I've seen this related question but something is bothering me. The accepted answer hints to the fact that we have to define a random variable on the sample space rather than on the field of events.

I'm struggling to see why is this the case.

In their book Probability, random variables, and stochastic processes Papoulis and Pillai discuss, in section 2-2, the Probability space. They say that, and I quote, "In the applications of probability theory to physical problems, the identification of experimental outcomes is not always unique." For a die rolling experiment, for example, we can define the sample space $S$ to be $S_1=\{f_1, \dots,f_6\}$ ($f_i$ faces of die) or $S_2=\{\text{even},\text{odd}\}$ (for even and odd number faces).

For me, $S_2$ can be seen as a subset of $S_1$ representing events, i.e. $S_2$ can be seen as a field $\mathcal{F}$ of events constructed from $S_1$. Which is to say that if we consider our sample space as $S_2$ we can interpret a random variable $X$ as being mapped from this sample space $S_2$ or from the field $\mathcal{F}$ equivalently!

am I wrong? Please correct me if I am. Any thoughts on that are appreciated.


closed as unclear what you're asking by Namaste, user223391, José Carlos Santos, kimchi lover, Did Nov 18 '17 at 10:55

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  • $\begingroup$ The elements of $S_2$ can be thought of as events on $S_1$, but the sigma algebra on $S_2$ is $\{ \{\}, S_2, \{even\}, \{odd\} \}$. $\endgroup$ – Jack M Nov 17 '17 at 20:45
  • $\begingroup$ @JackM Okay. suppose $\mathcal{F}=\{\{\}, S_2, \{even\}, \{odd\}\}$ the $\sigma$-algebra on $S_1$. Can we consider this $\sigma$-algebra as a sample space on which we define a random variable $X$? $\endgroup$ – Learn_and_Share Nov 17 '17 at 20:53
  • $\begingroup$ Sure. Any set can be made into a probability space by taking its power set for a sigma algebra. But that random variable $X$ may or may not have a meaningful interpretation. $\endgroup$ – Jack M Nov 17 '17 at 20:57
  • $\begingroup$ Good. Now my question is, if we can interpret $\mathcal{F}$, which is $\sigma$-algebra on $S_1$ as a sample space, then we can define a random variable $X$ on this $\sigma$-algebra (field). Then, why do we say that we can't define a random variable from the field of events (our $\sigma$-algebra). I think I'm missing something here! $\endgroup$ – Learn_and_Share Nov 17 '17 at 21:00
  • $\begingroup$ @MedNeit Take a probability space $(\Omega, \mathcal F)$. If you want, you can go ahead and define a random variable $X$ on the space $(\mathcal F, \mathcal P(\mathcal F))$. But that probably won't be very useful. I'm essentially forgetting about the fact that $\mathcal F$ is a sigma algebra and just thinking of it as a set, and putting a brand new sigma algebra on it. $\endgroup$ – Jack M Nov 17 '17 at 23:44

$S_2$ is not a $\sigma$-field, though. Rather, $S_2$ is a partition of the sample space, and the random variable $X$ which is $1$ if the die roll is odd and $0$ if the die roll is even is a function $X:S \to \{0,1\}$ which is constant on the sets in the partition $S_2$. That is, $X(f_1)=X(f_3)=X(f_5)=1$ and $X(f_2)=X(f_4)=X(f_6)=0$.

If $S_2$ were a $\sigma$-field we would have to include $\emptyset$ and $\{f_1,\dots,f_6\}$. How would you define $X(\emptyset)$? $X(\{f_1,\dots,f_6\})$?

  • $\begingroup$ Why is $S_2$ not a $\sigma$-field? Is it because the intersection of even and odd is not in the field I defined? If so, can we not consider the empty set as part of our field to make it a $\sigma$-field and work with that? $\endgroup$ – Learn_and_Share Nov 17 '17 at 20:38
  • $\begingroup$ Also how would you define $X(\emptyset)$ and $X(\{f_1, \dots, f_6\})$ of our sample space was $S_1 = \{f_1, \dots, f_6\}$? $\endgroup$ – Learn_and_Share Nov 17 '17 at 20:41
  • $\begingroup$ @MedNait $S_2$ is not a $\sigma$-field because it is not closed under intersections and unions. If the sample space is $S_1$ then $X$ is defined on the elements of $S_1$, i.e., $X(f_1), \dots,X(f_6)$. I defined these in my answer. $\endgroup$ – kccu Nov 18 '17 at 15:29
  • $\begingroup$ It took me a while, but I think I got it. Thanks. $\endgroup$ – Learn_and_Share Nov 19 '17 at 1:38

The idea you're trying to express is that you have two different probability spaces $S_1$ and $S_2$, and a measurable function $f : S_1 \to S_2$.

Some random variables $X : S_1 \to R$ can be written as a composite $X = Y \circ f$ for some other random variable $Y : S_2 \to R$.

By pushforward of measure, whatever probability measure $P$ you have on $S_1$ can be "pushed forward" to become a probability measure $f_*[P]$ on $S_2$; if $E$ is an event in $S_2$, then its probability is given by

$$ f_*[P](E) = P(f^{-1}(E)) = P(\{ x \in S_1 \mid f(x) \in E \}) $$

For such an $X$, questions about its distribution can often be answered in terms of the random variable $Y$. For example, given some event $I \subseteq R$, you'd have

$$ P(X \in I) = f_*[P](Y \in I) $$

  • $\begingroup$ I'm not sure this is what I meant, since I am imagining, say for an experiment, $S_2$ as another possibly sample space for this experiment instead of $S_1$ and at the same time a subset of $S_1$. Maybe this is what's wrong with my reasoning... $\endgroup$ – Learn_and_Share Nov 17 '17 at 20:56

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