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

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.

• Yes, what you have written is exactly equivalent to points 2 and 3 together. The proof should be straightforward; where did you get stuck? Commented Jul 24, 2020 at 19:08
• Do you know how to go about proving it? Commented Jul 24, 2020 at 19:25
• It's a bit tedious to write, but basically you just make a change of variables in your integral, letting $y_i = x_i - x_{i-1}$. This transforms the $X_{t_i}$ into independent random variables with known variances. Commented Jul 24, 2020 at 19:51

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.
• You are referring to this equation right, en.wikipedia.org/wiki/Chapman%E2%80%93Kolmogorov_equation? I see where the left hand matrix comes from in transforming $(X_{t_1}, X_{t_2}-X_{t_2}, \ldots X_{t_n}-X_{t_{n-1}}) \to (X_{t_1}, X_{t_2}, \ldots X_{t_n})$. I do not see where the relationship between the covariance matrices comes from or how the Chapman-Kolmogorov relation comes into play Commented Jul 30, 2020 at 18:07
• Do we need to use Chapman-Kolmogorov because I haven't covered it? Could we use that $X\sim \mathcal{N}(\mu, C)$ and then use positive definiteness to get the density function described here (en.wikipedia.org/wiki/…) and show that this density function is equal to the products $p(t_1,0,x_1)p(t_2-t_1, x_1, x_2)\ldots p(t_{n-1}, x_{n-1},x_n)$? Commented Jul 30, 2020 at 20:25
• This is just a reference for a proof that $C$ is the covariance vector, it is the fact you had described math.stackexchange.com/questions/332441/… Commented Jul 30, 2020 at 21:30