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Often, I encounter some problem were probability is involved. On an intuitive level, I understand I think what is going on but when I have to formalize it with probability spaces I am a bit lost.

For example :

"Pick an edge uniformly at random in a graph". I understand what this mean. It means that I pick any edge of the graph with probability $1/m$ where $m$ is the number of edges in the graph. But what is the probability space here? We have $\Omega = \{edges\}$ right? Is there a uniform random variable involved $X : \Omega \to \mathbb{R}$? I know that I should have $P(X=e_i)=\frac{1}{\mid \text{\{edges\}} \mid}$ but then it seems like the random variable goes from $\Omega$ to $\Omega$ which is kind of weird. What is this random variable actually?

roi_saumon
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  • "Is there a uniform random variable involved"? If you want to be incredibly pedantic. Random variables are a very specific type of object that we investigate in probability who are functions with codomain some measurable space, giving us a way to measure a particular event from our sample space with respect to this possibly different measure. If you are talking about real-valued random variables specifically, that measurable space for the codomain has to specifically be the real numbers. See wikipedia. – JMoravitz May 29 '20 at 12:12
  • You might as well say, "No, random variables aren't involved here (yet)" as an answer to your question. If you were to continue your question with "Pick an edge uniformly at random in a graph. What is the probability that the edge is incident to (i.e. shares a vertex with) at least 4 other edges" then we could talk about the random variable $X$ which counts the number of edges a particular edge is incident to. Otherwise, your function $P(X=e_i)=\frac{1}{|{\text{edges}}|}$ is just the probability measure, not a random variable. – JMoravitz May 29 '20 at 12:15
  • Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space. You want your random variable to take values in the set of edges, which I denote by $E$. So what you want is $X\colon \Omega \to E$ with $\mathbb{P}(X = e) = \frac{1}{|E|}$ for every $e \in E$. You don't have to specify what $(\Omega,\mathcal{F},\mathbb{P})$ is, what's important is that it exists. – Michh May 29 '20 at 12:16
  • If you want a concrete way of constructing $(\Omega,\mathcal{F},\mathbb{P})$, you can take $\Omega = E$, $\mathbb{P}$ the uniform probability measure on $E$ (i.e. $\mathbb{P}(\lbrace e\rbrace ) = \frac{1}{|E|}$ for every $e \in E$) and $X \colon E \to E$ the identity map. – Michh May 29 '20 at 12:17
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    It might be that this question and its answer(s) is interesting for you. There you find why it is not really needed to specify a probability space. As @Michh says: what's important is that it exists. – drhab May 29 '20 at 13:50

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