Suppose, if the price of a European option (say a put) can be shown to be monotone in volatility (say for any maturity), does it follow that American options has to be monotone in volatility?

CLARIFICATION: Monotonicity in volatility means, assuming all other paramters is fixed, the option price is increasing or decreasing in volatility level (at time 0)

^ Presumably, for Black-Scholes model, we can explicitly demonstrate this. (Though I have not tried myself)

what I am intersted in is, does anyone know any counter examples for this?

I work on optimal stopping problems. I am currently working on Barndoff-Nielson Shephard model and I am trying to prove the American put (or a more general pay off) under a pricing measure is monotone in volatility. While I think it should not be too difficult to show monotonicity in volatility if the option is European, it is a lot harder to do it for an American option. I am just trying to get some intuition for this.

  • 1
    $\begingroup$ define "monotone in volatility" please. Do you mean volatility being deterministic? $\endgroup$
    – Matt Wolf
    Commented May 7, 2013 at 10:35

2 Answers 2


For a standard American exercise option expiring at $T>0$, price is still monotically increasing in volatility under the Black-Scholes model (though obviously it is not strictly monotonic, due to early exercise rendering price insensitive to volatility in some regions of parameter space).

To see this, you can use one of three techniques:

  • Investigate the properties of a single step in the binomial tree pricing formulas
  • Formulate the price as an iterated solution in linear complementarity form at $t<T$, and decompose a single timestep's solution into a European option expiring at $t$ , another at $T$ and a forward contract, each of which is convex
  • Differentiate the Black-Scholes PDE with respect to volatility, and then prove it is positive on convex payoffs. Then use the property that applying early exercise conditions preserves convexity.
  • $\begingroup$ thank you for your input but maybe I have not phrased the question well enough. I would be more interested in a case when the European put is monotone in volatility at $t=0$ but American put isn't (or vice versa). At the moment, I think I can show it to be the case for European put for the model I am working on but not the American one. $\endgroup$
    – Lost1
    Commented May 7, 2013 at 14:17
  • $\begingroup$ What I'm saying is that you will not find such a case. $\endgroup$
    – Brian B
    Commented May 7, 2013 at 14:37
  • $\begingroup$ Even in models where volatility exhibits jumps? I don't quite follow why 'rendering price insensitive to volatility in some regions of parameter space' $\endgroup$
    – Lost1
    Commented May 8, 2013 at 20:52
  • $\begingroup$ When volatility (or anything) exhibits jumps then it is no longer the Black-Scholes model. One can still prove some things about families of models but that's much harder. $\endgroup$
    – Brian B
    Commented May 8, 2013 at 21:33

So to expand a bit further on what Brian had mentioned, you're going to get a different vol surface given american vs european. So this is something Brian already pointed out, but one very simple and practical way that you can prove this to yourself is just to think about how the implied forwards are generated.

In the European case we use the entire strip while for American options we only treat options which ul_last > strikes. So like I said, you'll get different forwards which will lead to a different vol surface.


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