How Volatility Smile Curvature Relates to Fat Tails and Option Prices
Summary
The discussion asks how a convex implied volatility smile relates to high kurtosis and the expected returns of options across strikes. Its answer says that assuming constant implied volatility alongside a symmetric, fat-tailed underlying distribution can imply an arbitrage: selling straddles and buying strangles would leave exposure to realized tail moves. A convex smile raises the relative cost of strangles, and at a sufficient level the proposed arbitrage disappears.
The answer points to the Breeden-Litzenberger relationship between option prices, smile curvature, and the implied probability distribution as the formal framework for analyzing the connection. It does not derive the formula, quantify a threshold, or provide empirical evidence. It also cautions that extrapolating expected returns at zero or infinite strikes is not meaningful. The post is therefore a conceptual pointer, not a complete pricing analysis; its conclusions depend on the assumptions and the precise construction of the option positions.
Key ideas
- A symmetric fat-tailed distribution paired with constant implied volatility may imply an arbitrage involving straddles and strangles.
- A convex implied volatility smile increases the cost of strangles and may remove that arbitrage at a sufficient level.
- Breeden-Litzenberger connects option prices and their strike curvature to the implied probability distribution.
- Expected returns extrapolated to extreme strikes are not considered meaningful in the answer.
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# How does a concave up volatility smile correct high kurtosis for ATM option contracts?
# How does a concave up volatility smile correct high kurtosis for ATM option contracts?
Theoretically speaking, if we are to assume the following:
- Constant implied volatility throughout all strike prices
- The underlying's prices change distribution is log-leptokurtic and symmetric
Then graphing the expected return of each strike price should generate some sort of quasi-exponential curve. For calls, as the strike prices tend toward zero, expected return approaches the expected return of the underlying. For puts, as the strike prices tend toward infinity, expected return approaches the expected return of the underlying (or the risk free rate). As the strikes tend toward the opposite direction of the previous example, expected return should approach infinity (again, I am speaking theoretically). Please correct me if my logic is wrong here. But if I am correct, how does a parabolic implied volatility curve correct this quasi-exponential return curve?
P.S.
When I say expected return I am assuming the integral of exponential returns:
\begin{equation} E[dS] = \int_{-\infty}^{+\infty}{[exp(dS)\times P(dS)]\:d^2S} \end{equation}
## Answer by Meph (score 0, accepted)
https://quant.stackexchange.com/a/59790
Your assumptions imply arbitrage: sell straddles buy strangles, you can build a portfolio with an exposure to the realized fat tails.
Moving to an implied volatility surface that is convex increases the cost of the strangle, at some breakeven level the arbitrage disappears.
Formally this is coded into the connection between smile curvature and the implied probability distribution - you should examine the Breeden-Litzenberger formula.
I don't think your 'expected return at infinite/zero strike' argument is meaningful.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.