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


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 $\rho$


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?


3 Answers 3


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


  • $\begingroup$ thanks a lot for helping me again, I really appreciate it:) $\endgroup$
    – M00000001
    Commented Dec 19, 2019 at 20:03

I propose a simpler answer: let $W_1(t)$ and $W_2(t)$ be two correlated Brownian motions, with:

$$\mathbb{E}\left[W_1(t)W_2(t) \right] := Cov\left(W_1(t),W_2(t) \right) = \rho_{1,2}t $$


$$\mathbb{E}\left[ -\sigma_1 W_1(t) + \sigma_2 W_2(t) \right] = 0$$


$$\mathbb{E}\left[ \left(-\sigma_1 W_1(t) + \sigma_2 W_2(t) \right)^2 \right] = \mathbb{E}\left[\sigma_2^2 W_2^2(t) - 2 \sigma_1 \sigma_2 W_1(t) W_2(t) + \sigma_1^2 W_1^2(t) \right] = \\ =\sigma_2^2t - 2 t \sigma_1 \sigma_2\rho_{1,2} + \sigma_1^2t = \\ = Var \left(-\sigma_1 W_1(t) + \sigma_2 W_2(t) \right) $$

We know that the two Brownians $W_1(t)$ and $W_2(t)$ are by definition Normally distributed with mean zero and variance $t$. We therefore know that the sum of these two Normally distributed variables will be Normally distributed, and we have just computed the moments above. We can therefore conclude that:

$$-\sigma_1 W_1(t)+\sigma_2 W_2(t) = (in distribution) = \left( \sqrt{\sigma_2^2t - 2 t \sigma_1 \sigma_2\rho_{1,2} + \sigma_1^2t} \right) Z = \\ = (in distribution) = \\ = \left( \sqrt{\sigma_2^2 - 2 \sigma_1 \sigma_2\rho_{1,2} + \sigma_1^2} \right) \sqrt(t)Z = \\ = (indistribution) = \\ = \left( \sqrt{\sigma_2^2 - 2 \sigma_1 \sigma_2\rho_{1,2} + \sigma_1^2} \right) W_3(t) $$

Where $W_3(t)$ is just another Standard Brownian motion.


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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