I have a silly question, tough not obvious to me. Most times we see in probability theory many of the definitions or theorems start with the following quotation: ``Let $X$ be a random variable that is uniformly distributed over $G$..." How are these two connected. Namely, how the variable is connected to the set $G$? Is it through the pdf of $G$ or does this have to do with the support or the domain of the random variable of $X$?
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2$\begingroup$ It means $X$ takes values on $G$. Uniformity means that the probability that $X$ takes a value in some subset $A$ of $G$ is proportional to the “size” of the subset $A$. Depending on what the structure/geometry of the set $G$ is, “size” will mean different things. $\endgroup$– user27182Commented Jan 3, 2022 at 11:44
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$\begingroup$ @user27182 when we say that $X$ is a random variable with support $G$ is it equivalent like saying that $X$ takes values on $G$? $\endgroup$– Hunger LearnCommented Jan 3, 2022 at 12:05
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1$\begingroup$ The short answer is yes, but there’s a caveat. The support of a random variable is the union of all the sets that have strictly positive (ie non zero) probability with respect to the distribution of the random variable. However, people might say that X~Unif[0,1] takes values on the reals even though it’s support is [0, 1]. $\endgroup$– user27182Commented Jan 3, 2022 at 13:37
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1 Answer
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As noted in the comments, "$X$ is uniformly distributed over $G$" means that $P(X\in I) \propto m(I)$ for some measure $m$ for and $\forall I \in \sigma(G)$ (the "sigma field" generated by $G$).
If $G \subset \mathbb{R}$ then $m$ is usually the Lebesgue measure and the measurable subsets are defined as the Borel sigma algebra $\mathcal{B}(\mathbb{R})$ (it's the sigma algebra defined by all the open sets of $G$)
For example, if $G = [0,1]$ then our "measurable sets" are the unions and intersections and complements of all the open subsets on $[0,1]$ (which basically includes all the sets one would care about, like $[0,0.5), [.1,.2], $etc.
The associated measure is the Lebesgue measure, which is boils down to the total length of the set for simple 1-d intervals like $(a,b)$
$$\text{Leb}([a,b]) = b-a$$
Technical point
A random variable is formally a function from an abstract probability space to a measurable space (usually some subset of the real numbers or integers):
$$X: \Omega \to W,\text{ where }S:=\left(\Omega, \mathcal{S}, \mathbb{P}\right), V:= \left(W,\mathcal{W}\right)$$
Unless one is doing theoretical work, we often "forget" about $S$ and just work with $V$ and the "induced probability measure" $\mathbb{P}_X := \mathbb{P}\left(X^{-1}(B)\right), \forall B \in \mathcal{W}$, which together forms another probability space:
$$V':=\left(W,\mathcal{W}, \mathbb{P}_X\right)$$
When we say "X is uniformly distributed over G" we are basically specifying $V'$ directly (i.e., instead of trying to derive from some underlying probability space that we don't really care about).
In this case, we don't really end up thinking of $X$ as a function anymore, but as the outcome of an experiment, where $V'$ is the probability space and $W=G$ is the sample space it is defined on.
I never understood why we teach Prob. 101 students the whole "RV as function" rule as its utility is very limited in non-theoretical contexts and often confuses more than helps (at least did so with me).
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$\begingroup$ what does this $\propto$ symbol stands for? $\endgroup$ Commented Jan 4, 2022 at 21:32
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$\begingroup$ @KingOdysseus "proportional to": $P(A) \propto m(A) \implies \frac{P(A)}{P(B)} = \frac{m(A)}{m(B)}$ or, equivalently, $\exists c: P(A) = cm(A)$ $\endgroup$– AnnikaCommented Jan 4, 2022 at 21:37