1

In this paper paper page 16-19 by Davis and this discussion derivation of the hedging error in a black scholes setup, the derivation of the delta hedging error in the Black Scholes model is discussed.

The result is strong and interesting but when I try go through the proof I don't quite understand several steps of it. He is using Ito a couple of times but doesn't really explain how and his definitions of processes could benefit from short explanations.

Can anyone in here provide a more thorough proof?

Sanjay
  • 1,657
  • 1
  • 12
  • 29

1 Answers1

5

The paper could be clearer indeed.

It is a slightly confusing topic, but the important step here is to understand the consequence of the derivative $C$ in the portfolio being priced at the assumed vol $\sigma$. This implies (by Black-Scholes) that it will by definition be true that:

$\theta_t + \frac{\partial{C}}{\partial{S}}rS_t+ \frac{1}{2}\frac{\partial^2{C}}{\partial{S^2}}\sigma^2S_t^2 = rC_t$ (Eq. 1)

That is, the theta of $C$ is linked to the known quantities $\sigma, S, r$ and to the two Greeks delta and gamma. (This is of course the starting point of the Black-Scholes formula).

Now, it is also true that $C_t$ dynamics in your portfolio must (by Ito) depend on the true dynamics of $S_t$ and in particular we have:

$dC_t = \theta_tdt+ \frac{\partial{C}}{\partial{S}}dS_t+ \frac{1}{2}\frac{\partial^2{C}}{\partial{S^2}}dS_t^2$ (Eq. 2)

This is where the important bit happens: you can replace $\theta_t$ in Eq. 2 by its value derived from Eq. 1. What have we got ?

$dC_t = rC_tdt + \frac{\partial{C}}{\partial{S}}(dS_t-rS_tdt)+ \frac{1}{2}\frac{\partial^2{C}}{\partial{S^2}}(dS_t^2-\sigma^2S^2dt)$ (Eq. 3)

And of course if $S_t$ follows a GBM with volatility $\beta$, the third term turns out as:

$ \frac{1}{2}\frac{\partial^2{C}}{\partial{S^2}}(\beta^2-\sigma^2)S^2dt$

This is where the hedging error comes from in a delta-hedged portfolio.

You should work this out starting with $\Pi_t = C_t - \Delta_t S_t$ to play with the mechanics for yourself but a crucial point to note is that all derivatives (theta, delta, gamma) in Eq. 1 and in Eq. 2 depend on the assumed (or pricing or implied) vol $\sigma$.

This is the reason why you can replace $\theta_t$ in Eq. 2. Once the call is in your portfolio, it must be valued at some $\sigma$, and this (and not $\beta$) determines its theta for hedging purposes. $\theta_t \equiv \theta_t(\sigma)$.

This is also why the hedging error is a function of your valuation gamma. $\frac{\partial^2{C}}{\partial{S^2}} \equiv \frac{\partial^2{C}}{\partial{S^2}}(\sigma)$

(Note they are not just a function of $\sigma$ but hopefully the point is clear).

Alex C
  • 9,372
  • 1
  • 21
  • 34
Ivan
  • 1,386
  • 8
  • 10
  • thanks, it makes sense. It's funny; TWO single sentences in your post made me understand the whole thing and suddenly everything was very clear :D I have now carried out all the derivation and it checks out. I think the interpretation of the model is a bit confusing as well :) thx – Sanjay Mar 26 '18 at 13:16
  • ... the way you define $\Pi_t$ means yo have a long Position in the Call and short position in the underlying, right? The opposite of what Davis does. Furthermore, feel free to comment on how/why why Davis define the value of the hedging portfolio as $X_0=C_0$ and $$dX_t=\Delta dS_t + (X_t- \Delta* S_t)rdt$$. This is still unclear to me. – Sanjay Mar 26 '18 at 14:15
  • 1
    That’s right, long call and short shares in my example. If you go back to sections 2.2-2.3 in the paper, he does a decent job of explaining what a hedging (or replicating) portfolio should be and do. Formally, because $X_T$ is supposed to replicate $C_T$, then $X_0$ must be worth $C_0$ otherwise an arbitrage opportunity exists. The $dX_t$ equation essentially states that since the hedging portfolio is composed of $\Delta S_t$ and a short bond it must hence evolve as $\Delta dS$ plus a cost of funding bit (the second term). – Ivan Mar 26 '18 at 16:17