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Measure, Integration & Real Analysis, 2021a

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Chapter 12 Probability <strong>Measure</strong>s 393<br />

Your understanding of the proof of the next result should be enhanced by Exercise<br />

13, which asserts that if the function H : R → (0, 1) appearing in the next result is<br />

continuous and injective, then the random variable X : (0, 1) → R in the proof is the<br />

inverse function of H.<br />

12.29 characterization of distribution functions<br />

Suppose H : R → [0, 1] is a function. Then there exists a probability space<br />

(Ω, F, P) and a random variable X on (Ω, F) such that H = ˜X if and only if<br />

the following conditions are all satisfied:<br />

(a) s < t ⇒ H(s) ≤ H(t) (in other words, H is an increasing function);<br />

(b)<br />

(c)<br />

lim H(t) =0;<br />

t→−∞<br />

lim H(t) =1;<br />

t→∞<br />

(d) lim<br />

t↓s<br />

H(t) =H(s) for every s ∈ R (in other words, H is right continuous).<br />

Proof First suppose H = ˜X for some probability space (Ω, F, P) and some random<br />

variable X on (Ω, F). Then (a) holds because s < t implies (−∞, s] ⊂ (−∞, t].<br />

Also, (b) and (d) follow from 2.60. Furthermore, (c) follows from 2.59, completing<br />

the proof in this direction.<br />

To prove the other direction, now suppose that H satisfies (a) through (d). Let<br />

Ω =(0, 1), let F be the collection of Borel subsets of the interval (0, 1), and let P<br />

be Lebesgue measure on F. Define a random variable X by<br />

12.30 X(ω) =sup{t ∈ R : H(t) < ω}<br />

for ω ∈ (0, 1). Clearly X is an increasing function and thus is measurable (in other<br />

words, X is indeed a random variable).<br />

Suppose s ∈ R. Ifω ∈ ( 0, H(s) ] , then<br />

X(ω) ≤ X ( H(s) ) = sup{t ∈ R : H(t) < H(s)} ≤s,<br />

where the first inequality holds because X is an increasing function and the last<br />

inequality holds because H is an increasing function. Hence<br />

( ]<br />

12.31<br />

0, H(s) ⊂{X ≤ s}.<br />

If ω ∈ (0, 1) and X(ω) ≤ s, then H(t) ≥ ω for all t > s (by 12.30). Thus<br />

H(s) =lim H(t) ≥ ω,<br />

t↓s<br />

where the equality above comes from (d). Rewriting the inequality above, we have<br />

ω ∈ ( 0, H(s) ] . Thus we have shown that {X ≤ s} ⊂ ( 0, H(s) ] , which when<br />

combined with 12.31 shows that {X ≤ s} = ( 0, H(s) ] . Hence<br />

as desired.<br />

˜X(s) =P(X ≤ s) =P( (0, H(s)<br />

] ) = H(s),<br />

<strong>Measure</strong>, <strong>Integration</strong> & <strong>Real</strong> <strong>Analysis</strong>, by Sheldon Axler

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