How a Change in Asset Drift Affects a Function’s Return

Summary

The document asks how the return of a function of an asset changes when the underlying asset’s drift rises by a constant, assuming the asset follows geometric Brownian motion. It applies Itô’s lemma to the function, separating the stochastic term from the drift term. The drift change contributes a term proportional to the asset sensitivity of the function, while the diffusion term determines the function’s volatility.

The accepted response points out that the volatility of the function must be defined explicitly as the underlying volatility multiplied by the asset level and the function’s derivative with respect to the asset. With that definition, the drift increase for the function is proportional to the ratio of its volatility to the underlying volatility. This is a local relationship, dependent on the function’s sensitivity and the model assumptions; the post’s derivation contains a sign inconsistency in its displayed Itô expansion, so the stated conclusion should be checked against the standard lemma.

Key ideas

• Itô’s lemma decomposes the change in a function of an asset into drift and diffusion components.
• A shift in the underlying asset’s drift changes the function’s drift through its derivative with respect to the asset.
• The function’s volatility is the underlying volatility times the asset level and the function’s asset sensitivity.
• Under this definition, the drift shift scales with the ratio of function volatility to underlying volatility.
• The derivation assumes geometric Brownian motion and its displayed expansion should be checked for a sign error.

Tags

# Volatility of a function of an asset


# 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}} $.

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?

## Answer by Kermittfrog (score 1, accepted)

https://quant.stackexchange.com/a/59806

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

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

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