Why Delta Hedging Does Not Treat Delta as Constant in Black–Scholes

Summary

The document raises a question about the delta-hedging step in the Black–Scholes derivation. It observes that option delta depends on both the underlying price and time, and asks why the change in a portfolio holding a delta-sized position in the asset appears to cancel the option’s price exposure without applying the full product rule.

This is a conceptual question about stochastic calculus and the construction of a locally riskless portfolio, rather than a worked explanation or empirical result. It usefully identifies that delta is a changing hedge ratio, but the document itself does not resolve how the hedge is adjusted or derive the relevant differential. Readers should treat it as a prompt to examine the assumptions and infinitesimal rebalancing behind the standard derivation, not as a complete account of the proof.

Key ideas

• The Black–Scholes hedge uses an asset position sized by the option’s delta.
• Delta depends on the underlying price and time, so it changes as the process evolves.
• The document asks how the hedge’s change is handled in the riskless-portfolio argument.
• It poses the issue but does not provide a derivation or resolution.

Tags

# Black Scholes derivation: Why treat Delta as a constant?


# Black Scholes derivation: Why treat Delta as a constant?












In the derivation of the Black-Scholes equation, it is argued (e.g. in the original paper and in Hull) that $$dV(S_t, t)=(…)dt + \frac{\partial V}{\partial S} dS_t,$$ where $V(S_t, t)$ is the value at time $t$ of an option on an asset with price process $S$. Thus, adding an amount of $-\frac{\partial V}{\partial S}$ assets to the option would make the resulting portfolio $P$ riskless since $$dP(S_t, t) \\ =(…)dt + \frac{\partial V}{\partial S} dS_t - d(\frac{\partial V}{\partial S} S_t) \\ =(…)dt + \frac{\partial V}{\partial S} dS_t - \frac{\partial V}{\partial S} dS_t \\ = (…)dt. $$

Why is the second equation true? We are treating $\frac{\partial V}{\partial S}$ like a constant when pulling it out of the derivative while it is actually a function of $t$ and $S$. Wouldn’t we have to use the Ito Product Rule here? (and I wouldn’t know how to do that since I don’t know the dynamics of $\frac{\partial V}{\partial S}$)

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.