Jump Diffusion Infinitesimal generator

Question

I have this difussion process

$dX(t)=\mu X(t)dt+\sigma X(t)dW(t)+u X(t) dN(t),\qquad X(0)=x > 0$

where $W(t)$ is a Brownian Motion and $N(t)$ is a Poisson process.

And I need to know the infinitesimal generator but I can't . Can someone help me?

$\mu,\sigma,u$ are constants.

Thank you so much

Answer 1

I use book Applied Stochastic Control and Jump Diffusion (2nd edittion) written by Bernt Oksendal and Agnes Sulem. You can find more details therein.

Let

$$dX_{t}=a(t,X_{t})dt+b(t,X_{t})dB_{t}+\int_{\mathbb{R}}\gamma(t,X_{t},z)N(dz,dt)$$

then infinitesimal generator is of the form

$$\mathcal{A}f(t,x) = \frac{\partial f}{\partial t}(t,x) + a(t,x)\frac{\partial f}{\partial x}(t,x) + \frac{1}{2}b^{2}(t,x)\frac{\partial^{2} f}{\partial x^{2}}(t,x)$$

$$+\int_{\mathbb{R}}\left(f(t,x+\gamma(t,x,z))-f(t,x)-\frac{\partial f}{\partial x}(t,x)\cdot\gamma(t,x,z)\right)\nu(dz)$$

where $\nu$ is Levy measure of jump.

In your case we have: $a(t,X_{t})=\mu X_{t},\ $ $b(t,X_{t})=\sigma X_{t}\ $ and $\ \gamma(t,X_{t},z)=u X_{t}$.

Note that $\gamma(t,X_{t},z)=u X_{t}$ doesn't depend on $z$, so in the last term in the infinitesimal generator we have to compute the following expression

$$\int_{\mathbb{R}}\nu(dz)=\nu(\mathbb{R})=\lambda$$

Taking everything into account the infinitesimal generator is of the form:

$$\mathcal{A}f(t,x)= \frac{\partial f}{\partial t}(t,x) + \mu x\frac{\partial f}{\partial x}(t,x) + \frac{1}{2}\sigma^{2}x^{2}\frac{\partial^{2} f}{\partial x^{2}}(t,x)$$

$$ + \left(f(t,x+ux) - f(t,x)-\frac{\partial f}{\partial x}(t,x)\cdot u x\right)\cdot \lambda $$

where $\lambda>0$ is Poisson process intensity and $f\in C^{1,2}(\mathbb{R}\times \mathbb{R})$.

Answer 2

I decided to post my answer because I always wanted to derive those things from scratch rather than using results from books I can hardly have access to. The exponential Levy jump diffusion model is defined as follows. $X_t = x \cdot \exp(L_t) $ where the process $L_t$ reads:

\begin{equation} L_t = \mu \cdot t + \sigma \cdot B_t + \sum\limits_{i=1}^{N_t} Y_i \end{equation}

Here $\mu \in {\mathbb R} $ and $\sigma > 0 $ are fixed numbers. Then $B_t = \sqrt{t} \cdot B_1 $ is a Brownian motion (in particular $B_1 = N(0,1)$), $N_t$ is a Poisson process and $(Y_i)_{i\ge 1}$ are iid random variables which are independent from $B_1$ and which have a common pdf $\omega_Y(y)$ such that the expectation $E[\exp(m Y)]$ exists for every $m=0,1,2,\cdots$.

The process $X_t$ is used as a model for the stock prices see 1.

In order to find the infinitesimal generator we will firstly compute the characteristic function (the Fourier transform of the pdf) of the process in question. We have:

\begin{eqnarray} &&\chi_{X_t}(k) := E\left[ e^{\imath k \cdot X_t}\right] = \\ &&\sum\limits_{n=0}^\infty E\left[\exp\left(\imath k x \cdot e^{\mu t} \cdot e^{\sigma \sqrt{t} \cdot B_1} \cdot e^{\sum\limits_{i=1}^n Y_i}\right)\right] \cdot \frac{(\lambda t)^n}{n!} \cdot e^{-\lambda \cdot t} \end{eqnarray} where in the last line above we have conditioned on the number of jumps. Note that the expectation is taken with respect to both the Brownian motion part and the jumps.

Now let us compute the $m$th moment (where $m \in {\mathbb N}$) of the process in question. We have:

\begin{eqnarray} &&E\left[ (X_t)^m \right] = \left. \frac{d^m}{d (\imath k)^m} \chi_{X_t}(k) \right|_{k=0} \\ && x^m e^{m \mu t} \sum\limits_{n=0}^\infty E \left[ e^{m \sigma \sqrt{t} \cdot B_1} \cdot e^{m \sum\limits_{i=1}^n Y_i} \right]\cdot \frac{(\lambda t)^n}{n!} \cdot e^{-\lambda \cdot t} \\ &&x^m e^{m \mu t} \sum\limits_{n=0}^\infty e^{m^2 \sigma^2 t/2} \cdot (E\left[ e^{m Y}\right])^n \cdot \frac{(\lambda t)^n}{n!} \cdot e^{-\lambda \cdot t} \\ && x^m e^{m \mu t} e^{m^2 \sigma^2 t/2} \sum\limits_{n=0}^\infty \frac{(\lambda t\cdot E\left[ e^{m Y}\right])^n}{n!} \cdot e^{-\lambda \cdot t} \\ && x^m e^{m \mu t} e^{m^2 \sigma^2 t/2} \cdot e^{\lambda t \left[ E\left[ e^{m Y}\right] - 1\right]} \end{eqnarray}

As we can see all the integer moments exist.

Now, we are ready to compute the infinitesimal generator ${\mathbb L}_X $. We take a smooth function $F(x)$ and we have:

\begin{eqnarray} &&{\mathbb L}_X F(x) := \lim\limits_{t \rightarrow 0} \frac{E\left[ F(X_t)\right] - F(x)}{t} = \\ && \sum\limits_{m=0}^\infty \frac{F^{(m)}(0)}{m!} \cdot \lim\limits_{t \rightarrow 0} \frac{E\left[ (X_t)^m \right]-x^m}{t} = \\ && \sum\limits_{m=0}^\infty \frac{F^{(m)}(0)\cdot x^m}{m!} \cdot \lim\limits_{t \rightarrow 0} \frac{e^{m \mu t} e^{m^2 \sigma^2 t/2} \cdot e^{\lambda t \left[ E\left[ e^{m Y}\right] - 1\right]}-1}{t} \\ && \sum\limits_{m=0}^\infty \frac{F^{(m)}(0)\cdot x^m}{m!} \cdot \left(m \mu + m^2 \frac{\sigma^2 }{2} + \lambda \left[ E\left[ e^{m Y}\right] - 1\right] \right) = \\ &&\left(\mu x \frac{d}{d x} + \frac{\sigma^2 }{2} x \frac{d}{d x} x \frac{d}{d x} \right) F(x) + \lambda \int\limits_{{\mathbb R}} \left[ F(x e^{\xi}) - F(x)\right] \cdot \omega_Y(\xi) d \xi \\ && \left( (\mu + \frac{\sigma^2}{2}) x \frac{d}{d x} + \frac{\sigma^2 x^2}{2} \frac{d^2}{d x^2}\right) F(x) + \lambda \int\limits_{{\mathbb R}} \left[ F(x e^{\xi}) - F(x)\right] \cdot \omega_Y(\xi) d \xi \end{eqnarray}

Now if we go to the body of the question and take $\omega_Y(y) = \delta(y- u) $ then we get ${\mathbb L}_X F(x) = \left( (\mu + \frac{\sigma^2}{2}) x \frac{d}{d x} + \frac{\sigma^2 x^2}{2} \frac{d^2}{d x^2}\right) F(x) + \lambda \left[ F(x e^{u}) - F(x)\right]$.

Unfortunately this is different from the answer given by Sean Meyn above. There are two differences, firstly the Ito correction, i.e. the coefficient at the first derivative and secondly the term in square brackets on the very right, which has $e^u$ instead of $1+u$.

1: Ernst Eberlein, Jump-type Levy processes