Dumb question: is risk-neutral pricing taking conditional expectation?
In trying to recall intuition for risk-neutral pricing, I think I read that we should price derivatives risk-neutrally because the risk is already incorporated in the stock or something. I also think I remember NNT saying something about how certain information is irrelevant in the expected price of oil if the information is public.
This made me think of risk-neutral pricing in terms of conditional expectation.
Simple I guess but it wasn't discussed in classes since conditional expectation and Radon-Nikodym was taught after one period model.
From what I recall of the one period model:
(Ω,F,P)=((u,d),2Ω,real world))
where t=0,1,u>1+R>d>0.
It can be shown that
Π(X,0)=11+REQ[X]=11+R(quX(u)+qdX(d))
where qu,qd are the risk-neutral probabilities under Q, equivalent to P
Also, I think σ(S1)=σ(X)={∅,Ω,{u},{d}}
Dumb question rephrased:
∃Z∈L1(Ω,F,P) s.t. EQ[X]=EP[X|Z]?
Well, the left hand side is a constant while the right hand side a random variable so I'm not sure that that would make sense
How about
∃Z∈L1(Ω,F,P) s.t. EQ[X]=EQ[EP[X|Z]]?
For (2),
One thing I did:
EQ[X] is constant and thus Z−measurable ∀ Z∈L1(Ω,F,P)
∫zEQ[X]dP=∫zXdP ∀ z ∈ σ(Z)
⟺E[EQ[X]1z]=E[X1z] ∀ z ∈ σ(Z)
There doesn't seem to be such a Z.
- Another thing I did: Well, I did think of Radon-Nikodym (duh)
EQ[X]=EP[XdQdP]
I guess Z=dQdP otherwise not sure how that's relevant but I guess since Q(z)=∫zdQdPdP ∀z∈σ(Z)⊆2Ω
dQdP is a version of E[dQdP|Z]
Gee how informative. Well, I think σ(Z) can be only either {∅,Ω}, in which case the Z is any (almost surely?) constant random variable or 2Ω=σ(X)=σ(S1), in which case qu=1u which would make sense iff qu is degenerate, which I guess violates equivalence assumption.
For (3),
I guess Z=S1? I'm not sure what that says. I was kinda expecting (lol) that real world probabilities E[1A] and risk-neutral probabilities E[1A|B]=E[1A1B]E[1B] would be like, respectively, prior P(A) and posterior P(A|B) probabilities.
Edit: dQdP=qupu1u+qdpd1d
Based on Section 4.5 of Etheridge's A Course in Financial Calculus, I guess
dQdP=(qupu)u(qdpd)1−u
This avoids indicator functions in favour of exponents as in the binomial theorem, binomial model or binomial distribution.
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