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I got a question and its partial solution, and have some doubts about the volatility of its geometric Brownian motion process:

Question:

How would you price an exchange call option that pays $max(S_{T,1}-S_{T,2},0)$ at maturity. Assume that $S_1$ and $S_2$ are non-dividend paying stocks and both follow geometric Brownian motions with correlation $\rho$

Solution:

The payoff of the exchange option depends on both $S_{T,1}, S_{T,2}$, so we need two geometric Brownian motions:

$dS_1 = \mu_1S_1dt+\sigma_1S_1dW_{t,1}$ $dS_2 = \mu_2S_2dt+\sigma_2S_2dW_{t,2}$

Yet if we use $S_1$ as the numeraire, we can convert the problem to just one geometric Brownian motion. The final payoff is $max(S_{T,1}-S_{T,2},0)=S_{T,1}max(S_{T,2}/S_{T,1}-1,0)$. When $S_{T,1} and S_{T,2}$ are geometric Brownian motions, $f=S_{T,2}/S_{T,1}$ is a geometric Brownian motion as well. More rigorously, we can the Ito's lemma to $f=S_{T,2}/S_{T,1}$:

$df = \frac{\partial f}{\partial S_1}dS_1+\frac{\partial f}{\partial S_2}dS_2+0.5*\frac{\partial^2 f}{\partial S_1^2}dS_1^2+0.5*\frac{\partial^2 f}{\partial S_2^2}dS_2^2+\frac{\partial^2 f}{\partial S_1\partial S_2}dS_1dS_2=(\mu_2-\mu_1+\sigma_1^2-\rho\sigma_1\sigma_2)fdt - \sigma_1fdW_{t,1}+\sigma_2fdW_{t,2} = (\mu_2-\mu_1+\sigma_1^2-\rho\sigma_1\sigma_2)fdt+\sqrt{\sigma_1^2-2\rho\sigma_1\sigma_2+\sigma_1^2}fdW_{t,3}$

So here is my doubt, why $- \sigma_1fdW_{t,1}+\sigma_2fdW_{t,2} = \sqrt{\sigma_1^2-2\rho\sigma_1\sigma_2+\sigma_1^2}fdW_{t,3}$ holds?

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Define the process $$X_t=\frac{1}{\sqrt{\sigma_2^2+\sigma_1^2-2\rho\sigma_1\sigma_2}}\left(\sigma_2W_{t,2}-\sigma_1W_{t,1}\right)$$ It is a martingale because it is a linear combinations of martingales.

Calculate $d<X_t,X_t>$ $$d<X_t,X_t>=\frac{1}{\sigma_2^2+\sigma_1^2-2\rho\sigma_1\sigma_2}\left(\sigma_2^2+\sigma_1^2-2\rho\sigma_1\sigma_2\right)dt=dt$$

By Levy's Characterization of Brownian Motion, $X_t$ is a Brownian motion, and we rename it $$X_t=W_{t,3}$$

Finally, differentiate the first question

$$\sqrt{\sigma_2^2+\sigma_1^2-2\rho\sigma_1\sigma_2}dW_{t,3}=\left(\sigma_2dW_{t,2}-\sigma_1dW_{t,1}\right)$$

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  • $\begingroup$ thanks a lot for helping me again, I really appreciate it:) $\endgroup$ – M00000001 Dec 19 '19 at 20:03
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Thanks a lot for the above great answer:). Here I also added the link for Levy's Characterization of Brownian Motion: http://individual.utoronto.ca/normand/Documents/MATH5501/Project-3/Levy_characterization_of_Brownian_motion.pdf

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