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!

  • $\begingroup$ The following notes provide a clean derivation of the backward and forward Kolmogorov equations first for $C^2$ functions of the solution of a SDE and then for transition probabilities=Markov kernels -whence the version for the density of the original SDE solution process follows. cims.nyu.edu/~holmes/teaching/asa19/handout_Lecture10_2019.pdf Also, passing from a general Markov process to the corresponding backward Kolmogorov equation can be seen as solving a martingale problem; you can find this in books on Markov provesses -like Bakry-Gentil-Ledoux. $\endgroup$
    – plm
    Commented Oct 28, 2023 at 18:11

1 Answer 1


I am not an expert but I think the Kolmogorov backward equation follows directly from Ito's lemma, and from there you can deduce the forward equation by integration by part.

The forward equation can also be understood directly as the second term in the rhs is just the diffusion induced by the Brownian part, and the first term is the divergence of the "current of matter" vector field induced by the drift, in physical terms. I don't know to what extent this can be made mathematically rigorous though.


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