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

Mathematics Asked by Andrew Shedlock on January 3, 2022

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 $$ninmathbb{N}$$ and any $$0 < t_0 we have that $$(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)$$ where $$sleq t$$
4. $$tto 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 $$0leq t_1 < ldots < t_n$$ and any finite collection of Borel sets $$F_1,ldots F_n$$ that
$$P(X_1in F_1, ldots X_nin F_n) = int_{F_1timesldots times F_n}p(t_1, 0, x_1)p(t_2 – t_2, x_1, x_2)ldots p(t_n-t_{n-1}, x_{n-1}, x_n) dx_1ldots dx_n$$
where $$p(t, x, y) = frac{1}{sqrt{2pi t}}exp(-frac{(x-y)^2}{2t})$$ 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.

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 $$C_{i,j}=t_{iwedge j}$$. 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.

Answered by julian on January 3, 2022

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