when is the expectation of an integral equal to the integral of the expectation
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When you computed the mean of the integral you changed the order of certain integrations. Formally, you did the following
E[X(t)]=E[∫0tT(s)ds]=∫0tE[T(s)]ds.
First
E[X(t)]=∫ΩXω(t)P(dω)=∫Ω∫0tT0(ω)(−1)Nω(s) dsP(dω)
At this point, it turns out that before we could talk about the expectation above we have to define the probability space [Ω,A,P] over which (−1)Nω(s) is a stochastic process. (−1)Nω(s) is neither continuous in s nor it is bounded or monotonous. So, defining such a stochastic process must be a hard task. Also, it is a question that (−1)Nω(s) can be integrated being an everywhere discontinuous function.
Assume for now that we are done with all this.
Second
Assuming that we have defined our stochastic process and our function can be integrated for any ω, so we can talk about the mean.
E[X(t)]=∫ΩXωP(dω)=∫Ω∫0tT0(ω)(−1)Nω(s) dsP(dω).
according to Fubini's theorem these integrals can be switched if the following integral exists
∬Ω×[0,t)∣T0(ω)(−1)Nω(s)∣d(P×λ)=t.
It does. So, switching the integrals seems to be all right. But for me the definition of this stochastic process and the integrability of the function are serious problems.
E[X(t)]=E[∫0tT(s)ds]=∫0tE[T(s)]ds.
First
E[X(t)]=∫ΩXω(t)P(dω)=∫Ω∫0tT0(ω)(−1)Nω(s) dsP(dω)
At this point, it turns out that before we could talk about the expectation above we have to define the probability space [Ω,A,P] over which (−1)Nω(s) is a stochastic process. (−1)Nω(s) is neither continuous in s nor it is bounded or monotonous. So, defining such a stochastic process must be a hard task. Also, it is a question that (−1)Nω(s) can be integrated being an everywhere discontinuous function.
Assume for now that we are done with all this.
Second
Assuming that we have defined our stochastic process and our function can be integrated for any ω, so we can talk about the mean.
E[X(t)]=∫ΩXωP(dω)=∫Ω∫0tT0(ω)(−1)Nω(s) dsP(dω).
according to Fubini's theorem these integrals can be switched if the following integral exists
∬Ω×[0,t)∣T0(ω)(−1)Nω(s)∣d(P×λ)=t.
It does. So, switching the integrals seems to be all right. But for me the definition of this stochastic process and the integrability of the function are serious problems.
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