Let $(\Omega,\mathcal F,P)$ be a probability space, and $(S,\mathcal S)$ a measurable space. A random variable $\mathcal F$-$\mathcal S$-measurable function $\Omega\longrightarrow S$. A random variable is called

  • a real-valued random variable if $S = \mathbb R$,
  • a $k$-variate real-valued random variable if $S=\mathbb R^k$, and
  • a real-valued stochastic process if $S = \mathbb R^{[0,\infty)}$.

Any random variable $X$ there exists a measure $P_X$, called the distribution of $X$, which gives raise to a probability space $(S,\mathcal S,P_X)$.

Especially in the context of stocahstic processes the notion of finite-dimensional distributions of a random variable $X$ is important. Clearly, in the case $S=\mathbb R$ and $S=\mathbb R^k$ the notion of distribution and finite-dimensional distributions coincide. But in the case $S=\mathbb R^{[0,\infty)}$ I get confused by the way finite-dimensional distributions are understood. Are the finite-dimensional distributions the same as $P_X$ restricted to all finite-dimensional subspaces of $\mathbb R^{[0,\infty)}$?


1 Answer 1


When $S=\mathbb{R}^{[0,\infty)}$, $\mathscr{S}$ is defined to be the $\sigma$-field $ \mathscr{B}(\mathbb{R^{[0,\infty)}})$ generated by all the cylinder sets. The sets in the form $\{u\in\mathbb{R}^{[0,\infty)}:(u(t_1),\cdots,u(t_n))\in A^{(n)}\}$, where $0\le t_1<\cdots<t_n$ and $A^{(n)}\in \mathscr{B}(\mathbb{R}^n)$, are called cylinder sets. For a stochastic process $X$, the distribution of $X$ is a measure $P_X$ on $(\mathbb{R}^{[0,\infty)},\mathscr{B}(\mathbb{R^{[0,\infty)}}))$, defined as $P_X(A)=\mathbb{P}(X\in A)$ for any $A\in \mathscr{B}(\mathbb{R^{[0,\infty)}})$.

For your question, the finite-dimensional distribution is exactly determined by the values $P_X(A)$ for cylinder set $A$. And these values determine all the values $P_X(A)$ for $A\in \mathscr{B}(\mathbb{R^{[0,\infty)}})$, i.e. the distribution of $X$. More precisely, if there exist two probability measures $\mathbb{P}, \mathbb{Q}$ on $(\mathbb{R}^{[0,\infty)},\mathscr{B}(\mathbb{R^{[0,\infty)}}))$ such that $\mathbb{P}(A)=\mathbb{Q}(A)$ for any cylinder set $A$, then $\mathbb{P}=\mathbb{Q}$. You can easily prove it using $\pi$-$\lambda$ lemma, since cylinder sets are closed under finite intersection. In the book of Billingsley, you can find a more general definition called seperating classes.

  • $\begingroup$ "generated by all the cylinder sets. The sets in the form $\{u\in\mathbb R^{[0,\infty)}:(u(t_1),\dots,u(t_n))\in A^{(n)}\}$, where $0\leq t_1<...<t_n$ and $A^{(n)}\in\mathcal B(\mathbb R^n)$" -- here I have problems regarding the notation. Would it be the same if I said that $\{u\in\mathbb R^{[0,\infty)}: \pi_Iu\in A^{I}\}$, where $\pi_Iu = (u(t_1),\dots,u(t_n))$ with $|I| = n$, and $A^I\in\mathcal B(\mathbb R^I)$: $\endgroup$
    – lmaosome
    Commented Apr 26, 2021 at 23:39
  • $\begingroup$ Exactly the same. $\endgroup$ Commented Apr 28, 2021 at 2:55

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