0

I am interested in the infinitesimal generator of the Brownian motion and its relation to the Laplacian. As explained here the infinitesimal generator is defined as

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

I would appreciate to answer my questions, by considering the fact that i have not that much knowledge in Stochastic calculus or probability theories.

1) What is $f(x)$? Can we say that: $Af(x)$ is the generator of the stochastic process $X$? I mean, $Af(x)=AX$? Also, if X is a function, how can we interpret $f(X)$?

2) Is there any simple way to explain what is $P_tf(x)$, without knowing about feller semigroup? Here i read that : "$\mathbb{P}^x$ is a conditional probability, which conditions on $X_0=x$". But this is not that much clear for me.

3) As it is written here, one can write:

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

How $\mathbb{E}^xf(x)$ became $f(x)$ in the equality above?

Thanks in advance.

Denis
  • 169
  • 11

1 Answers1

2

(1) The infinitesimal generator is an operator defined on a subspace of the space $C_0$ of continuous functions that vanish at infinity. More precisely, the subspace on which it is defined is $$D(A) = \left \{f \in C_0: \lim_{t \downarrow 0} \frac{\mathbb{E}^x(f(X_t))-f(x)}{t} \text{ exists} \right \}$$ and then $A:D(A) \to C_0$. This means that if $f \in D(A)$ then $Af$ is a function that vanishes at infinity. $A$ is the generator of the process $X$ but this does not mean that for $f \in D(A)$, $Af(X) = AX$. Indeed, you have no reason to believe that $X \in D(A)$ in the first place since in many situations $A$ will turn out to be a differential operator and $X$ need not have differentiable paths.

Finally, if $X$ is your process then $f(X)$ is the process that at a given time $t$ and for a given $\omega$ in your probability space is defined by $f(X)_t(\omega) = f(X_t(\omega))$.

(2) $P_t f(x) = \mathbb{E}^x[f(X_t)]$. This is simply the expected value of $f(X_t)$ is your process is started at $x$ at time $0$.

(3) $f(x)$ is deterministic and so $E^x[f(x)] = f(x) E^x[1] = f(x)$.

Rhys Steele
  • 19,671
  • 1
  • 19
  • 50
  • Thanks for the answer. About the third point, so why $\mathbb{E}^x(B_t-x)=0$. We know that $\mathbb{E}^x(B_t)=0$. Why $\mathbb{E}^x(x)=0$. Thanks in advance. – Denis Jun 01 '20 at 15:56
  • $\mathbb{E}^x$ is the expectation with respect to the law of the Brownian motion started from $x$ and not from $0$. This means that $\mathbb{E}^x[B_t] = x$. – Rhys Steele Jun 01 '20 at 16:01
  • We shouldn't have $f \in C^2_0$ ? – Denis Aug 18 '20 at 07:49
  • 1
    At which stage? If you are only interested in the standard Brownian motion on $\mathbb{R}$ then $D(A) = C_0^2$. This is a special case of what I wrote in point (1) since $D(A)$ is only a subspace of $C_0$. However $P_t$ is allowed to act on an arbitrary $f \in C_0$ without the assumption that $f \in C^2$. – Rhys Steele Aug 18 '20 at 13:59
  • @RhysSteele what does it mean that "if $X$ is your process then $f(X)$ is the process given at time t"? E.g. for Brownian motion, $B_t$ is itself the position at time t. – Kevin Nov 21 '23 at 20:06
  • 1
    @Kevin I think you are grouping words in that sentence incorrectly. The wording was unnecessarily confusing so I changed it, hopefully the edited answer is clearer. – Rhys Steele Nov 21 '23 at 21:01