In the Chapter 1 of the book:

Da Prato, Zabczyk - Stochastic equations in infinite dimensions, 1992,

we could find following general definitions:

  • If $(\Omega,\mathcal{F})$ and $(E,\mathcal{S})$ are two measurable spaces, then a mapping $X$ from $\Omega$ into $E$ such that the set $\{ \omega \in \Omega : X(\omega) \in A \}=\{ X \in A \}$ belongs to $\mathcal{F}$ for arbitrary $A \in \mathcal{S}$ is called a random variable from $(\Omega,\mathcal{F})$ to $(E,\mathcal{S})$.

  • If X is a random variable from $(\Omega,\mathcal{F})$ to $(E,\mathcal{S})$ and $P$ a probability measure on $\Omega$, then by $\mathcal{L}(X)$ we will denote the image of $P$ under the mapping $X$:

$\hspace{0.8cm} \mathcal{L}(X)(A)=P\{\omega \in \Omega: x(\omega) \in A \}, \forall A \in \mathcal{S}.$

The measure $\mu=\mathcal{L}(X)$ is called the distribution or the law of $X$.

Let's say that the set $E$ consist of functions that are given with:

$$U(x,t,\omega)= \begin{cases} u_1, x \in A, \\[2ex] u_2, x \notin A, \end{cases} $$

where $u_1,u_2$ are constants that could depend on $\omega$, and the set $A$ depends of variables $x$, $t$ and $\omega$. The easiest example I have in mind is $\forall \omega$ $A=\{x: x<t\}$ and $\forall \omega$ we have the same constants $u_1,u_2$. So I am interested in the discontinuous piecewise constant functions U.

I have two questions.

  1. Could we talk about distribution of a random variable U if the U is not continuous? If the answer is yes - and I am sure it is, what would be the law of the random variable U given above?
  2. Could we talk about convergence in the law of random variables $U_n$ to the random variable U, if the limit function is similar to the U given above (i.e. discontinuous) and the $U_n$ are some continuous or smooth random variables - even if they are not on the same spaces? When I say space I mean some Banach space-valued spaces such as: $L^p(\Omega;BV(\mathbb{R}^d))$, $L^p(\Omega;L^{\infty}(\mathbb{R}^d))$, $L^p(\Omega;C([0,T],L^{\infty}(\mathbb{R}^d)))$ and not the probability space. I assume that the answer is yes also but I need a little more details.

I hope that I am not asking the obvious things. Thank you all for the help in advance.


Yes, you can talk about the distribution of a non-continuous random variable. I think you are mistaken when you say "In my case set $E$ is some set of piecewise constant functions." Your random variables take on real values, so your $E$ is $\mathbb{R}$ or some subset of it. In any case, the measure $\mu$ associated with your $U$ is given by $$\mu(S)= \begin{cases} 0 & u_1,u_2 \notin S \\ P(U=u_1) & u_1 \in S, u_2 \notin S \\ P(U=u_2) & u_1 \notin S, u_2 \in E \\ 1 & u_1,u_2 \in S. \end{cases}$$

Yes, you can talk about convergence of random variables that are not on the same probability space. The Kolmogorov extension theorem essentially guarantees that you can build one probability space on which all of your random variables live and such that their joint distributions are all consistent.

  • $\begingroup$ Thank you for the answer. In the traditional definition $E=\mathbb{R}$ or some subset of it, but in the book I mentioned authors talk about generalization of random variables and stochastic processes on Banach spaces, so there $E$ is some function space - in my case $L^{\infty}$ or BV or some Sobolev space... But we agree, the answer to my first question is yes. For the second, this measure work fine when we talk about traditional definition but I have to check if it works in some Banach space valued space. $\endgroup$
    – Mark
    Mar 30 '19 at 15:38
  • $\begingroup$ See the Wikipedia page on the Kolmogorov extension theorem (general form of the theorem). $E=\mathbb{R}$ is not necessary for that theorem. $\endgroup$
    – kccu
    Mar 30 '19 at 15:40
  • $\begingroup$ For the third one: thank you for the reminder on Kolmogorov extension theorem. I forgot about it. In the problem I am working on I have some sequence of random variables/stochastic processes that are Sobolev space-valued and a limit random variable/stochastic process that is $L^{\infty}$ space-valued. In one case sample space is the same; in other I have different one. I think that I could use Kolmogorov theorem but I need to check some details. P.S. I saw your comment just now $\endgroup$
    – Mark
    Mar 30 '19 at 15:46

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