To show that a given process is Gaussian

Question

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

Answer 1

You have made a mistake considering a covariance function, that is not $W_{t-s}-W_{s}$, but $W_{t-s}-W_{t}$. However, using formulas $$\min\{a,b\}=\frac{a+b-|a-b|}{2}$$ $$\max\{a,b\}=\frac{a+b+|a-b|}{2}$$ that expression can be simplified to $$t-\frac{1}{2}\left(r+2t+s-2|t-s-r|+|s-r|\right)+\min\{r,s\}$$ which helps to see that it is obviously not $\min\{r,s\}$. Now the correct version looks like this: $$\min\{t-s,t-r\}-\min\{t-s,t\}-\min\{t,t-r\}+t$$ $$s+r-\max\{s,r\}$$ And it seems that you are correct about the Gaussian process part, but if you can assume that $W_t$ process is already Gaussian, then this could be simplified, look at my question.

Edit: Like I said, there are two ways of thinking about proving the second part.

(*) To prove that Brownian motion is Gaussian process, we have to show that all finite-dimensional distributions are Gaussian. Let $0<t_1<t_2<\dots<t_d$, $X=(W_{t_1},W_{t_2},\dots,W_{t_d})$, $Y=(W_{t_1},W_{t_2}-W_{t_1},\dots,W_{t_d}-W_{t_{d-1}})$. $Y$ is Gaussian since Brownian motion has independent increments, and $X=AY$ where $$A=\begin{pmatrix} 1 & 0 & \cdots & 0 \\ 1 & 1 & \cdots & 0 \\ \vdots & \vdots & \dots & \vdots \\ 1 & 1 & \cdots & 1 \end{pmatrix}$$ so $X$ is Gaussian vector.

First way. Assuming that you know (*) you can take $Y=(W_{t},W_{t-t_1},W_{t-t_2},\dots,W_{t-t_d})$ which was proved to be a Gaussian vector and then with some matrix $A$ that will be $X=AY$ where $X=(B_{t_1},B_{t_2},\dots,B_{t_d})$, or you can use no matrix and just say that any linear transformation applied to vector $Y$ results in a Gaussian vector.

Second way. Not using ( * ) you can take $Y=(W_{t-t_d},W_{t-t_{d-1}}-W_{t-t_d},\dots,W_{t-t_1}-W_{t-t_2},W_t-W_{t-t_1})$ (which seems similar to your chosen one), it is Gaussian since Brownian motion has independent increments and with some matrix $A$ you will get $X=AY$ where $X=(B_{t_1},B_{t_2},\dots,B_{t_d})$. This is similar to your solution but also I think that $\sum_{i=0}^na_i(W_{k_i}-W_{k_{i-1}})$ and $\sum_{i=0}^n b_iW_t$ are not independent..

I hope that I am not missing something.. have not been thinking about Brownian processes for a while so I might have became unobservant.

Edit 2: In the first way we added $W_t$ so that we could get $X$ with some matrix $A$, you can rearrange these variables anyhow you want, and take any indices $t_i$. Now $X$ has $d$ variables, $Y$ has $d+1$ (both column-vectors). So we need matrix $A$ to be $d\times d+1$, like this: $$ A=\begin{pmatrix} -1 & 1 & 0 & \cdots & 0 & 0\\ -1 & 0 & 1 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots\\ -1 & 0 & 0 & \cdots & 0 & 1 \end{pmatrix}$$ i.e. almost like in the comment, except we do not need the first row. And probably that would look a bit better to make $W_t$ as the last variable in the vector $Y$.