Formula for Merton jump diffusion call price

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

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}