I have read that the infinitesimal generator of Brownian motion is $\frac{1}{2}\small\triangle$. Unfortunately, I have no background in semigroup theory, and the expositions of semigroup theory I have found lack any motivation or intuition.

What is the infinitesimal generator of a process intuitively, and why is it interesting or useful to know that the generator of Brownian motion is $\frac{1}{2}\small\triangle$?


3 Answers 3


For a Markov process $(X_t)_{t \geq 0}$ we define the generator $A$ by

$$Af(x) := \lim_{t \downarrow 0} \frac{\mathbb{E}^x(f(X_t))-f(x)}{t} = \lim_{t \downarrow 0} \frac{P_tf(x)-f(x)}{t}$$

whenever the limit exists in $(C_{\infty},\|\cdot\|_{\infty})$. Here $P_tf(x) := \mathbb{E}^xf(X_t)$ denotes the semigroup of $(X_t)_{t \geq 0}$.

By Taylor's formula this means that

$$\mathbb{E}^xf(X_t) \approx f(x)+t Af(x)$$

for small $t \geq 0$. So, basically, the generator describes the movement of the process in an infinitesimal time interval. One can show that

$$\frac{d}{dt} P_t f(x) = A P_tf(x), \tag{1}$$

i.e. the generator is the time derivative of the mapping $t \mapsto P_tf(x)=\mathbb{E}^x(f(X_t))$. Reading $(1)$ as a (partial) differential equation we see that $u(t,x) := P_t f(x)$ is a solution to the PDE

$$\frac{\partial}{\partial t} u(t,x) = Au(t,x) \qquad u(0,x)=f(x).$$

This is one important reason why generators are of interest. Another, more probabilistic, reason is that the process

$$M_t^f := f(X_t) - f(X_0)- \int_0^t Af(X_s) \, ds, \qquad t \geq 0 \tag{2}$$

is a martingale. This means that we can associate with $(X_t)_{t \geq 0}$ a whole bunch of martingales, and this martingale property comes in handy very often, for example whenenver we deal with expectations of the form $\mathbb{E}^x(f(X_t))$. This leads to Dynkin's formula.

Generators are also connected with the martingale problem which in turn can be used to characterize (weak) solutions of stochastic differential equations. Futhermore, generators of stochastic processes are strongly related to Dirichlet forms and Carré du champ operators; it turns out that they are extremely helpful to carry over results from probability theory to analysis (and vica versa). One important application are heat-kernel estimates.

Example: Brownian motion In the case of (one-dimensional) Brownian motion $(B_t)_{t \geq 0}$, we see that

$$\mathbb{E}^x(f(B_t)) \approx f(x)+ \frac{t}{2} f''(x)$$

for small $t$. This formula can be motivated by Taylor's formula: Indeed,

$$\mathbb{E}^x(f(B_t)) \approx \mathbb{E}^x \left[f(x)+f'(x)(B_t-x)+\frac{1}{2} f''(x)(B_t-x)^2 \right]= f(x)+0+\frac{t}{2} f''(x)$$

using that $\mathbb{E}^x(B_t-x)=0$ and $\mathbb{E}^x((B_t-x)^2)=t$.

From $(1)$ we see that $u(t,x) := \mathbb{E}^x(f(B_t))$ is the (unique) solution of the heat equation

$$\partial_t u(t,x) = \frac{1}{2}\partial_x^2 u(t,x) \qquad u(0,x)=f(x).$$

Moreover, one can show that the solution of the Dirichlet problem is also related to the Brownian motion. Furthermore, $(2)$ yields that

$$M_t^f := f(B_t)-f(B_0) - \frac{1}{2} \int_0^t f''(B_s) \, ds.$$

is a martingale. Having Itô's formula in mind, this is not surprising since

$$f(B_t)-f(B_0) = \int_0^t f'(B_s) \, dB_s+ \frac{1}{2} \int_0^t f''(B_s) \,ds = M_t^f + \frac{1}{2} \int_0^t f''(B_s) \,ds.$$

The above-mentioned results (and proofs thereof) can be found in the monograph Brownian Motion - An Introduction to Stochastic Processes by René L. Schilling & Lothar Partzsch.

  • 1
    $\begingroup$ Thank you. Just what I was looking for. $\endgroup$
    – Potato
    Mar 3, 2014 at 17:18
  • 2
    $\begingroup$ @ThomasE I agree that it is an interesting question, but I wouldn't expect that any martingale has such a representation, no. If you know that the filtration is generated by a Lévy process, then you can write the martingale as a stochastic integral (driven by the Lévy process)... that's a nice representation result, but a different story. $\endgroup$
    – saz
    Apr 12, 2017 at 18:26
  • 1
    $\begingroup$ @user1 In some sense, yes. If you consider $t>0$, then you first have to apply the semigroup $P_t$ (in your words "change the $x$") and then take the derivative. The Markov property is indeed very crucial for all of this. $\endgroup$
    – saz
    Oct 16, 2017 at 15:07
  • 1
    $\begingroup$ Could you tell me why $(P_t)$ is a semi group ? You set $P_t f(x)=\mathbb E^x[f(X_t)]$, so why $P_{t+s}=P_tP_s$ ? I don't get how to prove it. $\endgroup$
    – user621345
    Jan 11, 2019 at 14:55
  • 2
    $\begingroup$ @user657324 Well, yes, the "approximation" for $E^x f(X_t)$ holds only for small $t$. Re your 2nd question: The distribution of any random variable $Y$ is uniquely characterized by expectations of the form $\mathbb{E}[f(Y)]$ for a class of functions $f$ which is "large enough" (e.g. $f \in C_b^2$) Thus, if we know $\mathbb{E}^x f(X_t)$ for a sufficiently large class of functions $f$, then we have information on the distribution of $X_t$. $\endgroup$
    – saz
    Mar 31, 2019 at 8:25

In fact, there is a deeper relation between the Laplacian and Brownian motion.

Let $(M, g=\langle\cdot, \cdot\rangle)$ be a smooth Riemannian manifold without boundary. The Laplace-Beltrami operator is defined as the contraction of the covariant derivative of the differential of any smooth function on $M$

$$\forall f \in C^\infty(M): \Delta_M f := \mathrm{tr} \nabla \mathbf df = \mathrm{div}\ \mathrm{grad} \ f \in C^\infty(M),$$

where the well-known definition can be recovered with suitable generalisations of the divergence and the gradient. This means, for any orthonormal basis $E_1,...E_n$ for $T_pM$ ($p \in M$),

$$\forall f \in C^\infty(M): \Delta_M f(p) = \sum_{i=1}^n \nabla\mathbf d f(E_i,E_i) = \left\langle \nabla_{E_i}\mathrm{grad} \ f, E^i \right\rangle,$$

where we used Einstein notation. Moreover, we can generalise the term of a continuous semimartingale as follows: Every adapted $M$-valued stochastic process $X$ is a semimartingale on $M$ if, for all $f \in C^\infty(M)$, the composition map is $f(X)$ a real-valued semimartingale.

Then we can define Brownian motion on $M$ by the usual martingale problem (this is known as the extrinsic definition):

Let $X$ an adapted $M$-valued process. A process $X$ is called Brownian motion on $(M,g)$ if, for all $f \in C^\infty(M)$, the real-valued process

$$f(X) - \frac 12 \int \Delta_M f(X) \mathrm dt$$

is a local martingale.

In particular, we can prove Lévy's characterisation also for BM$(M,g)$. But this requires a reasonable definition of the quadratic variation.

The problem with this definition lies in the manifold itself: There does not exist a Hörmander-type representation of the Laplace-Beltrami operator if $M$ is not parallelizable, i.e. the tangent bundle $TM \overset\pi\longrightarrow M$ is not trivial. But it holds the fundamental relation

$$\Delta_{\mathcal O(M)} \pi^* = \pi^* \Delta_M,$$

more precisely,

$$\Delta_{\mathcal O(M)}(f \circ \pi)(u) = \Delta_M f(x),$$

for all $u \in \mathcal O(M)$ with $x = \pi(u)$. Moreover, there exist $n$ well-defined unique horizontal vectors $L_i(u) \in H_u\mathcal O(M)$, $\pi_* L_i(u) = ue_i$, $(e_i)$ basis for $\mathbb R^n$, the so called fundamental horizontal vector fields and the we define

$$\Delta_{\mathcal O(M)} := \sum_{i=1}^n L_i^2,$$

where $\mathcal O(M)$ denotes the orthonormal frame bundle, the prototypical example of a smooth principal fibre bundle whose structure group is given by the orthogonal group.

Using this relation, it is due to Malliavin, Eells and Elworthy that there always exists a lifted Brownian motion as solution of the globally defined SDE

$$\mathrm d U = L_i(U) \circ \mathrm d B^i,$$

on $\mathcal O(M)$, where $B$ is a real $n$-dimensional Brownian motion and we used Einstein notation. A solution is a diffusion generated by $\frac 12\Delta_{\mathcal O(M)}$. The idea is to solve the SDE in $\mathcal O(M)$ and $X = \pi(U)$ is the projection of the lifted Brownian motion $U$ on the manifold $M$ via $\mathcal O(M) \overset\pi\longrightarrow M$. It follows that $X$ is a Brownian motion on $M$ starting from $X_0 = \pi(U_0)$.

In geometrical terms, the idea is to "roll" our manifold $M$ by means of the (stochastic) parallel displacement along the paths of an $\mathbb R^n$-valued Brownian motion ("rolling without slipping"), known as stochastic development.


  • Hsu, Elton P. Stochastic analysis on manifolds. Vol. 38. American Mathematical Soc., 2002.
  • (in german) Hackenbroch, Wolfgang, and Anton Thalmaier. Stochastische Analysis. Vieweg+ Teubner Verlag, 1994.
  • Elworthy, Kenneth David. Stochastic differential equations on manifolds. Vol. 70. Cambridge University Press, 1982.
  • Malliavin, Paul. Géométrie différentielle stochastique. Montreal, Presses de l’universite de Montreal, 1978.
  • 1
    $\begingroup$ Beautiful answer. It might be useful to tell who $\mathcal O (M)$ and $\Delta _{\mathcal O (M)}$ are. $\endgroup$
    – Alex M.
    Jun 23, 2018 at 6:22

The generator is $A f (x) = \lim_{t \downarrow 0} \frac{\mathbf{E}^{x} [f(X_{t})] - f(x)}{t}$. If $X_{t}$ were a degenerate stochastic process say just given by an ODE then the generator would just give you an ODE for $f(X_t)$.

You can use a generator to for example derive PDEs relevant to the stochastic process. For a simple example say you wanted to find a PDE for the stationary distribution of $X$. Assume this distribution is given by $\pi(x)$. Take the expectation of both sides against $\pi(x)$, since it's a stationary distribution the right hand side will be $0$. On the left hand side do essencially integration by parts to move the differential opperator $A$ from $f$ to $\pi$ and think of $f(x)$ as a test function. Then you get that $A^* \pi(x) = 0$ where $A^*$ is the adjoint of $A$.

So in this example the steady state will solve $\Delta \pi = 0$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .