Merton Jump-Diffusion European Call Price

Options Pricing · Hard · Free problem

A stock follows jump-diffusion dynamics under the risk-neutral measure:

$dS_t = S_t\bigl[(r - \lambda\kappa)\,dt + \sigma\,dW_t\bigr] + S_{t^-}(J - 1)\,dN_t$

where $r$ is the risk-free rate, $\sigma$ is the diffusion volatility, $N_t$ is a Poisson process with intensity $\lambda$, and each jump multiplier $J$ is drawn independently from $\ln J \sim N(\mu_J,\, \sigma_J^2)$. Define $\kappa = E[J - 1] = e^{\mu_J + \sigma_J^2/2} - 1$, so the drift compensates for the expected jump.

  1. Derive the European call price $C(S_0, K, T)$ as a Poisson-weighted infinite series of Black-Scholes prices. Write the full series representation.
  1. For each term in the series, specify the adjusted risk-free rate $r_n$ and adjusted volatility $\sigma_n$ that feed into the $n$-th Black-Scholes component.
  1. Explain the economic intuition: why does the price decompose as a mixture over the number of jumps?

Hints

  1. Condition on the number of jumps $N_T = n$. What does the terminal stock price distribution look like when you know exactly how many jumps occurred?
  2. A sum of $n$ independent $N(\mu_J, \sigma_J^2)$ random variables is $N(n\mu_J, n\sigma_J^2)$. Combined with the diffusion, the conditional log-price is Gaussian -- so you can apply Black-Scholes with adjusted parameters.
  3. Define $\sigma_n^2 = \sigma^2 + n\sigma_J^2/T$ and $r_n = r - \lambda\kappa + n\ln(1+\kappa)/T$. Average the resulting BS prices over $n$ with Poisson weights using intensity $\lambda' = \lambda(1+\kappa)$.

Worked Solution

How to Think About It: The Merton model adds random jumps on top of geometric Brownian motion. The key insight is that if you condition on exactly $n$ jumps occurring during $[0, T]$, the stock price is just a GBM with a modified drift and volatility -- so you can price it with plain Black-Scholes. The unconditional price is then a weighted average over all possible jump counts, with the weights being Poisson probabilities. This is a beautiful example of the law of total expectation: condition on the "hard" part (jumps), solve the "easy" part (GBM), then average.

Quick Sanity Checks:

  • When $\lambda = 0$ (no jumps), only the $n = 0$ term survives, and you recover the standard Black-Scholes formula.
  • As $\lambda \to \infty$ with small jumps, the jump component contributes additional variance, so the effective volatility should increase.
  • Each Poisson weight $e^{-\lambda' T}(\lambda' T)^n / n!$ sums to 1, so the call price is a proper convex combination of BS prices -- it must lie between the cheapest and most expensive BS component.

Derivation:

The terminal stock price is:

$S_T = S_0 \exp\!\left[\left(r - \lambda\kappa - \tfrac{\sigma^2}{2}\right)T + \sigma W_T\right] \prod_{i=1}^{N_T} J_i$

Condition on $N_T = n$. The product of $n$ independent log-normal jumps gives:

$\prod_{i=1}^{n} J_i = \exp\!\left(\sum_{i=1}^{n} \ln J_i\right)$

where $\sum_{i=1}^{n} \ln J_i \sim N(n\mu_J,\, n\sigma_J^2)$. So, conditional on $n$ jumps, the log stock price is Gaussian:

$\ln S_T \mid N_T = n \;\sim\; N\!\left(\ln S_0 + \left(r - \lambda\kappa - \tfrac{\sigma^2}{2}\right)T + n\mu_J,\;\; \sigma^2 T + n\sigma_J^2\right)$

This is the same distribution as a GBM with adjusted parameters. Define:

$\sigma_n^2 = \sigma^2 + \frac{n\sigma_J^2}{T}$

$r_n = r - \lambda\kappa + \frac{n(\mu_J + \sigma_J^2/2)}{T} = r - \lambda\kappa + \frac{n\ln(1+\kappa)}{T}$

Note: the second equality uses $\mu_J + \sigma_J^2/2 = \ln E[J] = \ln(1+\kappa)$.

Conditional on $n$ jumps, the call price is just $C_{BS}(S_0, K, T, r_n, \sigma_n)$ -- the standard Black-Scholes formula evaluated at the adjusted rate and volatility.

The unconditional price uses the law of total expectation over $N_T \sim \text{Poisson}(\lambda T)$. However, there is a subtlety: we also adjust the Poisson intensity. Define $\lambda' = \lambda(1 + \kappa) = \lambda E[J]$. Then the Merton formula is:

$C = \sum_{n=0}^{\infty} \frac{e^{-\lambda' T}(\lambda' T)^n}{n!}\; C_{BS}(S_0, K, T, r_n, \sigma_n)$

The reason $\lambda$ shifts to $\lambda' = \lambda(1+\kappa)$ is that the risk-neutral drift compensation absorbs the expected jump: the factor $e^{-\lambda\kappa T}$ in the drift, combined with the jump size expectation, effectively changes the Poisson parameter from $\lambda T$ to $\lambda' T$ when you reorganize the series.

Practical Interpretation:

In practice, you truncate the series at $n = 10$-

0$ terms, since the Poisson weights decay rapidly. The Merton model produces fatter tails than Black-Scholes -- short-dated options show a pronounced implied volatility smile because with few jumps, the distribution is a mixture of normals with very different variances. For long-dated options, the central limit theorem kicks in (many small jumps average out), and the smile flattens.

Traders use this framework to understand why out-of-the-money puts are expensive: the jump component adds crash risk that a pure diffusion model cannot capture. The decomposition also makes calibration tractable -- you fit $\lambda$, $\mu_J$, and $\sigma_J$ to the observed smile.

Answer:

The Merton jump-diffusion European call price is:

$C = \sum_{n=0}^{\infty} \frac{e^{-\lambda' T}(\lambda' T)^n}{n!}\; C_{BS}(S_0,\, K,\, T,\, r_n,\, \sigma_n)$

where $\lambda' = \lambda(1+\kappa)$, and the $n$-th Black-Scholes component uses:

$r_n = r - \lambda\kappa + \frac{n\ln(1+\kappa)}{T}, \qquad \sigma_n = \sqrt{\sigma^2 + \frac{n\sigma_J^2}{T}}$

The price decomposes as a mixture because, conditional on exactly $n$ jumps, the terminal log-price is Gaussian, reducing to a standard Black-Scholes setting. The Poisson weights average over all possible jump scenarios.

Intuition

The Merton formula is really just the law of total expectation dressed up in finance clothing. Jumps are the "hard" part of the model -- they make the terminal distribution non-Gaussian. But if you condition on exactly how many jumps happened, the remaining randomness is purely Gaussian (diffusion plus a known sum of normals), and you are back in Black-Scholes territory. The unconditional price is then a weighted average over all possible jump counts, with Poisson weights. This "condition and conquer" trick appears everywhere in quantitative finance: mixture models for volatility, regime-switching pricing, and stochastic volatility approximations all follow the same logic.

The practical punchline is that jump-diffusion explains the implied volatility smile for short-dated options. A single large jump creates a bimodal-ish distribution that no amount of diffusion volatility can replicate. This is why OTM puts on equity indices are expensive -- the market prices in discrete crash risk that the Merton model captures through its jump component. As maturity grows, the many-jumps terms dominate, the CLT smooths things out, and the smile flattens toward the Black-Scholes flat-vol world.

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