Suppose that $ G $ is a function of the underlying asset $ S $, which follows a geometric Brownian motion. Suppose that $ \sigma_{S} $ and $ \sigma_{G} $ are the volatilities of $ S $ and $ G $, respectively. Show that if $ \mu $, the rate of return of $ S $, increases by a constant $ \lambda $, then the rate of return of $ G $ increases by $\lambda \frac{\sigma_{G}}{\sigma_{S}} $.

This is my answer:

Since $S$ follows a geometric brownian motion:

$$dS_{t}=\mu_{S} S_{t}dt + \sigma_{S} S_{t}dW_{t}$$

Using Ito for the function $G(s)$ we get:

$$dG(S_{t})=G^{'}(S_{t})dS_{t} + \frac{1}{2}G^{''}(S_{t})S_{t}^{2}\sigma^{2}_{S}dt$$

$$=G^{'}(S_{t})(\sigma_{S} S_{t}dW_{t} + \mu_{S} S_{t}dt) - \frac{1}{2}G^{''}(S_{t})S_{t}^{2}\sigma_{S}^{2}dt$$

$$=G^{'}(S_{t})\sigma_{S} S_{t}dW_{t} + (\mu_{S} S_{t}G^{'}(S_{t}) - \frac{1}{2}G^{''}(S_{t})S_{t}^{2}\sigma_{S}^{2})dt$$

When $\mu_{S}$ increases by a constant $\lambda$

$$(\mu_{S}+\lambda) S_{t}G^{'}(S_{t}) - \frac{1}{2}G^{''}(S_{t})S_{t}^{2}\sigma_{S}^{2}=\mu_{S} S_{t}G^{'}(S_{t}) + \lambda S_{t}G^{'}(S_{t}) - \frac{1}{2}G^{''}(S_{t})S_{t}^{2}\sigma_{S}^{2}$$

\Rightarrow $\mu_{G}$ increases by $\lambda S_{t}G^{'}(S_{t})=\lambda \frac{\sigma_{G}}{\sigma_{S}}$

Is this ok?


1 Answer 1


What you've written looks comprehensible. What is missing is the definition of $\sigma_G \equiv \sigma_G(S_t,t)$, i.e.

Let $\sigma_G\equiv \sigma_S S \frac{\partial G(S_t,t)}{\partial S_t}$

  • $\begingroup$ The problem doesnt say that. Can I assume that is $\sigma_{G}$? $\endgroup$ Dec 7, 2020 at 16:04
  • $\begingroup$ Is there another way of answering the question? $\endgroup$ Dec 7, 2020 at 16:08
  • $\begingroup$ I think you can say that, yes. $\endgroup$ Dec 7, 2020 at 16:52

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