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Leland Morrow

Leland Morrow

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The following is the definition of a random variable:
Let ( Ω , F , P ) be a probability space. A random variable is a real-valued function X on Ω such that for all x R , { ω : X ( ω ) B } F , B B ( R )
I don't understand how this makes sense if our choice of F can be arbitrary.
Let's say we are rolling a biased dice such that each side has a different probability, and the information we know about the system is F = { , { 1 } , { 2 , . . . , 6 } , Ω } -- then for this F, according to the definition, X is not a random variable (because you could construct a (disjoint) Borel set which covers the probabilities of e.g. { 1 , 3 , 5 }, and { 1 , 3 , 5 } is not in F). How does this make sense?

Answer & Explanation



Beginner2022-06-25Added 14 answers

Well sure, then the biased dice roll would not be a random variable in the sense you're used to thinking about it. However, it is quite meaningless to claim that it should be a traditional random variable in the first place, since F , or often: F or Σ, is (intuitively) the sigma-algebra of possible measurable events which we can consider.
With your choice of F , only the events: nothing, 1, {2⋯6} or everything, can be considered in terms of probability. It is meaningless to say that the probability of 3 is different to that of 5, since neither 3 nor 5 are measurable events and as such do not have any sense of probability. You can only ask questions such as: what is the probability X 2, or the probability X = 1. You cannot talk of the probability X 3 even, since { 3 , 4 , 5 , 6 } is not a measurable event and as such has no probability defined for it.
So: yes, the sigma-algebra of events is arbitrary, but so is the probability space itself. You keep the theory general, and make specific definitions and examples when it suits.

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