Random variable, X, has: support S, image I, intersection of support and image U and image closure T. Recall that S is contained in T since T is closed with probability 1. The question is: Can all four sets S-U, U, I-U & T-(S union I) be non-empty?
How finely can a random variable's support and image be used to partition the closure of the image?
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probability-theory
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0No restrictions on X in mind u$p$front. Taking X formally to be "a function with domain a probability space with image in the reals such that pre-images of Borel sets are events in the probability space". – 2012-09-17
1 Answers
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Let $q_1, q_2, q_3,\ldots$ be an enumeration of the positive rationals and $p_1,p_2,\ldots$ an enumeration of the negative rationals. Let $r_1, r_2, r_3,\ldots$ be an enumeration of the rationals in $[0,1]$. Let $(\Omega,\Sigma,\mu)$ be the unit interval with Borel $\sigma$-algebra and uniform (Lebesgue) measure.
Define the random variable $X:\Omega\to\mathbb{R}$ in the following way: Map irrational numbers in the interval $[0,1/2)$ to $q_1$, irrational numbers in $[1/2,3/4)$ to $q_2$ and so on. Map the rational number $r_n$ to $p_n$. Then we have:
$S=\mathbb{R}_+$
$I=\mathbb{Q}$
$U=\mathbb{Q}_+$
$T=\mathbb{R}$
And all the sets listed in the question are nonempty.