Heat equation with brownian motion
$ \left\{ \begin{array}{rcr} \frac{\partial u}{\partial t} &= & \frac{\partial^2 u}{\partial x^2} \\ u(x,0) & = & \phi(x) \\ \end{array} \right. $
admits as solution $ \mathbb{E}[ \phi(x+ X_t)]$
Heat equation for Poisson Process
We want to show an equivalent of the Heat equation expressed with a Brownian motion, for a Poisson process.
$ (P_t)_{ t \geq 0}$ is a Poisson process of intensity $\lambda$
$(Z_n)_n$ are iid with distribution $ \nu(dz)$ on $\mathbb {R}$
$Y_t =\sum_{k=1}^{P_t} Z_k$ It is a Compound Poisson process
$u : [0,T] \times \mathbb{R} \to \mathbb{R}$ is $\mathcal{C}^2$ on $[0,T] \times \mathbb {R} $ with bounded derivatives.
initial condition $u(0,x) = g(x)$
$$ \frac{\partial u}{\partial t} = \int_{ \mathbb{R} } \left( u(t,x+z)- u(t,x) \right) \lambda \nu(dz) $$
- We fix $t \in [0,T]$ and $ x \in \mathbb{R}$
We want to show that
$$ \boxed{ u(t,x)= \mathbb{E} g(x+Y_t) }$$
Show that there is a random Poisson measure $N$ such that $Y_t= \int_{\mathbb{R} } z N(dr,dz)$
- $\tilde{N}$ is the compensated Poisson measure.
- $Z_t= \int_{0}^t \int_{\mathbb{R} } [ u(t-r, x + Y_r +z) - u(t-r, x +Y_r)] \tilde{N} (dr, dz) $
- Prove that the process $Z_t$ is integrable and with null expectation.
- Show that $u(t,x)= \mathbb{E} g(x+Y_t)$. Apply Ito formula to $(r,y) \mapsto u(t-r,y)$ and Levy process $ (r, x+ Y_r)$
\begin{align*} Y_t &=& \sum_{0 < s \leq t} \Delta Y_s \\ &=& \int_{0}^t z J_Y(dr,dz) \\ \end{align*}
$J_Y$ is the Poisson random measure of the jumps, andhas for intensity $ \rho(dt \times dx) = \lambda dt \nu(dx)$
- According to Financial modelling with jump processes , proposition 2.16,
- $A_t= \int_{0}^t \int_{ \mathbb{R} } f(s,y) \tilde N(ds,dy)$ is a martingale
- if $ \mu(f)= \int_{0}^t \int_{ \mathbb{R} } |f(s,y) |\rho(dt \times dx) < \infty $
Or we may just use that $u$ has bounded derivatives.
$ \begin{align*} Z_t &=& \int_0^t \frac{\partial u}{\partial t} (t-r, x+ Y_r) dr \\ |Z_t| &\leq & \int_0^t | \frac{\partial u}{\partial t} (t-r, x+ Y_r) | dr \\ |Z_t| &\leq & K t \\ |Z_t| &\leq &K T \end{align*} $
The question seems closed to this one and to this article, equation (10).
- The Ito formula for Jump processes is given in this online course page 69 and 70 theorem 6.4 and in Financial modelling with jump processes proposition 8.18
We follow the hint and consider $ (r,y) \mapsto u(t-r,y)$ and the Levy process $(r,x+Y_r)$
$\begin{align*} f(s,Y_s) &=u(s,x+Y_s) \\ f(s,Y_s) - f(0,X_0) &= \int_{0}^{s} \frac{ \partial f } { \partial w } (w,Y_w) dw + \int_{0}^{s}\frac{ \partial f } { \partial y_1 } (w, Y_{w-}) dY_w + \sum_{0<w \leq t} \left[ f(w,Y_w) -f(w,Y_{w-}) - \Delta Y_s \frac{ \partial f } { \partial y_1 }(w,Y_w-) \right] \\ u(s,x+Y_s) -u(t,0) &= - \int_{0}^t \frac{ \partial u } { \partial t}u(w,Y_w-) +\int_{0}^{s}\frac{ \partial u } { \partial x } (w, x+Y_{w-}) dY_w + \sum_{0<w \leq t} \left[ u(w,x+Y_w) -u(w,x+Y_{w-}) - \Delta Y_s \frac{ \partial u } { \partial x }(w,Y_{ {w^-} +x }) \right] \end{align*} $
I hope the derivative of the composite function is correct. And $ \sum_{0<w \leq t}\Delta Y_s \frac{ \partial u } { \partial x }(w,Y_w+x) = \int_{0<w \leq t} \frac{ \partial u } { \partial x }(w,Y_{ {w^-}}+x) dY_w$
so we should have one cancellation of term ,and the rest should look like the martingale in the previous question.
