Deriving Delta-Hedge Error from Black–Scholes Pricing and True Volatility
Summary
The note explains how delta-hedging error arises when an option is valued using an assumed volatility that differs from the underlying asset's true volatility. The option's Black–Scholes pricing equation links its theta to its price, delta, gamma, interest rate, and assumed volatility. Applying Itô's formula to the option value under the actual asset dynamics and substituting the pricing equation for theta isolates the terms that drive the hedged position's gains and losses.
Under geometric Brownian motion with true volatility different from the pricing volatility, the gamma term contains the variance difference multiplied by the option's gamma and the elapsed time. This identifies volatility mismatch as the source of the local hedging error, with the relevant gamma itself determined by the volatility used to value the option. The derivation assumes the Black–Scholes framework and continuous-time dynamics; it does not cover discrete rebalancing, transaction costs, or jumps.
Key ideas
- The option's theta is constrained by its Black–Scholes pricing volatility and other pricing inputs.
- Applying Itô's formula under the true asset dynamics exposes deviations from the pricing model.
- A difference between true and assumed variance generates a gamma-weighted term in the hedging error.
- The hedge error uses Greeks calculated from the option's valuation volatility.
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# Deriving Delta Hedge error in the B-S setup (part 2)
# Deriving Delta Hedge error in the B-S setup (part 2)
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?
## Answer by Ivan (score 5, accepted)
https://quant.stackexchange.com/a/38992
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).Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.