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If a discrete random variable $X$ is defined as a mapping of subsets of a sample space $\mathcal{S}=\{s_1,s_2,s_3\ldots\}$ to particular values $x_i$ which together make up its range $R_X=\{x_1,x_2,x_3,\ldots\}$, then must the mapped subsets be disjoint? I am asking this because the general formula for the probability of an event $\mathcal{A}\subset{R}_X$ is $P[A]=\sum\limits_{x\in\mathcal{A}}P[X=x]$ which assumes they are disjoint since it adds up the probabilities without considering their intersections. Is this because the outcomes in $R_X$ must be disjoint?

Asaf Karagila
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Undertherainbow
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1 Answers1

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You have a map $X$ (which happens to be measurable) from some set $\mathcal{S}$ (which happens to be a probability space) onto some set $R_X$ (which happens to be a subset of say, the reals). Then, as a map, if $A$ and $B$ are disjoint, then $X^{-1}(A)\cap X^{-1}(B)=\emptyset$. $(X=x)$ is just the notation used in probability theory for the set $X^{-1}(\{x\})$, so, following the above, these sets are disjoint, because $X$ is a map.

WoolierThanThou
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  • Do $A$ and $B$ represent values of the random variable or are they sets? I understood from your answer that they are sets but then what would $X^{-1}(A)$ mean? – Undertherainbow Sep 09 '19 at 15:18
  • I'm saying the way to think about is: $X$ is a map. $X^{-1}(A)$ is the pre-image of the set $A$. It lives in some abstract sample space, so there isn't really much to say about how it looks. – WoolierThanThou Sep 09 '19 at 15:21
  • So you mean the members of $R_X$ are disjoint so since $X$ is a map it must be that the pre-images in the sample space of these members are disjoint? – Undertherainbow Sep 09 '19 at 15:27
  • That's exactly the case. – WoolierThanThou Sep 09 '19 at 15:28