Solution of Merton's Jump-Diffusion SDE

Question

In many textbooks and also in the original Merton's paper the solution of the SDE

$$ dS_t = S_t\,\mu\,dt+S_t\,\sigma\,dW_t+S_{t^-}\,d\left(\sum_{j=1}^{N_t}V_j-1\right) $$

is written as

$$ S_t = S_0\,\exp\left(\left(\mu-\frac{1}{2}\,\sigma^2\right)\,t+\sigma\,W_t\right)\,\prod_{j=0}^{N_t}V_j. $$

Can someone suggest me a textbook or a paper where the solution is explicitly derived? I am pretty confident that this is an application of the Ito lemma for semimartingale.

Answer 1

Let $$ dS_t = \mu S_t dt + \sigma S_t dW_t + S_{t^-} dJ_t $$ where $$ J_t = \sum_{j=1}^{N_t} (V_j - 1) $$ is a compound Poisson process, with $V_j$ i.i.d. jump sizes (positive random variables) whose statistical properties are not relevant for what needs to be proven and $N_t$ a standard Poisson process of intensity $\lambda$. The processes $W_t$, $N_t$ and the random jump sizes $V_j$ are all assumed to be independent of each other and defined over the same probability space.

Applying Itô's formula for semi-martingales with jumps to the function $f(t,S_t) = \ln(S_t)$ yields (see here) $$\ln(S_t) = \ln(S_0) + \left(\mu - \frac{\sigma^2}{2} \right)t + \sigma W_t + \int_0^t ( \ln(S_u) - \ln(S_{u^-}) )dN_u $$ From the SDE we then have that, at a jump time $u$ $$ S_u - S_{u^-} = S_{u^-} (V_j - 1) \iff S_u = S_{u^-} V_j $$ such that $$ \ln(S_u) - \ln(S_{u^-}) = \ln\left(\frac{S_u}{S_{u^-}}\right) = \ln(V_j) $$ and therefore $$ \ln(S_t) = \ln(S_0) + \left(\mu - \frac{\sigma^2}{2} \right)t + \sigma W_t + \sum_{j=1}^{N_t} \ln(V_j) $$ Finally, because $$\sum_{j=1}^{N_t} \ln(V_j) = \ln \left( \prod_{j=1}^{N_t} V_j \right) $$ we get \begin{align} S_t &= S_0 \exp \left( \left(\mu - \frac{\sigma^2}{2} \right) t + \sigma W_t \right) \prod_{j=1}^{N_t} V_j \\ &= F(0,t) \mathcal{E}(\sigma W_t) \prod_{j=1}^{N_t} V_j \end{align} with $\mathcal{E}(X_t) := \exp(X_t - 1/2 \langle X \rangle_t)$ denoting the stochastic exponential of a process $X_t$ (Doléans-Dade exponential).

More info on jump processes (and better mathematical treatment because what I wrote is not always rigorous) in this excellent document

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Note that because \begin{align} E_0[S_t] &= F(0,t) E_0\left[\prod_{j=1}^{N_t} V_j\right] \\ & \ne F(0,t) \end{align} the above dynamics cannot be used for risk-neutral pricing purpose.

To obtain a proper risk-neutral framework, the compound Poisson process needs to get compensated by a drift term (so that the whole emerges as a martingale). The resulting SDE writes

$$ dS_t = (\mu - k) S_t dt + \sigma S_t dW_t + S_{t^-} dJ_t $$

where one can show that $$ k = \lambda (E(V_1) - 1) $$

and where the solution in that case reads $$ S_t = F(0,t) \mathcal{E}(\sigma W_t) e^{-kt} \prod_{j=1}^{N_t} V_j $$

Answer 2

Use Ito for jumps $$ dS_t = \frac{\partial S_t}{\partial t} dt + \frac{\partial S_t}{\partial W_t}dW_t + \frac{1}{2}\frac{\partial^2 S_t}{\partial W_t^2} dt + \frac{\partial S_t}{\partial N_t}d N_t $$

The first part is pretty straight forward

$$ \frac{\partial S_t}{\partial t} dt = S_t(\mu - \frac{1}{2}\sigma^2) $$ $$ \frac{\partial S_t}{\partial W_t}dW_t = S_t \sigma dW_t $$ $$ \frac{1}{2}\frac{\partial^2 S_t}{\partial W_t^2} dt = \frac{1}{2} S_t\sigma^2 dt. $$

Now we have to calculate the derivative according to the jump process $N_t$

Denote $ M_k = \prod_{j=1}^{k} V_j$ and write

$$ M_t = \prod_{j=1}^{N_t} V_j = \sum_{k=1}^\infty\{N_t = k \} \prod_{j=1}^{k} V_j = \sum_{k=1}^\infty\{N_t = k \} M_k. $$

Now if $N_t$ is currently $k$ and jumps during the timestep $t$ to $t+dt$ from $k$ to $k+1$ the process $M_t = M_k$ changes to $M_{t+dt} = M_{k+1}$ and the change is $$ M_{k+1} - M_k = M_k(V_{k+1} - 1). $$

Therefore, and since $N_t$ has no influence on $\exp((\mu + \sigma^2)t + \sigma W_t)$

$$ \frac{\partial S_t}{\partial N_t} = S_{t-}(V_t - 1)$$

Now note that $$(V_t - 1)dN_t = d\sum_{j=1}^{N_t} (V_j - 1)$$ and you are finished.