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@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 $\tau$?

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    $\begingroup$ I think that your reasoning should hold. The jump arrive independent of the underlying motion, and the jump sizes are independent as well. This should suffice for us to be able to specify the option price as a weighted combination of parameter-adjusted American option prices. I would even go so far and assume that the two adjustments (vol and risk neutral drift) carry over. I am not able to prove this formally, though. $\endgroup$ – Kermittfrog Oct 29 '20 at 6:49
  • $\begingroup$ @Kermittfrog I applied the law of total expectation to American options, see edit. In principal, this seems to agree with your intuition. The only question is how to deal with the stopping time featuring the probability of $n$ jumps occurring? $\endgroup$ – Alex Oct 31 '20 at 8:53

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