Let $B=(B_t)_{t\ge 0}$ be a Brownian motion on a probability space $(\Omega,\mathcal A,\operatorname{P})$, i.e. $B$ is a real-valued stochastic process with
- $B_0=0$ almost surely
- $B$ has independent and stationary increments
- $B_t\;\tilde\;\mathcal{N}_{0,\;t}$
- $B$ is almost surely continuous
I would like to show, that $$B_t-B_s\;\tilde\;\mathcal{N}_{0,\;t-s}\;\;\;\text{for all }0\le s<t.\tag{1}$$
I know, that the sum of two independent normally distributed random variables $Y\;\tilde\;\mathcal{N}_{\mu_Y,\;\sigma^2_Y}$ and $Z\;\tilde\;\mathcal{N}_{\mu_Z,\;\sigma^2_Z}$ is normally distributed, too, with $$Y+Z\;\tilde\;\mathcal{N}_{\mu_Y+\mu_Z,\;\sigma^2_Y+\sigma^2_Z}$$
Now, by definition of $B$, $B_t-B_s$ and $B_s$ are independent. However, I don't know the distribution of $B_t-B_s$. I only know, that each $B_s\;\tilde\;\mathcal{N}_{0,\;s}$.
So, how can we show $(1)$?
Best Answer
The increments are stationary. Since $B_t - B_s$ is the increment over the interval $[s, t]$, it is the same in distribution as the incremeent over the interval $[s-s, t-s] = [0,t-s]$. Hence,
$$B_t-B_s \sim B_{t-s}-B_0.$$
But $B_0 = 0$ almost surely, so that:
$$B_t-B_s \sim B_{t-s}.$$
Finally, $B_{t-s} \sim \mathcal{N} (0,t-s)$.
We didn't use the Markov property nor the continuity of the trajectories (so this argument generalizes to Lévy processes).