Let $B_t$ be a Brownian motion on $\mathbb{R}^d$. I often need to compute $\mathbb{P} ( B_t \in A) $ for certain (measurable) sets. I see two main ways to provide a solution:
- since $B_t$ basically distributes as $N(0,t)$, integrate a gaussian over $A$;
- simulate $B_t$ and count.
(or use both for double-checking and exploiting mutual advantages).
Let's now replace the Brownian motion with a generic Ito diffusion process. The "algorithm" number 2 certainly provides a solution (fine to implement, e.g. via Euler's discretization). On the other hand, I am wondering if there is an equivalent solution by using the method number 1.
In other words: do Ito-diffusions admit a transition kernel with known density?
I consulted various books in my library but didn't manage to find a clear answer. Rather there was a lot of (interesting) material and focus on the generator, theoretical investigation on existence and regularity (nice and hard, do not misunderstand me), but for my viewpoint, before going deeper in the theory, it would be better to have at first a clear idea about the question above. Thanks in advance.
Ps: I am trying to study the relationships between Monte Carlo integration, SDE simulation and some class of PDE solver, if the context can help.
In general, no. Identifying the transition kernel comes down to solving the corresponding Fokker-Planck or Kolmogorov forward equation, which is a partial differential equation for the transition kernel, and one cannot write down a general solution. For a diffusion $$dX_t = \mu(X_t) dt + \sigma(X_t) dB_t,$$ the forward equation is $$ \frac{\partial p(x,t)}{\partial t} = -\frac{\partial}{\partial x}(\mu(x)p(x,t))+\frac{1}{2}\frac{\partial^2}{\partial x^2}(\sigma(x)^2 p(x, t)).$$
As you point out, in the case $\mu = 0$ and $\sigma = 1$, $p(x, t) = \frac{1}{\sqrt{2\pi t}} e^{-x^2/2t}$ solves the forward equation with boundary conditions determined by $p(x, t) \rightarrow 0$ as $t\rightarrow 0$ for any $x \neq 0$.
The second most popular example of processes for which the transition kernel can be identified are the Ornstein-Uhlenbeck processes, where $\mu(x) = -ax$ and $\sigma = c$ for some constant $c>0$. For various reasons, one usually writes $\sigma$ in terms of the diffusion coefficient, $\sigma =\sqrt{2D}$. In this case, $$p(x, t) = \sqrt{\frac{a}{2\pi D(1-e^{-2at})}} \exp\left(-\frac{a x^2}{2D(1-e^{-2at})}\right),$$ with appropriate initial conditions.
Possibly the only other thing one can say for "general" systems is that gradient systems with $\mu(x) = -\nabla V(x)$ and $\sigma = c$ admit an explicit steady-state solution to the forward equation, namely $$\lim_{t\rightarrow\infty} p(x,t) \propto e^{-2V(x)/c^2}.$$ Observe that this applies to the Ornstein-Uhlenbeck process above, with $V(x) = ax^2/2$ (so the stationary distribution is Gaussian).