# Delta neutrality (derivation)

I'm confused about the math for the delta-neutral portfolio.

Assume we have a short position in a European call option with price $$p(t,S_t)$$ and want to hedge it with the stock with price $$S_t$$. The portfolio value is $$X(t,S_t)=-p(t,S_t)+\Delta\times S_t$$. To make the portfolio delta neutral we require the portfolio to be insensitive to changes in $$S_t$$, thus, we have $$\frac{\partial X}{\partial S}=-\frac{\partial p}{\partial S}+\Delta=0$$ (assuming $$\Delta$$ does not depend on $$S$$). But somehow from here all textbooks give $$\Delta=\frac{\partial p}{\partial S}$$ which, in general, violates the assumption that $$\Delta$$ does not depend on $$S$$.

To see this more clearly, the portfolio $$Y(t,S_t)=-p(t,S_t)+\underbrace{\frac{\partial p}{\partial S}}_{=\Delta}\times S_t$$ is not delta neutral because $$\frac{\partial Y}{\partial S}=-\frac{\partial p}{\partial S}+\frac{\partial^2 p}{\partial S^2}S+\frac{\partial p}{\partial S}\neq 0$$ (unless it is gamma neutral). What is the mistake? What do I miss in the derivation?

Update: I was able to show that if one applies Ito's lemma to portfolio $$Y$$, then $$dY_t = -\left(\frac{\partial p}{\partial t}+\frac{1}{2}\frac{\partial^2 p}{\partial S^2}\sigma^2 S_t^2 \right)dt$$ which is independent of $$dS_t$$. But now my question is: where does the idea of gamma-hedging come from? Again, rigorous way of getting the fact that gamma is needed.

• What? Delta is the partial derivative of option value wrt to change in underlying value. Why are you assuming delta does not depend on the underlying. – Chris Oct 15 '20 at 17:27
• Could you please show that portfolio $Y$ is delta-neutral? It does not make this assumption that $\Delta$ does not depend on the underlying. – Qwerty Oct 15 '20 at 17:30
• Delta is constantly changing. Rather than writing just $\Delta$ we might write $\Delta(t,S_t)$. – rubikscube09 Oct 15 '20 at 17:45
• Good question. It is an intricate topic- the world of infinitesimals! The argument relies on the ability to adjust delta very very rapidly, so the gamma etc are negligible.This is generally true (not just specific to Black Scholes): if you go infinitesimals, even non linear functions look linear. – Magic is in the chain Oct 15 '20 at 17:57
• @Magicisinthechain It is a bit dangerous to claim that any sufficiently zoomed in function looks linear. This is only true for differentiable functions (and is basically the definition of differentiability), and is not true for say - Brownian motion, which drives much of the Black-Scholes theory. – rubikscube09 Oct 15 '20 at 18:43

The Black Scholes hedge portfolio is given by: $$\Pi_t = \frac{\partial V}{\partial S}(t,S_t)S_t + \left[1 - \frac{\partial V}{\partial S}(t,S_t)\right]B_t$$ where $$B_t$$ is the risk-free asset. Differentiating with respect to $$S$$ as usual, we have that the portfolio delta is: $$\frac{\partial^2 V}{\partial^2 S}(t,S_t) + \frac{\partial V}{\partial S_t}(t,S_t) - \frac{\partial^2 V}{\partial^2S} (t,S_t) = \frac{\partial V}{\partial S}(t,S_t)$$ meaning that this combined (opposite signs) with one unit of $$V$$ is a locally risk-free portfolio.
There are most likely some rigor that is being missed here in terms of taking derivatives of nowhere differentiable functions like $$S_t$$ - any posters can and should feel free to fill in the blanks.
• You are on the right track and a rigorous argument is given here. This is related to the fact that "simple derivations" of the Black-Scholes PDE always write for the delta-hedge portfolio $\Pi_t = -V_t + \Delta_t S_t$ the step $d\Pi_t = -dV_t + \Delta_t dS_t$. People always struggle with why it is not $d\Pi_t = -dV_t + \Delta_t dS_t + S_t d\Delta_t$. They are missing the self-financing aspect of the replicating portfolio. – RRL Oct 15 '20 at 18:51