Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In many books and derivations of the Black-Scholes PDE one sees that

$$\Pi=V-\Delta F \Rightarrow d\Pi=dV-\Delta dF$$

which implicitly assumes that $d\Delta=0$. Somewhere down the road one then deduces that

$$\Delta=\frac{\partial V}{\partial F}$$

to simplify the equation. Doesn't this contradict the initial assumption that $d\Delta=0$? If one performs a full differentiation

$$\Pi=V-\Delta F \Rightarrow d\Pi=dV-\Delta dF - F d\Delta$$

the rest of the story goes wrong. Isn't it true that $\Delta = \Delta(t, S)$, i.e. is depending on time and the underlying stochastic process and hence has to be differentiated?

share|improve this question
The delta hedge ensures that the portfolio is self-financing, i.e. there is not change in portfolio value. There is an explanation on the Wikipedia page: en.wikipedia.org/wiki/Black-Scholes_equation –  user13247 May 21 at 11:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.