MARTINGALE REPRESENTATION THEOREM PROOF

[dushyant kochar]

Now if Yt is also a martingale with respect to Q (ie based on the same probabilities q and 1 - q ) then, first, Yt must also follow a binomial model with:

Y +u(t, Y ) with probability q​

Y t+1 =

Yt +d(t,Yt) with probability 1-q​

since both Xt and Yt must be measurable with respect to the same sigma- algebra Ft for all t .

can anyone explain how come Xt and Yt both are measurable by the same filtration? and why they have the same probability "q" for ups and the remaining for down?

 

[Mark Mitchell]

It is part of the theorem that Xt and Yt are both martingales with respect to the same probability measure. In the context of this illustration of the theorem in discrete time (ie the binomial model) this means we have only the probabilities q and 1-q. There are no other choices. When specifying Yt then, q is the probability of going in one direction and 1-q is the probability of going in the other. We choose q as the probability of going up and 1-q as the probability of going down, but if we swap these, the definitions of u~ and d~ will swap, so the choice is arbitrary.

That the processes are measurable with respect to the same sigma algebra again is part of the theorem, I think. The theorem (lazily?) states that Yt is another martingale with respect to the measure P, but doesn't go on to say that this means E(Yt|Fs) = Ys under P, ie the filtrations are the same. Certainly, where the martingale representation theorem is used later in the chapter to prove the general risk-neutral pricing formula, the processes Dt and Et are both martingales based on the sigma algebra Ft generated by the share price process St.

 

[dushyant kochar]

Thanks sir .your explanation was very helpful,it cleared all my doubts.thanks.