Friday, March 4, 2016

integration - Expectation of mathbbE(Xk+1)


I have difficulties with an old exam problem :



Let X be a positive random variable defined on a probability space (Ω,F,P). Show that
0tkP(Xt)dt=0Ωtk{X(ω)t}dtdP(ω) Infer from this the integral expression of E(Xk+1) (where E is the expectation)



We have Fubini theorem, which we can apply to a B(R)F-measurable function because the Lebesgue measure is σ-finite and P is also σ-finite because it is a probability. I think we can write P(Xt) as {X(w)t}dP(ω) but I don't know how to proceed next. Especially I don't see how to introduce the Ω.


Edit


From the comments, there must be an error in the description of the exam problem. It should have been the following :




Let X be a positive random variable defined on a probability space (Ω,F,P). Show that
0tkP(Xt)dt=0Ωtk1{X(ω)t}dtdP(ω) Where 1{X(ω)t} is the characteristic function of {X(ω)t}


Infer from this the integral expression of E(Xk+1) (where E is the expectation)



Answer



By Fubini's theorem, we have


0tkP(Xt)dt=0tkE[1{ω:X(ω)t}]dt=0tkΩ1{ω:X(ω)t}dPdt=ΩX(ω)0tkdtdP=Ω1k+1Xk+1(ω)dP(ω)=1k+1E[Xk+1].


Hence E[Xk+1]=(k+1)0tkP(Xt)dt.


The crucial part here is that 1{ω:X(ω)t}(ω)=1{t:tX(ω)}(t).


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