Prove that the increments of the Brownian motion are normally distributed

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 $(1)$?

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).

• @Calculon My definition of stationary increments is as follows: A real-valued stochastic process $(X_t)_{t\in I}$ has stationary increments $:\Leftrightarrow$ $$X_{s+t+r}-X_{t+r}\sim X_{s+r}-X_r\;\;\;\text{for all }r,s,t\in I\;.$$ If $0\in I$, then we only need to consider $r=0$. – 0xbadf00d Jul 3 '15 at 11:50
• @Calculon: that depends on the context. AFAIK, for Lévy processes, stationarity is usually understood as strong stationarity. – D. Thomine Jul 3 '15 at 12:00
• I'm sorry, but that's exactly the definition I found in many textbooks. However, a stochastic process $(X_t)_{t\in I}$ is called stationary $:\Leftrightarrow$ $$(X_{s+t})_{t\in I}\sim (X_t)_{t\in I}\;\;\;\text{for all }s\in I\;.$$ Maybe we/you are confused by the definitions of stationarity and stationary increments. By stationary increments I don't mean, that the process $(X_{s+t}-X_t)_{t\in I}$ is stationary, for all $s\in I$. – 0xbadf00d Jul 3 '15 at 12:01
• You should write $$B_{t-s}\sim\mathcal{N}(0,t-s)\;,$$ since I refer to $\mathcal{N}(\mu,\sigma^2)$ in my question as being the distribution with mean $\mu$ and variance $\sigma^2$. – 0xbadf00d Jul 3 '15 at 12:14
• @0xbadf00d: my mistake. That's fixed. – D. Thomine Jul 3 '15 at 12:15