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Say $X_{1} = \{ -3,-2,-1,0,1,2,3 \}$ is a random variable; say we are asked to find the expectation $E(X_{1})$. Mathematically $X_{1}$ is just a set. To calculate the expectation we need the probability distribution function $p_{1}$ that maps set $X_{1}$ to $P_{1}$: $$E(X_{1}) = \sum_{i=1}^{|X_{1}|} \Big( x_{i} \cdot p_{1}(x_{i}) \Big)$$. Set $X_{1}$ is just the domain of the function $p_{1}$ (its attribute). Would it therefore not be more correct to write $E(p_{1})$? Also we can have two identical sets (random variables) $X_{1}$ and $X_{2}$, but they could have entirely different probability distributions $p_{1}$ and $p_{2}$. Would it therefore not be more correct to refer to "random variables" as $p_{1}$ and $p_{2}$ rather than $X_{1}$ and $X_{2}$?

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    $\begingroup$ A random variable is a measurable mapping, not a set. $\endgroup$ – user223391 Jun 12 '17 at 7:28
  • $\begingroup$ A set on its own is not a random variable (assuming it has more than one element; if not then it is not very random) $\endgroup$ – Henry Jun 12 '17 at 7:30
  • $\begingroup$ @ZacharySelk well... any function can be described as a set. Of course in the context of this question the provided set seems not the representation of a measurable mapping. $\endgroup$ – Masacroso Jun 12 '17 at 7:33
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    $\begingroup$ @Masacroso Not every set is a function though. $\endgroup$ – YoTengoUnLCD Jun 12 '17 at 7:35
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Your idea is on the right track, but there are more or less important points that is missing. But basically a random variable is just about prescribing probabilities and not as much random as their name indicates.

The most important point that you're missing is that a random variable also involves the concept of interdependence. This is not entirely covered by the probability distribution of the variable. To cover this you would have to have a mapping from the entire event space to value space of the variable in question. Consider for example when rolling two dice and the variables corresponding to their sum and difference (these are interdependent as they are both either odd or both even).

Another detail is that a random variable don't have to map from a discrete or finite set of outcomes. The theory will allow for continuous distribution yet assign a non-zero probability for certain individual results.

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  • $\begingroup$ I believe random variables are not random in the formal axiomatic theory only because otherwise the theory needs to define formally what 'random' means. The intuition behind random variable is a function of a random output of a probabilistic experiment. $\endgroup$ – kludg Jun 12 '17 at 8:21
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    $\begingroup$ "basically a random variable is just about prescribing probabilities" Well, no (and the irony is that the rest of your post briefly touches upon some of the reasons why this is not so). $\endgroup$ – Did Jun 12 '17 at 8:59
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    $\begingroup$ @Did What more than prescribing probabilities do you think random variables are about? That I don't dig into details is just because I didn't think the OP would benefit from those (judging from the level of the question)... $\endgroup$ – skyking Jun 12 '17 at 9:04
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    $\begingroup$ First, because as soon as several random variables are involved, one must be careful to consider their joint distributions. Second, because properties of a random process are not always apparent from its finite marginals (see the proof of the almost sure continuity of Brownian paths for a salient example). Note also that your first sentence "Your idea is on the right track, but there are more or less important points that (are) missing" might be inspired by a desire to avoid hurting the OP's feelings but is factually false: the OP is not on the right track. $\endgroup$ – Did Jun 12 '17 at 10:19

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