Deriving the Gamma Approximation for Delta-Hedged Option PnL
Summary
The document derives the local profit approximation for a long option position hedged with the underlying. After buying an option and offsetting its initial delta, a small spot move creates portfolio PnL equal to the option value change less the gain or loss on the hedge. A second-order Taylor expansion cancels the first-order price term, leaving approximately one half of gamma times the squared spot change, with higher-order terms omitted.
The answer also describes plots comparing the option, underlying hedge, combined portfolio, and gamma approximation under a stated Black-Scholes setup. The relationship explains why a long gamma position benefits from sufficiently large moves in either direction before other effects are considered. It is an instantaneous, local approximation: for finite moves, higher-order terms matter, and the document does not account for rehedging costs, time decay, volatility changes, or financing. The example’s later rehedging trade is not itself the source of the initial move’s gamma PnL.
Key ideas
- Delta hedging removes the option’s first-order exposure to a small underlying price move.
- A second-order expansion leaves gamma as the leading contribution to hedged PnL.
- The gamma approximation is local and becomes less accurate for larger price moves.
- Rehedging updates exposure after the move and may involve costs outside the approximation.
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# Proof of gamma profit formula
# Proof of gamma profit formula
http://www.volcube.com/resources/options-articles/gamma-hedging-trading-strategies-part-i/
I would like to have proven to me the above formula, mostly because I don't quite understand it. The formula is an approximation of the profit from gamma trading/gamma hedging, $$0.5 \Gamma (\Delta S)^2$$ So, my questions are, how to prove that, and secondly, what does it mean exactly by "profit"?
Example:
Today, an ATM 1-year 25 % volatility call is bought for 10, and we short $\Delta = 0.5$ in the underlying, which is worth 100. So working that out, we get portfolio value $\Pi = 10 - 50 = -40$, our portfolio value.
Some time later, the spot goes up to 105. The call goes up in value, from 10 to 13.
Currently we have shorted $0.5$ of the underlying, so we owe $0.5 \cdot 105 = 52.5$, so we have $\Pi = -39.5$.
So profit is 0.5.
Then we perform our re-hedge: if delta moved from 0.5 to 0.6, then we need to short 0.1 of the underlying. So, we add $-10.5$ to $\Pi$, i.e, $\Pi = -50$.
Where does the formula from above come into the picture here?
## Answer by LocalVolatility (score 7, accepted)
https://quant.stackexchange.com/a/32975
Assume you buy a plain vanilla call option at the price $V$ and the spot $S$. You immediately delta hedge buy selling $\partial V / \partial S$ units of the underlying asset.
The underlying asset now instantaneously jumps form $S$ to $S' = S + \Delta S$. The new value of the call option is $V'$. Your total p&l is
\begin{equation} \text{P&L} = V' - V - \frac{\partial V}{\partial S} \Delta S. \end{equation}
You can expand the change in the option price to the second order as
\begin{equation} V' = V + \frac{\partial V}{\partial S} \Delta S + \frac{1}{2} \frac{\partial^2 V}{\partial S^2} (\Delta S)^2 + \mathcal{O} \left( (\Delta S)^3 \right). \end{equation}
Substituting back yields
\begin{equation} \text{P&L} = \frac{1}{2} \frac{\partial^2 V}{\partial S^2} (\Delta S)^2 + \mathcal{O} \left( (\Delta S)^3 \right). \end{equation}
This is visualized in the below plots. They are based on $T = 1 / 12$, $K = 100$, $S = 100$, $r = 0\%$, $\sigma = 20\%$. The blue (green) line is the p&l of holding a long (short) position in the call option (underlying asset). The red line is the actual net portfolio p&l and the yellow one is the second order approximation of the latter using the gamma.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.