What is the importance of the infinitesimal generator of Brownian motion?

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?

Answer

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.

Attribution
Source : Link , Question Author : Potato , Answer Author : saz

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