$W(t)_{t>0}$ is a Brownian motion. $V(t)=W(s+t)-W(s). \text{ } s,t>0$.

Is $V$ also a Brownian motion?

It is clear that $E[V]=0$. I would argue that the variance is $$\operatorname{var}[V]=\operatorname{var}[W(t+s)]-\operatorname{var}[W(s)]-2\operatorname{cov}[W(s+t),W(s)]=t+s-s-2\min(t+s,s)=t-2s$$

Here I wonder if I have computed the variance of $V$ correctly and how I use the variance to conclude whether $V$ is Brownian motion.


For a Brownian motion $W_t-W_s, 0<s<t$, is the distribution resulting at time $s$ later (or at $t+\Delta t$) if the state is $W_t$ at $t$, after the elapsing of a time $s-t$ (until time $t)$.

The other way to think of it is that it is the distribution of the increment of the process between times $s$ and $t$. By definition, a Brownina motion has independent increments of this form.

|In the question, is n$s$ supposed to be fixed?

By the way, it looks to me like in your mat the $t+s-s$ equals t - i.e., giving you $W_t$


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