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Under the assumption of no arbitrage without vanish risk, in an incomplete market $(\Omega,{\cal F}, P)$, the set of equivalent martingale measure is NOT empty, i.e. $\mathcal{P} = \{Q: Q \sim P\}\neq \emptyset$

My question is: in the following simplified market with one stock which is driving by two independent Brownian Motions and one bond, i.e.

$$dS_t = S_t(\mu dt + \sigma_1 dW_1(t) + \sigma_2 dW_2(t))$$ $$dB_t = rB_tdt, \mbox{ } B_0 = 1$$

How to calculate all the equivalent martingale ${\cal P}.$ We suppose that $\mu,\sigma_1,\sigma_2, r$ are constants.

One approach in my mind is using another stock to complete the market, i.e. we suppose there is another stock $\tilde{S}$ with parameters, $\tilde{\mu}, \tilde{\sigma_1},\tilde{ \sigma_2}$ such that $$d\tilde{S_t} = \tilde{S_t}(\tilde{\mu} dt + \tilde{\sigma_1} dW_1(t) + \tilde{\sigma_2} dW_2(t)).$$

Then, following the classic method, we could get the equivalent martingale measures described by parameters, $\mu,\sigma_1,\sigma_2, r, \tilde{\mu}, \tilde{\sigma_1},\tilde{ \sigma_2}.$

But, how could I know the equivalent martingale measure obtained by above approach are the set of all the equivalent martingale measures in this financial market?

Any suggestion, reference books, or papers are welcome. Thanks.

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The question might be suitable for the sister stackexchange site devoted to Quantitative Finance: – Andrey Rekalo Feb 5 '12 at 21:21

One possible approach is to use the fact that the density process $\left. Z_t =\frac{d\mathcal{Q}}{d\mathcal{P}} \right\vert_{\mathcal{F}_t}$ for every equivalent local martingale measure $\mathcal{Q}$ is a true martingale in the Brownian filtration and you can characterize it via the martingale representation theorem. As a guiding example you may look e.g. on Appendix A of R. Frey's paper "Derivative Asset Analysis in Models with Level Dependent and Stochastic Volatility", CWI Quaterly 10, no 1 (special issue on the Mathematics of Finance) p 1-34 which he links on his webpage:

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Thanks. It seems that the approach works for my problem. – Jun Feb 7 '12 at 16:17

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