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The stock price is modeled by $$S_t = S_0 e^{bt +\sigma B_t + \sum_{k=1}^{N_t} Y_k}$$

with $B_t$ Brownian motion, $Y_k$ iid $N(\mu,\delta^2)$, $N_t$ a Poisson process independent of $(B_t)$ and $Y_k$ for all $k$, and all parameters are given. We assume a constant risk free rate $r > 0$.

Let's say we compute the parameters of the model after a measure change only modifying the drift and leading to a risk neutral measure. (If I understood correctly, this will only change $b$.)

Then we want to compute the price of European calls with strike $K$, and time to maturity $T$ using that risk neutral measure with 2 different methods: series approximation and Monte Carlo methods.

What does that mean? What's the price of a European call under this model? And how can we simulate it on a computer? What does "using that risk neutral measure" mean?

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  • $\begingroup$ You have a lot of different but also relatively broad questions. I suggest you do some more reading and then ask a more specific question. A nice introduction that covers most of what you are looking for is Chapter 15 "Incomplete markets and jump-diffusion models" in Joshi's "The Concepts and Practice of Mathematical Finance". $\endgroup$ – LocalVolatility Jan 3 '18 at 11:27
  • $\begingroup$ Thank you for the reference. I found every answer but I still don't understand what "using that risk neutral measure" mean. Does it simply mean that we use the model with the newly computed $b$ ? $\endgroup$ – W. Volante Jan 3 '18 at 14:50
  • $\begingroup$ Pricing corresponds to taking the expectation under a risk-neutral probability measure. Thus, you also need to use the corresponding spot price dynamics which in your case corresponds to using the adjusted drift $b$. $\endgroup$ – LocalVolatility Jan 3 '18 at 15:02
  • $\begingroup$ I don't see why doing it under a risk-neutral probability measure changes anything for the series representation (see Merton 1976). The series representation uses the price under black-scholes model, which has nothing to do with $b$, right ? So whatever drift changing measure we working with, the series will be the same ? $\endgroup$ – W. Volante Jan 4 '18 at 2:39
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I'll decompose your big question into smaller questions and answer them in (hopefully) simple terms.

1. What is meant by the risk neutral measure?

This is how I understand the risk-neutral measure (commonly denoted by $\mathbb{Q}$): It is the probability measure under which the current value of all financial assets at a time, say $t$, are equal to the expected future payoff of the asset discounted at the risk-free rate, $r$. It's used heavily in the the prices of financial derivatives because of the Fundamental Theorem of Asset Pricing (see: Wikipedia).

This theorem implies that in a complete market (i.e., a market that allows the hedging of the risk inherent in any investment strategy), a financial derivative's price is the discounted expected value of the future payoff under $\mathbb{Q}$.

It's well-known that in the case of a geometric Brownian motion model a unique risk-neutral measure exists. However, the introduction of jumps, as in Merton's 1976 paper, destroys this notion of completeness and so we no longer have a unique risk-neutral measure $\mathbb{Q}$.

Finally, the risk-neutral pricing formula of a European call at time $t$ with the parameters you mentioned is

$$C = C(t, S_t)=\mathbb{E}_{\mathbb{Q}}[e^{-rT}(S_t-K)^+|\mathcal{F}_t],$$

where for now just read $\mathcal{F}_t$ as the all the information known at time $t$.

2. What's the price of European call in Merton's model?

A closed-form solution for European options under Merton's jump-diffusion model exists. Let $C_{BS}$ denote the price of your European call under the Black-Scholes model. You'd like $C_{JD}$, its value under Merton's jump-diffusion model, where your jump size follows a log-normal distribution with average jump size $m$ and jump size volatility $\nu$. The formula for $C_{JD}$ can be written as:

$$C_{JD}(S, K, \sigma, r, T, \lambda, m, \nu)=\sum_{k=0}^{\infty}\frac{\exp{(-m\lambda T)(m\lambda T)^k}}{k!}C_{BS}(S, K, \sigma_k, r_k, T),$$

where $\sigma_k = \sqrt{\sigma^2 + k\nu^2/T}$ and $r_k = r - \lambda(m-1)+k\log(m)/T$.

Each term in the infinite series corresponds to every possible jump frequency scenario.

3. How do we simulate Merton's jump-diffusion model on a computer?

This is possibly the broadest question of them all and (correct me users if I'm wrong) depends on a variety of factors. In my opinion, the simplest way to do so is follows (I won't go into much detail here):

i. Get the Euler discretisation of the Merton jump-diffusion model;

ii. Get your parameters (a major topic in its own right);

iii. Generate three sets of independent random numbers corresponding to the three random variables in your discretisation scheme;

iv. Get the values for the simulated stock path using these;

iv. Use Monte Carlo integration to get the price of your call option.

I hope this helps and excuse any mistakes I've made along the way.

Thanks, Vladimir

Extra: Here's a thesis and book which provide great introductions (and more) to this topic.

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  • $\begingroup$ Thank you very much ! If we use the series using BS model to compute the price, then the result do not depend on the drift $b$ in Merton's model, right ? So any change of measure that only affects the drift won't change the price, I find this really counter-intuitive. $\endgroup$ – W. Volante Jan 4 '18 at 14:36
  • $\begingroup$ You're correct. $b$, or as it is often referred to in literature $\mu$, doesn't feature in the BS model. Consequently, it doesn't feature in the closed form above. Chapter 4 should provide you with the intuition you might find interesting behind risk-neutral measures in jump-diffusion models. If you like my answer please accept it by clicking on the check mark :-) $\endgroup$ – Vladimir Nabokov Jan 4 '18 at 16:49
  • $\begingroup$ Once again, thank you. A final question about the simulation: what's the "euler discretisation of the Merton jump-diffusion model" ? Can you give more details or a link to a code ? $\endgroup$ – W. Volante Jan 4 '18 at 21:11
  • $\begingroup$ It's all about moving from the continuous-time to discrete-time space. See: en.wikipedia.org/wiki/Euler%E2%80%93Maruyama_method $\endgroup$ – Vladimir Nabokov Jan 4 '18 at 22:33
  • $\begingroup$ Not sure I'm understanding this, I posted another question regarding this here quant.stackexchange.com/questions/37589/… $\endgroup$ – W. Volante Jan 4 '18 at 23:04

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