Download Solved Questions on Non Degenerate Random of Intervals | STA 6467 and more Study notes Probability and Statistics in PDF only on Docsity! Billingsley (3rd ed), Exercise 21.8: (a) Suppose that X and Y have first moments, and prove E[Y ]− E[X] = ∫ ∞ −∞ ( P [X < t ≤ Y ]− P [Y < t ≤ X] ) dt. (b) Let (X,Y ] be a nondegenerate random interval. Show that its expected length is the integral with respect to t of the probability that it covers t. Note: Let (Ω,F , P ) be the probability space under consideration. In doing this prob- lem, you will end up looking at integrals (with respect to P ×Leb. meas.) of functions like IB(ω, t), where, say, B = {(ω, t) : X(ω) < t ≤ Y (ω)}. Of course you need IB to be measurable F ×R1, or equivalently, B ∈ F ×R1. Since B = {(ω, t) : X(ω) < t} ∩ {(ω, t) : t ≤ Y (ω)} = {(ω, t) : t ≤ X(ω)}c ∩ {(ω, t) : t ≤ Y (ω)}, it suffices to show that {(ω, t) : t ≤ X(ω)} ∈ F ×R1 for a general random variable X. Consider the mappings f : Ω× R −→ R f(ω, t) = X(ω) and g : Ω× R −→ R g(ω, t) = t. Note that for any H ∈ R1, f−1(H) = X−1(H)× R ∈ F ×R1 and g−1(H) = Ω×H ∈ F ×R1. Thus f and g are measurable (F × R1)/R1. Now by a theorem proven in class last semester (see below), {(ω, t) : t ≤ X(ω)} = {(ω, t) : g(ω, t) ≤ f(ω, t)} ∈ F ×R1. I certainly do not claim that the argument above is the best one, but it is at least illustrative of the ideas that one might use. I do not really expect you to verify these kind of measurability details in all exercises, but it is easy to make the mistake of assuming that a function or set is measurable when in fact it is not. Thus you should always be aware of measurability questions and be able to construct the necessary arguments if you have to. Note: Recall that for a general measurable space (Ω,F ), if f, g : Ω→ R are measurable F/R1, then {ω : f(ω) < g(ω)} = ⋂ r∈Q {ω : f(ω) < r < g(ω)} = (⋂ r∈Q {ω : f(ω) < r} )⋂(⋂ r∈Q {ω : r < g(ω)} ) ∈ F (Q denotes the rationals). This proof has the advantage that it also works for extended real valued functions (considering f−g is problematic at points where both are∞ or −∞). 1
