0
$\begingroup$

As you know, the definition of random vector is the following:

Let $(\Omega, \mathcal{F})$ be a measurable space. Then function $\mathbf{X}: \Omega \to \mathbb{R}^n$ is called random vector on $(\Omega, \mathcal{F})$ if it is measurable-$\mathcal{F}$, i.e. $~\{\omega: \mathbf{X}(\omega) \in B\} \in \mathcal{F}, ~~ \forall B \in \mathcal{B}(\mathbb{R}^n)$.

Here $\mathcal{B}(\mathbb{R}^n)$ is Borel sigma-algebra on $\mathbb{R}^n$.

And it is well-known fact that every component $X_i: \Omega \to \mathbb{R}$ of random vector $\mathbf{X} = (X_1, \ldots, X_n)$ is also measurable-$\mathcal{F}$ function on $(\Omega, \mathcal{F})$.

But there are no words about measure in the aforementioned definition. Let's define measure $P$ on $(\Omega, \mathcal{F})$ and get a probability space $(\Omega, \mathcal{F}, P)$. After that let's define random vector $\mathbf{X}$ on this "domain" probability space.

Is it correct to assume that components $X_1, \ldots, X_n$ must be defined on the same probability space $(\Omega, \mathcal{F}, P)$ as vector $\mathbf{X} = (X_1, \ldots, X_n)$ ?

I said above that $\mathbf{X}$ and its components $X_1, \ldots, X_n$ must be defined on the same measurable space $(\Omega, \mathcal{F})$. But if we consider probability space instead of measurable space, then should this space be the same for $\mathbf{X}$ and every component $X_i$?

P.s. Keep in mind that I don't talk about distributions of $\mathbf{X}$ and $X_i$ (i.e. about induced measures). I know that they can be different. I am talking about the "domain" measure $P$ on $(\Omega, \mathcal{F})$.


EDIT
Indeed, in few probability textbooks and wikipedia there is a requirement that components $X_i$ must be defined on the same probability space as each other, $(\Omega, \mathcal{F}, P)$.

Also it is interesting to look at the well-known formula for the random vector $\mathbf{X} = (X_1, \ldots, X_n)$ with independent components. This formula has the following form:

$$P(\mathbf{X} \in B) = \prod_{i=1}^n P(X_i \in B_i).$$

Note that the same measure $P$ is used in both left and right sides of the expression. I think this means that random vector $\mathbf{X}$ and every its component $X_i$ are defined on the same "domain" probability space $(\Omega, \mathcal{F}, P)$.

What do you think about it? Should we always assume that components $X_1, \ldots, X_n$ must be defined on the same probability space $(\Omega, \mathcal{F}, P)$ as vector $\mathbf{X} = (X_1, \ldots, X_n)$ ?

$\endgroup$
  • 2
    $\begingroup$ In probability theory (I believe) it is customary to use the term probability rather than measure. When discussing several random variables, there is implicitly assumed to be only one underlying probability space, otherwise it would be very difficult to discuss relationships. $\endgroup$ – herb steinberg Jun 4 '18 at 21:53
1
+50
$\begingroup$

I'm not sure if my understanding will unravel your confusion. Hope it'll help you.

As I think, it doesn't matter whether you define the components on different probability spaces, because you can always build a big one to include them and divide the big one into several parts.

  1. Consider a random vector $\textbf{X}=(X_1,X_2,\cdots,X_n)$, whose components are defined on different probability spaces $(\Omega_i,\mathcal{F}_i,P_i),\ 1\leq i\leq n$. Define a product probability space on $\Omega=\Pi_{i=1}^n\Omega_i$ by setting $$\mathcal{F}=\sigma\{\Pi_{i=1}^nA_i|A_i\in\mathcal{F}_i\}$$ and we know that there exists a unique probability measure on $(\Omega,\mathcal{F})$ such that $$P(\Pi_{i=1}^nA_i)=\Pi_{i=1}^nP_i(A_i),\ A_i\in\mathcal{F}_i\ .$$ Now we redefine the variables by letting $$Y_i(\omega_1,\omega_2,\cdots,\omega_n)=X_i(\omega_i),\quad \omega_j\in\Omega_j.$$ Here we shall see the value of $Y_i$ is only dependent on $\omega_i$, so actually we can treat $Y_i$ the same as $X_i$. Thus we now have a new probability space $(\Omega,\mathcal{F},P)$. Note that $Y_i(1\leq i\leq n)$ are measurable on $(\Omega,\mathcal{F},P)$, for $$Y_i^{-1}(\mathcal{B}(\mathbb{R}))=\{\Omega_1\times\cdots\times\Omega_{i-1}\times A\times\Omega_{i+1}\times\cdots\times\Omega_n|A\in\mathcal{F_i}\}\subset\mathcal{F}.$$

  2. And we can divide the big probability space into smaller ones by setting $$\Omega_i=\Omega,\ \mathcal{F}_i=\sigma(X_i),\ P_i=P|_{\mathcal{F_i}}\ .$$

I think your definition will contribute to intuitive understanding, but it's all the same in mathematical sense.

$\endgroup$
  • $\begingroup$ Thanks for the answer. I have two questions. 1) It seems a bit strange for me that in (1.) we "redefine" the domain measurable space $(\Omega_i, \mathcal{F}_i)$ of $X_i$ to another measurable space. Because usually in mathematics when we change the domain of a function then we get a different function. Let's denote the "old" function $X_i(\omega_i) = f(\omega_i)$. Should we redefine $X_i$ in a such a way that "new" $X_i(\omega) = f(\omega_i), \forall \omega \in \Omega$ ? If yes then this procedure will not seem so strange for me. $\endgroup$ – Rodvi Jun 11 '18 at 15:40
  • $\begingroup$ 2) Also is it true that in (1.) random vector $\mathbf{X} = (X_1, \ldots, X_n)$ will be measurable on $(\Omega, \mathcal{F}, P)$? $\endgroup$ – Rodvi Jun 11 '18 at 15:41
  • 1
    $\begingroup$ @Rodvi Added up more details. I think now will be better. $\endgroup$ – XIAODA QU Jun 11 '18 at 16:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.