# Black-Scholes derivation assumption contradiction

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?

• What is $F$ ? also could you provide a link to one of the "many books" you mention ;) - cheers Apr 8, 2014 at 6:54
• You are using rather non-standard notation. However, you can look at $\Pi$ as the value of a delta-hedged portfolio (option plus a short position in $\Delta$ underlying). This is why $\Delta$ is not differentiated. In this context, $\Delta$ is a given quantity, by definition equal as the partial derivative of V wrt F. Apr 8, 2014 at 7:34
• @selfTaught as I see it this is an actual answer to the question at hand - don't be afraid to post it as such ;) Apr 8, 2014 at 8:16
• An undergraduate introduction to financial mathematics, J.Buchanan, Chapter 7. The mathematics of derivatives, R.Navin, Chapter 5. Apr 8, 2014 at 8:32
• Whether given or derived, the fact that $\Delta$ is the partial derivative of the (stochastic) option $V$ means it's stochastic and time dependent. Trying to deduce that it's constant (i.e. $d\Delta=0$) also fails. Apr 8, 2014 at 8:36

$\Pi$ is the value of a delta-hedged portfolio (option plus a short position in Δ underlying). The notation for $\Delta$ is overloaded. Here it represents the number of underlying contracts (f.ex shares) in your delta hedged portfolio, equal to the greek $\Delta$ when the portfolio is created. Therefore in the calculation of $d \Pi$, $\Delta$ (the number of shares in your portfolio) is treated as a constant.
Yes, the greek $\Delta$ evolves as the option approaches maturity and wrt $F$ and you will have to rebalance your portfolio. But this is not contemplated in the infinitesimal $d \Pi$.
• I think that the misunderstanding is in the way things are formulated. The $\Delta$ is indeed constant but in the calculation people forget to tell that the 'simplification' $\Delta=\frac{\partial V}{\partial F}$ is not defining $\Delta$ but is a (first order) constraint on $V$. Later on the constraint is inserted into the derivation and forgotten about. It suggests to me that some sort of Lagrangian or path-integral formulation would be more appropriate, but I haven't looked into it. Apr 8, 2014 at 14:01
• Is there a derivation with non-constant $\Delta$ of Black-Scholes? Does it make semantically sense? Apr 8, 2014 at 14:04