# calculate all the equivalent martingale measure

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. ${\cal P} = \{Q: Q \sim P\}\neq \emptyset.$

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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– learningmath Feb 5 '12 at 22:16

## 2 Answers

I'm afraid your market is complete here at least if all assets can only involve $S$ and $r$, so here only 1 Equivalent Martingale measure exists.

You can see this by seeing that you can express the dynamic of $S$ with only one Brownian motion as you have : $dS_t/S_t=\mu dt + \sqrt{\sigma_1^2+\sigma_2^2}dB_t$

where $B_t=\frac{\sigma_1 W^1_t+\sigma_2 W^2_t}{\sqrt{\sigma_1^2+\sigma_2^2}}$ is Brownian motion.

So you are back in a Black Scholes framework which is a complete market framework.

The situation would have be different if you had supposed that there exist assets that would involve only $W_1$ and/or $W_2$ separatly in there dynamics.

Best Regards

Ps : I agree with learningmath that this question would have been suitable for quantstackexchange forum.

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It is not complete since the martingale measure is NOT unique. – Jun Feb 7 '12 at 1:35
Jun : Yes it is, as long as you don't have assets driven by W^1 or W2 serparately as I have shown in my post. – TheBridge Feb 7 '12 at 7:54

The question is ambiguously worded. If the "martingale" in "equivalent martingale measure" is understood relative to the filtration generated by the stock price process $S$, then there is but one equivalent martingale measure (EMM), as noted by @TheBridge. If the filtration $({\mathcal F}_t)_{t\ge 0}$ is taken to be that generated by $W_1$ and $W_2$, then there are many EMMs. Suppose (for simplicity) that $\mu=0$ and $\sigma_1=\sigma_2=1$. Then for each real $\alpha$ the measure $Q_\alpha$ defined on ${\mathcal F}_t$ by $$dQ_\alpha/dP := \exp(\alpha W_1(t)-\alpha W_2(t)- \alpha^2t)$$ is an equivalent martingale measure. And there are more: If $H$ is a bounded $({\mathcal F}_t)$-predictable process then, on ${\mathcal F}_t$, $$dQ/dP :=\exp\left(\int_0^t H_s dW_1(s) - \int_0^t H_s dW_2(s) -\int_0^t H_s^2 ds\right)$$ defines an EMM.

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Yes. The filtration is generated by $W_1, W_2.$ Thanks. – Jun Feb 9 '12 at 3:22
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