Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In Lamperti's Stochastic Processes, given

  • a time-homogeneous Markov process $X(t), t\geq 0$ with Markov transition kernel $p_t(x,E)$ and state space being a measurable space $(S, \mathcal{F})$,
  • a Banach space $M(S)$ of finite signed measures on $(S, \mathcal{F})$ with total-variation norm,

he defines an operator $U_t$ on $M(S)$ as, $\forall \nu \in M(S), $ $$ (U_t \nu)(E) = \int_S \nu(dx) p_t(x,E). $$ Then $\{U_t, t \geq 0\}$ forms a contraction semigroup of operators on $M(S)$.

The general theory of a contraction semigroup of operators implies that we can

  • first find the generator $A: [0, \infty) \to L(M(S))$ of $\{U_t, t \geq 0\}$,
  • then $u(t) = U_t\nu$ is the solution to the abstract Cauchy problem: $$u'(t)=Au(t),$$ $$u(0)=\nu, $$ for $\nu \in M(S)$. Lamperti said that this abstract Cauchy problem is the Kolmogorov forward equation of $X(t)$.

My questions are for a time-homogeneous Ito diffusion process $X(t), t\geq 0$ evolving according to the stochastic differential equation $$ dX(t) = \mu(X(t))dt + \sigma(X(t))dW(t) $$

  1. I was wondering how to derive the generator $A$ of its semigroup $\{U_t, t\geq0\}$ of operators, and its Kolmogorov forward equation based on the above abstract Cauchy problem?

  2. Standard textbooks say that, if $X_t$ has a density function $f(x,t)$, its Kolmogorov forward equation can be expressed in terms of the density function $f(x,t)$ of $X_t$: $$ \frac{\partial}{\partial t}f(x,t)=-\frac{\partial}{\partial x}[\mu(x)f(x,t)] + \frac{1}{2}\frac{\partial^2}{\partial x^2}[\sigma^2(x)f(x,t)], t \ge 0 $$ $$f(x,0)=g(x),$$ where $g$ is a density function of a probability measure. The above version of KFE is also called the Fokker–Planck equation.

    I was wondering how the Fokker–Planck equation is derived from the abstract Cauchy problem?

References are also appreciated!

Thanks and regards!

share|improve this question
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.