One of the conditions for a family of random variables
$W =(W_t, t \geq 0)$ to be Brownian motion is that the increments are normally distributed, specifically $W_t – W_s \sim N(0, t-s)$, for every $0 \leq s \leq t$.
This Wikipedia article https://en.wikipedia.org/wiki/Wiener_process states that
$W_t \sim N(0, t)$ (in the section titled "Properties of a one-dimensional Wiener process"), so we will also have $W_s \sim N(0, s)$.
Therefore, I am struggling to understand why it is not the case that
$W_t – W_s \sim N(0, t+s)$, by addition of variances of random variables? Is it because
$W_s$ and $W_t$ are not independent?
Best Answer
Yes, it is because $W_t$ and $W_s$ are not uncorrelated for $t,s>0$ (and so in particular not independent):
You can actually find the variance of $W_t-W_s$ from the covariance function of the Brownian motion: $$\mathbb V(W_t-W_s)=\mathbb V(W_t) -2 \operatorname{Cov}(W_t, W_s)+\mathbb V(W_s) = t -2\min(t,s)+s,$$ which is $t-s$ when $t\ge s$. (Note that if $W_t$ and $W_s$ were independent, we would have in particular that $\operatorname{Cov}(W_t, W_s)=0$.)