Why Volatility Skew Does Not Guarantee Delta-Hedged Arbitrage
Summary
The document asks whether a volatility skew creates expected profit from buying an at-the-money option and delta hedging it. Its answer separates implied volatility from the path of realized prices: a move to a particular terminal level does not reveal how turbulent the path was, and an option’s payoff depends on the terminal underlying price relative to its strike and premium.
The discussion treats skew as evidence that the market’s pricing distribution differs from the constant-volatility, normally distributed returns assumed by Black–Scholes. It does not provide a quantitative arbitrage test, hedge strategy, or empirical evidence that the skew predicts realized volatility. The example’s assumed probabilities and implied volatilities therefore do not establish a profitable trade; additional assumptions about the price process and risk-neutral versus realized outcomes would be needed.
Key ideas
- A terminal price change does not determine the realized volatility along the path.
- An option’s expiration payoff depends on the terminal underlying price relative to its strike and premium.
- Implied volatility describes market pricing of an expiration distribution under a model, rather than the realized path itself.
- A volatility skew indicates that constant-volatility, normal-return assumptions may not fit market prices.
Tags
Full text
# Are there arbitrage possibilities on the volatility skew? # Are there arbitrage possibilities on the volatility skew? In the real world, it is observed that the BS model does not hold due to the constant volatility assumption. Accordingly, we observe a volatility skew, where the prices of OTM options are higher/lower than ATM. What confuses me is the following: Say you have an ATM option with IV = 10. Then, say the skew is V’d, so the down 1% option is IV = 20, and up 1% is IV = 20. Now, if you were to buy and delta hedge the ATM option, wouldn’t the skew imply that you make money in expectation? For example, say there is a 1/3 chance of the stock staying at ATM, 1/3 chance of moving down 1% and 1/3 chance of moving up 1%. In the cases that it moves +- 1%, the skew implies that realized vol is 20. If we stay ATM, the realized vol is expected to be 10. Thus, isn’t the average realized vol > 10, hence giving you profit on the delta hedged option you bought for 10 IV? ## Answer by D Stanley (score 1) https://quant.stackexchange.com/a/81856 Somewhat non-quantitative answer: I think you're correlating the actual movements and the realized volatility too much. In your scenario, a 1% change in the stock does NOT mean that the realized volatility was exactly what the skew would predict. It just means that the stock moved up 1%. It could have done that smoothly (low realized vol) or wildly (high realized vol). If you buy an ATM call option (any call option really, but using ATM from your scenario), you only profit if the stock ends up higher than the strike price plus the premium. It does not matter how it got there. The implied volatility is solely used to quantify the probability distribution of the price of the stock at option expiry, and in turn the probability of the option expiring in-the-money (and by how much). The actual payout of the option is in no way related to the realized volatility. The skew is a signal that the market does not believe that the returns of a stock are normally-distributed, so the probability quantified using a constant volatility and normal return distribution does not hold.
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