(1 \leq j \leq N)$ are independent.

• For a certain natural number $N \geq 1$ and $j \in \left\{0, 1, 2, ..., N\right\}$, we define $\Delta t = T/N$ en $t_j = j \cdot \Delta t$. Further, let $W_j = \tilde{W(t_j)}$, the numerical approximation of $W(t_j)$. Now let $Z_j \sim \mathcal{N}(0, 1)$ independent vaiables for $j \in \left\{1, 2, ..., N\right\}$ and let $W_0 = 0$ and $W_j = W_{j-1} + \sqrt{\Delta t} \cdot Z_j$ for $j \in \left\{1, 2, ..., N\right\}$. We call $W_j$ $(j = 0, 1, 2, ..., N)$ a discretized Brownian motion.

A discretized Brownian motion is considered a special case of the standard Brownian motion. It is clear that properties (1) (by definition of the discretized Brownian motion) and (3) hold since $W_j$ can be written as $\sqrt{\Delta t} \cdot \sum_{n = 1}^{j} Z_n$ and all $Z_n$'s are independent. However, it is not clear to me that $W_t - W_s$ is normally distributed with mean 0 and variance $t-s$. Since $W_t - W_s = \sqrt{\Delta t} \cdot \sum_{n = s + 1}^{t} Z_n$, shouldn't $W_t - W_s$ be normally distributed with mean 0 and variance $(t - s) \cdot \Delta t$?

Answer 1

You have an incorrect statement in your summation: you should have $$\displaystyle W_t - W_s = \sqrt{\Delta t} \cdot \sum_{n = 1+s/\Delta t}^{t/\Delta t} Z_n$$ and the variance of this is $$\Delta t \cdot \dfrac{t-s}{\Delta t} = t-s$$ as you would hope