Brownian motion start value

Question

Suppose the definition of a Wienerprocess without the assumption of $W_{0}=0$. Now I want to show that $Var[W_{t}-W_{s}]=E[(W_{t}-W_{s})^2]=t-s$ and $E[W_{t}-W_{s}]=0$ but I'm a bit stuck. Could anyone help me out? So far I have for the first one: $E[(W_{t}-W_{s})^2]=E[(W_{t}-W_{0})^2]+E[(W_{s}-W_{0})^2]-2E[(W_{t}-W_{0})(W_{s}-W_{0})]=t+s- ?$

Answer 1

Note that $B_t := W_t-W_0$ is a Brownian motion started at $0$. Since $W_t-W_s = B_t-B_s$, we find

$$\mathbb{E}(W_t-W_s) = \mathbb{E}(B_t-B_s) = 0$$

and

$$\mathbb{E}((W_t-W_s)^2) = \mathbb{E}((B_t-B_s)^2)=t-s$$

where we have used that $B_t-B_s \sim B_{t-s} \sim N(0,t-s)$ for any $s \leq t$.