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Why a Delta-Hedged Straddle Can Have Path-Dependent P&L

Article Quant Q&A · Author: snoreBore
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Summary

The document explains why selling an at-the-money straddle at one implied volatility and continuously delta hedging with that same volatility does not guarantee a fixed profit when realized volatility is lower. Under an idealized lognormal Brownian stock process, hedging with the true realized volatility would remove path dependence and leave a gain equal to the difference between option values at the implied and realized volatilities.

Using the higher implied volatility to calculate delta instead gives a mismatched hedge. The response says this produces zero expected P&L under its assumptions, while actual P&L remains path dependent. The discussion is theoretical: it assumes perfect continuous hedging and a constant-volatility diffusion, and explicitly cautions that real stocks do not follow such idealized dynamics. It supplies intuition about hedge-model error, but no empirical evidence or practical transaction-cost analysis.

Key ideas

  • Continuous delta hedging does not eliminate path dependence when the hedge uses a volatility different from the stock’s realized volatility.
  • Under the stated idealized assumptions, hedging at realized volatility removes path dependence.
  • Using implied volatility for delta produces a mismatched hedge with path-dependent P&L and zero expected P&L in the example.
  • The result relies on a constant-volatility lognormal diffusion and perfect hedging.

Tags

OptionsVolatilityDerivatives pricingRisk management
Full text
# PnL of a delta-hedged straddle


# PnL of a delta-hedged straddle












On Twitter, this question has been making the rounds:

> If you sold a 30 vol for a one year out at the money straddle, have access to free, perfect, and continuous delta hedging, and stock realizes a 28 vol, have you made money, lost money, or can't be sure?

One user commented that your "pnl is the integral of (implied vol - realized vol) so it depends on the price trajectory." Now, I imagine he's saying that the PnL is like $\int_a^b iv-rv$, so those $a$ and $b$ bounds changes the equation up. However, I thought the entire point of continuously delta hedging was to remove the path dependency? If you're continuously doing this, how is there any price dependency at all? Moreover, why is that the PnL at the end of it all? I lack intuition for this problem, in general. Any help would be greatly appreciated.

## Answer by dm63 (score 1)

https://quant.stackexchange.com/a/78737

Ok I’ll assume a couple of extra details: the option is sold for 30 lognormal vol, then delta hedged according to a Black Scholes model at a 30 vol. The stock then moves with perfect lognormal Brownian motion with 28 vol.

It is known that if the option were hedged at 28 vol, then indeed all path dependency is theoretically removed and you would realize a profit of $BS(30)-BS(28)$, where $BS()$ refers to the black scholes value at a given vol. When you instead calculate Greeks using 30vol, you introduce a ‘wrong’ delta which will have an expected pnl of zero but it will be path dependent.

In practice this is all irrelevant because stocks do not follow nice diffusion processes with constant volatilities, but that’s not what the question asks.

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.