# Volatility of Exchange Option

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?

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

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

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

• thanks a lot for helping me again, I really appreciate it:) – M00000001 Dec 19 '19 at 20:03

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

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

Then:

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

And:

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

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