Taylor Expansion of Delta-Hedged Option P&L
Summary
This document asks how a Taylor expansion produces a short-time P&L formula for a delta-hedged option. The setup combines the change in option value with financing and underlying-position cash flows, including interest and repo or dividend yield. Choosing the hedge ratio as the option’s delta cancels the first-order underlying-price move.
The response expands the option value in time and spot price, retaining first-order time terms and the second-order spot term, then substitutes that expansion into the P&L expression. The result separates financing and carry contributions from curvature exposure, represented by gamma times the squared relative spot move. The derivation assumes a pure diffusion and a small time step. The displayed answer contains a sign inconsistency: its intermediate line gives a negative gamma term, while its final line gives a positive one, so the stated final formula should be checked against the original P&L convention.
Key ideas
- Delta hedging cancels the first-order P&L contribution from a small spot move.
- A second-order Taylor expansion introduces gamma exposure through the squared spot change.
- Interest, repo, and dividend yield contribute to the time-dependent P&L terms.
- The derivation assumes diffusion dynamics and a short time interval.
- The response's intermediate and final expressions disagree on the gamma term's sign.
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Full text
# taylor expansion of PnL
# taylor expansion of PnL
I have a question about the following derivation in this pdf (sample chapter from Bergomi - Stochastic Volatility Modeling). He derives the PnL for a delta hedged position as
$$PnL = -[P(t+\delta,S+\delta S)-P(t,S)] + rP(t,S)\delta t + \Delta(\delta S - rS \delta t + q S\delta t)$$ where $r$ is the interest rate, $q$ the repo rate including dividend yield and $P(t,S)$ denotes the price of the option with underlying $S$ at time $t$. He chooses $\Delta = \frac{\partial P}{\partial S}$. Then he wants to expand the PnL in powers of $\delta S$ and $\delta t$ just looking at $\delta t, \delta S$ and $\delta S \delta S$ terms. His result is
$$ PnL = -(\frac{dP}{dt}- rP +(r-q)S \frac{dP}{dS})\delta t - \frac{1}{2}S^2\frac{d^2P}{dS^2}(\frac{\delta S}{S})^2$$
I'm not sure how he gets that one. If I expand $P$ in the terms mentioned above I find
$$P(t+\delta,S+\delta S) = P(t,S) + \frac{dP}{dt}\delta t + \frac{dP}{dS}\delta S + \frac{1}{2}*\frac{d^2P}{dS^2}(\delta S)^2$$
Doing this for the three terms of $P$ above I dont get the correct result. So how do you get his formula
## Answer by Quantuple (score 4, accepted)
https://quant.stackexchange.com/a/31069
$\require{cancel}$ $$\text{PnL} = -[P(t+\delta t,S+\delta S)-P(t,S)] + rP(t,S)\delta t + \Delta(\delta S - rS \delta t + q S\delta t)$$ Assuming a pure diffusion, at the order 1 as $\delta t \to 0$ $$P(t+\delta,S+\delta S) = P(t,S) + \frac{\partial P}{\partial t}\delta t + \frac{\partial P}{\partial S}\delta S + \frac{1}{2}\frac{\partial^2P}{\partial S^2}(\delta S)^2$$ Plugging that back in the first equation and using the identity $\Delta = \frac{\partial P}{\partial S}$ gives: \begin{align} \text{PnL} &= -\frac{\partial P}{\partial t}\delta t \cancel{- \frac{\partial P}{\partial S}\delta S} - \frac{1}{2} S^2 \frac{\partial^2P}{\partial S^2}\left(\frac{\delta S}{S}\right)^2 + rP\delta t + \cancel{\Delta\delta S} - (r - q) S \Delta \delta t \\ &= -\left( \frac{\partial P}{\partial t} - rP + (r - q) S \frac{\partial P}{\partial S} \right) \delta t + \frac{1}{2} S^2 \frac{\partial^2P}{\partial S^2}\left(\frac{\delta S}{S}\right)^2 \end{align}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.