# Volatility of a function of an asset

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 $$\lambda \frac{\sigma_{G}}{\sigma_{S}}$$.

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?

What you've written looks comprehensible. What is missing is the definition of $$\sigma_G \equiv \sigma_G(S_t,t)$$, i.e.
Let $$\sigma_G\equiv \sigma_S S \frac{\partial G(S_t,t)}{\partial S_t}$$
• The problem doesnt say that. Can I assume that is $\sigma_{G}$? Commented Dec 7, 2020 at 16:04