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Why European Jump-Model Pricing Does Not Directly Extend to American Options

Article Quant Q&A · Author: Alex
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Summary

The document asks whether the Poisson-mixture formula used to price European options under Merton’s jump-diffusion model can also price American options. For European contracts, the cited formula weights Black–Scholes prices adjusted for the number of jumps by the probability of each jump count. The question is whether the same adjustment and weighting can be applied to American calls and puts.

The proposed argument uses conditional expectation, but American exercise introduces a choice of stopping time. The jump-count probabilities depend on that stopping time, and the maximization over exercise times cannot simply be treated as a fixed mixture of American Black–Scholes values. The document poses this issue rather than resolving it, so it provides no pricing result or numerical evidence. It highlights that handling early exercise in a jump model requires care beyond the European formula; the precise valuation method remains unanswered here.

Key ideas

  • European Merton jump-diffusion prices can be expressed as a weighted sum over possible jump counts.
  • American option valuation maximizes discounted payoff over eligible stopping times.
  • The jump-count probabilities vary with the chosen exercise time.
  • A European pricing mixture therefore does not by itself establish an equivalent American pricing formula.
  • The document raises the valuation question but does not supply a solution.

Tags

OptionsDerivatives pricingVolatilityStatistics
Full text
# American Options in Merton's (1976) Jump Model


# American Options in Merton's (1976) Jump Model












@LocalVolatility proves in this stellar answer that European call option prices in the Merton jump diffusion model are given by $$ C_{Merton}(S_0,r,q,\sigma,K,T) = \sum_{n=0}^\infty e^{-\lambda T}\frac{(\lambda T)^n}{n!} C_{BS}(S_0^{(n)},r,q,\sigma^{(n)},K,T). $$

Question: Does the same reasoning apply to American options, where we take the maximum over all stopping times into account? Is the price of American call (put) options in the Merton model equal to the sum of Black-Scholes American calls (puts), weighted by the probability of $n$ jumps occurring? Do we need to adjust $S_0$ and $\sigma$ for all $n$ in the same way as for European calls (puts)?

The argument, based on the law of iterated expectation, could begin like this \begin{align*} c_{American}&=\max\limits_{\tau\;stopping\;time\;on\;[0,T]}\mathbb{E}^\mathbb{Q}[e^{-r\tau}(S_\tau-K)^+] \\ &=\max\limits_{\tau\;stopping\;time\;on\;[0,T]}\mathbb{E}^\mathbb{Q}\big[e^{-r\tau}\mathbb{E}^\mathbb{Q}[(S_\tau-K)^+|N_\tau=n]\big] \\ &= \max\limits_{\tau\;stopping\;time\;on\;[0,T]} \sum_{n=0}^\infty \mathbb{Q}[N_\tau=n] \mathbb{E}^\mathbb{Q}[e^{-r\tau}(S_\tau-K)^+|N_\tau=n] \\ &= \max\limits_{\tau\;stopping\;time\;on\;[0,T]} \sum_{n=0}^\infty e^{-\lambda \tau}\frac{(\lambda \tau)^n}{n!} \mathbb{E}^\mathbb{Q}[e^{-r\tau}(S_\tau-K)^+|N_\tau=n]\\ \end{align*} The latter part looks like a standard American call option in the BS world (with the same adjustments to $S_0$ and $\sigma$ as for European options) but how to deal with the jump probabilities that depend on the stopping time $\tau$?

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.