Why the Black–Scholes Delta Position Has More Than One Differential Term

Summary

The document examines a step in the Black–Scholes hedging argument: whether the change in the value of a delta position can be written as delta multiplied by the change in the underlying price. The response says this equality does not generally hold for the Black–Scholes call delta.

For a call, delta is expressed as the normal cumulative distribution evaluated at d1. As both the underlying price and delta vary, applying stochastic calculus to their product introduces a change in delta and a covariation term, in addition to the delta-weighted change in the underlying. The displayed calculation therefore contradicts the simplified differential in the question. This is a focused correction about stochastic differentials, not a complete derivation of the Black–Scholes equation or a treatment of the assumptions needed for a self-financing hedge. The excerpt offers no numerical example or further discussion of how the omitted terms enter a full hedging derivation.

Key ideas

• The differential of delta times the underlying is not generally equal to delta times the underlying’s differential.
• In the Black–Scholes call example, delta changes with the underlying price and time.
• The product differential includes a term for the change in delta and a covariation term.
• The correction addresses one step in a derivation rather than presenting the full hedging argument.

Tags

# Assumption in black scholes solution


# Assumption in black scholes solution












Under the usual notations,

In most textbooks on Quantative Finance, for deriving the Black-Scholes solution I find that authors, while setting up the riskless portfolio, assume that,

$$\text{d} (\frac{\partial V}{\partial S} S_t) = \frac{\partial V}{\partial S} \text{d} S_t $$

At least can we prove this post facto, as in, does this equation hold true for famous Black Scholes equation

The same issue is also pointed out here.

## Answer by Gordon (score 2)

https://quant.stackexchange.com/a/35712

This is not true. Note that $\frac{\partial C}{\partial S_t} = N(d_1)$. Then \begin{align*} d\left(\frac{\partial C}{\partial S_t}S_t\right) &= \underbrace{S_t dN(d_1) + d\langle N(d_1), S\rangle_t} + N(d_1) dS_t\\ &\ne N(d_1) S_t. \end{align*} That is, \begin{align*} d\left(\frac{\partial C}{\partial S_t}S_t\right)\ne \frac{\partial C}{\partial S_t}dS_t. \end{align*}

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