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Volatility Monotonicity for American Options Under Black–Scholes

Article Quant Q&A · Author: Lost1

Summary

The document asks whether American option prices must increase with volatility whenever European option prices do, and seeks intuition for proving the result in more general models. The answer states that a standard American option under Black–Scholes is nondecreasing in volatility for a convex payoff, though early exercise can make its value locally insensitive to volatility.

Three proof routes are outlined: examine a step of a binomial tree, express a time step through a linear complementarity formulation involving European options and a forward, or differentiate the Black–Scholes PDE with respect to volatility and use convexity. A further response notes that American and European implied volatility surfaces may differ because their forward estimates use different option data. These arguments are tied to the stated Black–Scholes setting; they do not establish the same result for the questioner's Barndorff–Nielsen–Shephard model.

Key ideas

  • A standard American option with a convex payoff is nondecreasing in volatility under Black–Scholes.
  • Early exercise can leave the option value insensitive to volatility in some regions.
  • Binomial trees, linear complementarity formulations, and the Black–Scholes PDE offer proof approaches.
  • Convexity of payoffs is central to the PDE-based argument.
  • American and European implied volatility surfaces can differ when their forward estimates use different option data.

Tags

Full text
# Relationship between European, American options volatility


# Relationship between European, American options volatility












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.

## Answer by Brian B (score 6)

https://quant.stackexchange.com/a/7922

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.

## Answer by ast4 (score 1)

https://quant.stackexchange.com/a/7924

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.

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.