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?


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}

  • $\begingroup$ Hi, just my two cents : I think that constant (Black Scholes) implied volatility implies normal log returns of the underlying. If I am correct, then you might need to reformulate the assumptions. (I think) $\endgroup$ Commented Nov 27, 2020 at 20:23
  • $\begingroup$ It was a cause and effect example. If we make these assumption, this is what will happen. These were the assumptions prior to 1987, my question was just asking how does a concave up volatility smile correct the expected return curve we see under these assumptions. $\endgroup$
    – brickbobed
    Commented Nov 27, 2020 at 20:34
  • $\begingroup$ Ok; i think i misread your Post; sorry $\endgroup$ Commented Nov 27, 2020 at 20:54

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.