The finite-dimensional distributions for a Wiener process are given by this formula?

Question

A stochastic process $X = \{X_t\}$ on is Wiener Process if the following properties hold

1. $X_0 = 0$
2. $X$ has independent increments: for any $n\in\mathbb{N}$ and any $0 < t_0<\ldots < t_m$ we have that $(X_{t_1}-X_{t_0}), (X_{t_2}-X_{t_1}), \ldots (X_{t_n}-X_{t_n-1})$ are independent
3. $X_t - X_s \sim \mathcal{N}(0, t-s)$ where $s\leq t$
4. $t\to X_t(\omega)$ is continuous for almost all sample paths $\omega$

I am looking for a proof that these properties imply that the finite-dimensional distributions are given by the following formula. Letting $0\leq t_1 < \ldots < t_n$ and any finite collection of Borel sets $F_1,\ldots F_n$ that $$P(X_1\in F_1, \ldots X_n\in F_n) = \int_{F_1\times\ldots \times F_n}p(t_1, 0, x_1)p(t_2 - t_2, x_1, x_2)\ldots p(t_n-t_{n-1}, x_{n-1}, x_n) dx_1\ldots dx_n $$ where $p(t, x, y) = \frac{1}{\sqrt{2\pi t}}\exp(-\frac{(x-y)^2}{2t})$ when $t > 0$ and $p(0,x,y) = \delta_x(y)$? We know from Kolmogorov's Extension Theorem that the family of all these finite-dimensional distributions will give us a Wiener process, I am curious if the properties go the other way and for a proof.

Answer 1

The easiest way to prove this `Chapman-Kolmogorov' relation is by noting that it suffices to prove that it is equivalent to the fact that $(X_{t_1},\dots,X_{t_n})$ is multinomial normal distributed with mean $0$ and convariance matrix $C_{i,j}=t_{i\wedge j}$. For this observe that by 2. and 3., $(X_{t_1},X_{t_2}-X_{t_1},\dots,X_{t_n}-X_{t_{n-1}})$ is multinomial normal distributed with mean $0$ and covariance matrix $$ \bar{C}=\mathrm{diag}(t_1,t_2-t_1,\dots,t_n-t_{n-1}). $$ Now $$ \begin{pmatrix} X_{t_1}\\\vdots\\X_{t_n} \end{pmatrix}=\begin{pmatrix} 1 & 0 &\cdots & 0\\ 1 & 1 & \cdots & 0\\ \vdots& & \ddots & \vdots\\ 1 & 1 & \cdots & 1 \end{pmatrix}\begin{pmatrix} X_{t_1}\\X_{t_2}-X_{t_1}\\\vdots\\X_{t_n}-X_{t_{n-1}} \end{pmatrix} $$ and therefore $$ C=\begin{pmatrix} 1 & 0 &\cdots & 0\\ 1 & 1 & \cdots & 0\\ \vdots& & \ddots & \vdots\\ 1 & 1 & \cdots & 1 \end{pmatrix}\bar{C}\begin{pmatrix} 1 & 1 &\cdots & 1\\ 0 & 1 & \cdots & 1\\ \vdots& & \ddots & \vdots\\ 0 & 0 & \cdots & 1 \end{pmatrix} $$ and you can easily check that this indeed gives you the required covariance matrix.