Given the two jump-diffusions: \begin{equation} \begin{aligned} dX_{1,t} &= a_1 dt + b_1 dW_t + c_1 dN_t(\lambda) \\ dX_{2,t} &= a_2 dt + b_2 dW'_t + c_2 dN_t(\lambda) \\ corr(dW,dW') &= \rho \\ dN & \mbox{: Poisson process, of intensity } \lambda \end{aligned} \end{equation} which SDE $$df_t = ? $$ satisfies the function $$ f_t = f(X_{1,t}, X_{2,t}) \mbox{ ???} $$ Thanks in advance for help and/or references.
$\begingroup$
$\endgroup$
0
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
0
Answer
Assuming the Poisson process $N_t$ is independent from the Brownian motions $(W_{1,t},W_{2,t})$, you'll have \begin{align} df(X_{1,t},X_{2,t}) &= \frac{\partial f}{\partial X_{1,t}} dX_{1,t}^c + \frac{\partial f}{\partial X_{2,t}} dX_{2,t}^c + \dots \\ &+ \frac{1}{2} \frac{\partial^2 f}{\partial X_{1,t}^2 } d\langle X_{1,t} \rangle_t^c + \frac{1}{2} \frac{\partial^2 f}{\partial X_{2,t}^2} d\langle X_{2,t} \rangle_t^c + \frac{\partial^2 f}{\partial X_{1,t} \partial X_{2,t}} d\langle X_{1,t} X_{2,t} \rangle_t^c + \dots \\ &+ \left( f(X_{1,t},X_{2,t}) - f(X_{1,t^-},X_{2,t^-}) \right) dN_t \end{align} where the superscript $c$ denotes the continuous part of the semi-martingales $(X_{i,t})_{t \geq 0}$ i.e. $$ d X_{i,t}^c = a_i dt + b_i dW_{i,t},\,\,\, i=1,2$$ such that $$ d\langle X_{i,t} \rangle_t^c = b_i^2 dt,\,\,\, i=1,2 $$ $$ d\langle X_{1,t}, X_{2,t} \rangle_t^c = \rho b_1 b_2 dt $$
Remark 1 : All the derivatives above should be evaluated at $t^-$;
Remark 2 : I've used the notation $\langle X \rangle_t$ to refer to the optional quadratic variation (sometimes denoted by $[ X ]_t$ in the literature) rather than previsible quadratic variation (previsible quadratic variation is the compensator of optional quadratic variation). In case the process has continuous paths the 2 concepts coincide but here it's not the case so I hope this clarifies things.
Background
Consider a non-continuous semi-martingale $(X_t)_{t \geq 0}$ solution of $$ dX_t = a dt + b dW_t + c dN_t $$ Anticipating on what comes next, we also introduce its continuous counterpart $(X_t^c)_{t \geq 0}$ verifying $$ dX_t^c = a dt + bdW_t $$
In differential form, the generalised Itô formula for non-continuous semi-martingales reads (cf. equation (2) in this great blog + demonstration), \begin{align} df(X_t) &= \frac{\partial f}{\partial X_t} dX_t + \frac{1}{2} \frac{\partial^2 f}{\partial X_t^2} d\langle X \rangle_t \dots \\ &+ \left( \underbrace{\left( f(X_t)-f(X_{t^-}) \right)}_{\Delta f(X_t)} - \frac{\partial f}{\partial X_t} \underbrace{c}_{\Delta X_t} - \frac{1}{2} \frac{\partial^2 f}{\partial X_t^2} \underbrace{c^2}_{\Delta X_t^2} \right) dN_t \tag{1} \end{align}
The quadratic variation of the non-continuous semi-martingale $(X_t)$ computes as $$ d\langle X \rangle_t = b^2 dt + c^2 dN_t = d\langle X \rangle_t^c + c^2 dN_t $$ assuming the Poisson process is independent from the Brownian motion under our working probability space (cf. section 15.4). Along with the definition of the SDE satisfied by $(X_t)_{t \geq 0}$ this result allows us to rewrite $(1)$ as \begin{align} \require{cancel} df(X_t) &= \frac{\partial f}{\partial X_t} \left(a dt + b dW_t + \cancel{c dN_t} \right) + \dots \\ &\frac{1}{2} \frac{\partial^2 f}{\partial X_t^2} \left( b^2 dt + \cancel{c^2 dN_t} \right) + \dots \\ &+ \left( (f(X_t)-f(X_{t^-}) \cancel{- \frac{\partial f}{\partial X_t} c} \cancel{- \frac{1}{2} \frac{\partial^2 f}{\partial X_t^2} c^2} \right) dN_t \end{align}
$$ df(X_t) = \frac{\partial f}{\partial X_t} dX_t^c + \frac{1}{2} \frac{\partial^2 f}{\partial X_t^2} d\langle X_t\rangle_t^c + \left( f(X_t)-f(X_{t^-}) \right) dN_t $$
Now you can repeat the experiment starting from the multivariate counterpart of $(1)$ i.e. \begin{align} df(X_t) &= \frac{\partial f}{\partial X_t} dX_t + \frac{\partial f}{\partial Y_t} dY_t + \dots \\ &\frac{1}{2} \frac{\partial^2 f}{\partial X_t^2} d\langle X \rangle_t + \frac{1}{2} \frac{\partial^2 f}{\partial Y_t^2} d\langle Y \rangle_t + \frac{1}{2} \frac{\partial^2 f}{\partial X_t Y_t} d\langle X, Y \rangle_t \dots \\ &+ \left( \Delta f(X_t,Y_t) - \frac{\partial f}{\partial X_t} \Delta X_t - \frac{\partial f}{\partial Y_t} \Delta Y_t \dots \\ - \frac{1}{2} \frac{\partial^2 f}{\partial X_t^2} \Delta X_t^2 - \frac{1}{2} \frac{\partial^2 f}{\partial Y_t^2} \Delta Y_t^2 - \frac{\partial^2 f}{\partial X_t \partial Y_t } \Delta X_t \Delta Y_t \right) dN_t \tag{2} \end{align} to end up on the above mentioned answer.
