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?