The following is the definition of a Wiener process that I am following:
I am confused regarding the multivariate Brownian motion which is defined as follows:
My question is, does $\mathbf{W}_t$ follow the same conditions 1-4 for a univariate Wiener process? Obviously, condition 1. is satisfied since $\mathbf{W}_0 = (W_0^1, \cdots, W_0^k)^T = (0, \cdots, 0)^T$, but how are conditions 2. and 3. satisfied? For example, is it true that $\mathbf{W}_u – \mathbf{W}_t$ independent of $\mathbf{W}_s – \mathbf{W}_r$? If i write out the vectors, I get:
$\mathbf{W}_u – \mathbf{W}_t =\begin{bmatrix} W_u^1 – W_t^1 \\ \vdots \\ W_u^k – W_t^k \end{bmatrix}$ and $\mathbf{W}_s – \mathbf{W}_r =\begin{bmatrix} W_s^1 – W_r^1 \\ \vdots \\ W_s^k – W_r^k \end{bmatrix}$. How are these two vectors independent of each other? If we look element by element, clearly $W_u^1 – W_t^1$ is independent of $W_s^1 – W_r^1$ but is $W_u^1 – W_t^1$ independent of $W_s^2 – W_r^2$?
Finally, for condition 3. is it true that $\mathbf{W}_t – \mathbf{W}_s \sim N(0, (t-s)\mathbf{I}_k)$ where $\mathbf{I}_k$ is the $k \times k$ identity matrix? If so, why is it true?
Best Answer
your first definition is the definition of a standard one-dimensional Brownian motion. The second definition is of a non-standard $k$-dimensional Brownian motion. In particular $$ Z_t - Z_s \sim N(\mu (t-s), (t-s) \Sigma ). $$ Therefore, if you set $\mu = 0$ and $\sigma = I_k$, then $\Sigma = I_k$ and $$ W_t - W_s \sim N(0, (t-s)I_k). $$
Note that in the second definition it is assumed that $W_t$ consists of $k$ independent (one dimensional) standard Brownian motion. In particular, $W^2_u$ is independent of $W^1_t$ for all $u$ and $t$.