Formula for Merton jump diffusion call price

Question

What is the formula for a call price in Merton's jump diffusion model?

I am asking because I was taught: $$BS \left[ S = S_0e^{ n(m + \frac{v}{2}) - C \cdot T} , vol = \sqrt{\sigma^2 + nv/T} \ \right]$$ i.e., we use the BS formula where spot and volatility are given as above. Here, $$ C = \lambda (e^{ (m + v/2) } -1 ).$$

And yet, the prices I get from this do not align with some that I found online.

Is the formula right? What formula do people usually use?

Answer 1

The solution that you provided in your question is conditional on the number of jumps being equal to some fixed $n$. To get the option price, you need to take the probability weighted sum over all values of $n \in \mathbb{N}$.

Starting from the standard risk-neutral pricing formula, you use the tower law to condition on the total number of jumps until maturity $N_T$, i.e.

\begin{eqnarray} V_0 & = & e^{-r T} \mathbb{E}_{\mathbb{Q}} \left[ \left( S_T - K \right)^+ \right]\\ & = & e^{-r T} \mathbb{E}_{\mathbb{Q}} \left[ \mathbb{E}_{\mathbb{Q}} \left[ \left. \left( S_T - K \right)^+ \right| N_T = n \right] \right]\\ & = & \sum_{n = 0}^\infty e^{-r T} \mathbb{E}_{\mathbb{Q}} \left[ \left. \left( S_T - K \right)^+ \right| N_T = n \right] \mathbb{Q} \left\{ N_T = n \right\}. \end{eqnarray}

Let the jump size in the logarithmic asset price be normal with mean $\mu$ and variane $\nu^2$. Conditional on $N_T = n$, the logarithmic terminal asset price $\ln \left( S_T \right)$ is normally distributed with

\begin{equation} \mathcal{N} \left( \ln \left( S_0 \right) + \left( r - \frac{1}{2} \sigma^2 - \lambda \left( \exp \left\{ \mu + \frac{1}{2} \nu^2 \right\} - 1 \right) \right) T + n \mu, \sigma^2 T + n \nu^2 \right). \end{equation}

Now imagine a Black-Scholes model with an initial spot of $\hat{S}_0(n)$ and a volatility of $\xi(n) = \sqrt{\sigma^2 + n \nu^2 / T}$. The logarithmic terminal asset price $\ln \left( \hat{S}_T(n) \right)$ in this model would be normally distributed with

\begin{equation} \mathcal{N} \left( \ln \left( \hat{S}_0(n) \right) + \left( r - \frac{1}{2} \xi^2(n) \right) T, \xi^2(n) T \right). \end{equation}

While the variances of the two distributions match by construction, we can solve for $\hat{S}_0(n)$ such that their means do as well. We get

\begin{equation} \ln \left( \hat{S}_0(n) \right) = \ln \left( S_0 \right) - \lambda \left( \exp \left\{ \mu + \frac{1}{2} \nu^2 \right\} - 1 \right) T + n \left( \mu + \frac{1}{2} \nu^2 \right). \end{equation}

I.e. denoting by $V_{\text{BS}} \left( S_0, \sigma \right)$ the Black-Scholes solution for the same plain vanilla call with initial spot $S_0$ and volatility $\sigma$, we get

\begin{equation} \ldots = \sum_{n = 0}^\infty e^{-\lambda T} \frac{(\lambda T)^n}{n!} V_{\text{BS}} \left( \hat{S}_0(n), \xi(n) \right). \end{equation}