In the Merton model, dS/S = (mu - lambda k) dt + sigma dW + (J - 1) dN with ln J ~ N(a, b^2) and k = E[J] - 1. With sigma = 20%, lambda = 1 per year, a = 0 and b = 0.10, compute the excess kurtosis of the one year log return.

In the Merton model, dS/S = (mu - lambda k) dt + sigma dW + (J - 1) dN with ln J ~ N(a, b^2) and k = E[J] - 1. With sigma = 20%, lambda = 1 per year, a = 0 and b = 0.10, compute the excess kurtosis of the one year log return.

Approach: Write the log return as an independent sum of a Gaussian part and a compound Poisson part, then add cumulants, since the fourth cumulant of the Gaussian part is zero.

0.12. The log return over T is (mu - lambda k - sigma^2/2)T + sigma W_T + sum_{i=1}^{N_T} ln J_i, an independent sum, so the cumulants add. The Merton jump diffusion has second cumulant sigma^2 T + lambda T (a^2 + b^2) = 0.04 + 1(0 + 0.01) = 0.05 and fourth cumulant equal to that of the compound Poisson part alone, lambda T E[(ln J)^4], since a Gaussian has no fourth cumulant. With a = 0, E[(ln J)^4] = 3 b^4 = 3(0.0001) = 0.0003, so the fourth cumulant is 0.0003. Excess kurtosis is the fourth cumulant divided by the square of the second, 0.0003/0.0025 = 0.12. The jumps contribute a fifth of the total variance here, and half again as much as that when measured against the diffusion alone, and produce a modest fat tail that decays as 1/T, which is why the implied smile from this model flattens with maturity.

Follow-up: How does the excess kurtosis behave as lambda grows with lambda b^2 held fixed, and what does that limit say about calibrating jumps to a short dated smile?

Key concepts: Merton jump diffusion, cumulants, excess kurtosis, compound Poisson.