Thank you all for answering my question. I wanted to know what influence has the exponential-Lévy model on a call price (how the curve changes).

If we add Merton jumps, we get an EDPID like this one: enter image description here


enter image description here

What is your financial interpretation of the curve transformation ?




1 Answer 1


While you are asking about the call price curve, the effect of adding compound Poisson jumps to a diffusion is more clearly observable when looking at either the implied probability density or the implied volatility smile. We can also prove that the excess kurtosis of the logarithmic returns is always non-negative.

Excess Kurtosis is Non-Negative

First, we can show that any jump-diffusion model generates a non-negative excess kurtosis. Let $X_t = \ln \left( S_t / S_0 \right)$ be the logarithmic asset price following

\begin{equation} X_t = \gamma t + \sigma W_t + \sum_{i = 1}^{N_t} Y_i. \end{equation}


  • $\gamma \in \mathbb{R}$ is the drift such that the discounted asset price is a martingale under the risk-neutral probability measure $\mathbb{P}^*$,
  • $W$ is a standard Brownian motion,
  • $N$ is a compound Poisson process with intensity $\lambda \in \mathbb{R}_+$ and
  • $\left( Y_i \right)_{i = 1}^\infty$ is a sequence of i.i.d. normal random variables with characteristic function $\phi_Y(\omega)$.

The cumulant generating function of $X_t$ is given by

\begin{eqnarray} \psi_{X_t}(\omega) & = & \ln \left( \mathbb{E}_{\mathbb{P}^*} \left[ e^{\mathrm{i} \omega X_t} \right] \right)\\ & = & \left( \mathrm{i} \omega \gamma - \frac{1}{2} \omega^2 \sigma^2 + \lambda \left( \phi_Y(\omega) - 1 \right) \right) t. \end{eqnarray}

This is a standard result, see e.g. Proposition 3.4 in Cont and Tankov (2004). The $n$-th cumulant is given by

\begin{eqnarray} c_n \left( X_t \right) & = & \frac{1}{\mathrm{i}^n} \frac{\partial^n \psi_{X_t}}{\partial \omega^n}(0)\\ & = & \left( \gamma \mathrm{1} \{ n = 1 \} + \sigma^2 \mathrm{1} \{ n = 2 \} + \lambda \frac{1}{\mathrm{i}^n} \frac{\partial^n \phi_Y}{\partial \omega^n}(0) \right) t \end{eqnarray}

We recognize that

\begin{equation} \frac{1}{\mathrm{i}^n} \frac{\partial^n \phi_Y}{\partial \omega^n}(0) = \mathbb{E} \left[ Y^n \right], \end{equation}

given that derivative at zero/expectation exists. The normalized fourth cumulant is equal to the excess kurtosis

\begin{equation} \mathcal{K} \left( X_t \right) = \frac{c_4 \left( X_t \right)}{c_2 \left( X_t \right)^2}, \end{equation}

see e.g. Section 2.4 in Lukacs (1970). From the above we can conclude that $\mathcal{K} \left( X_t \right) \geq 0$ for jump-diffusion models.

Implied Probability Density and Volatility Smile

Given the positive excess kurtosis, the implied probability density becomes more fat-tailed. The below plot is for a Merton jump-diffusion model with $Y \sim \mathcal{N} \left( \alpha, \beta^2 \right)$. I used $T = 1$ month, $r = 0\%$, $\sigma = 20\%$, $\lambda = 25$, $\alpha = 0$ and the lines correspond to $\beta = 0\%$ (blue) $\beta = 1\%$ (green) and $\beta = 2\%$ (red).

Merton Implied Densities

The corresponding implied volatility smiles become more convex ($S_0 = 100$).

Merton Implied Volatility Smiles

Finally, here are two plots for the skewness as a function of $\alpha$ when $\beta = 2\%$ and for the excess kurtosis as a function of $\beta$ when $\alpha = 0\%$.

Skewness vs. Jump Mean Excess Kurtosis vs. Jump Volatility

I put the Jupyter notebook used to generate these plots on GitHub so you can play around with it yourself.


Cont, Rama and Peter Tankov (2004) Financial Modelling With Jump Processes: Chapman & Hall

Lukacs, Eugene (1970) Characteristic Functions: Griffin London, 2nd Edition


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.