Suppose I have given a Brownian Motion $W$, this is a Gaussian process, and I define:
$B_s:=W_{t-s}-W_t$
for $0\le s\le t$. Clearly this random variable has expectation zero. For the covariance function I did this:
$Cov(W_{t-s}-W_s,W_{t-r}-W_r) = (t-s)\wedge(t-r) - ((t-s)\wedge r)-((t-s)\wedge s) +(r\wedge s)$
The first term is equal $r\vee s$. But how can I simplify the second and third one to get a covariance function $s\wedge r$?
The last point which is to show is that $B_s$ is a Gaussian process. So let $t_0,\dots,t_n$ be given time points. I have to show that $(B_{s_0},\dots,B_{s_n})$ is multivariate normal distributed, that is equivalent to
$\sum_{i=0}^n a_i B_{s_i}$
should be normal distributed for any scalars $a_i$. My idea was: Write $(B_{s_0},\dots,B_{s_n})$ as $(B_{s_0}-B_0,\dots,B_{s_n}-B_{t_{s-1}})$ using a linear transformation $\phi$ which is a $(n+1)$ matrix. Then the above equation leads to
$\sum_{i=0}^n a_i (W_{t-s_i}-W_{t})-a_i(W_{t-s_{i-1}}-W_{t})=\sum_{i=0}^na_i(W_{k_i}-W_{k_{i-1}})+\sum_{i=0}^n b_iW_t$
where $k_i:=t-s_i$ and $b_i:=-2a_i$ and $W_{k_{-1}}:=W_0=0$
Am I correct that the first sum is normal distributed, since the increment are independent and they are also independent to $\sum_{i=0}^nb_i W_t$ hence the whole thing is normal distributed. Therefore $(B_{s_0},\dots,B_{s_n})$ is multivariate normal distributed, since this remains true under linear transformations. Are my thoughts correct?
hulik