3

Let $(X_t)$ an diffusion Itô process, i.e. a solution of $$dX_t=b(X_t)dt+\sigma (X_t)dB_t.$$ The infinitesimal generator of $(X_t)$ is $$Af(x)=\lim_{t\to 0^+}\frac{\mathbb E^x[f(X_t)]-f(x)}{t},$$ where $\mathbb E^x$ is the expectation wrt $\mathbb P^x$.

Q1) What represent exactly $Af(x)$ for $X_t$ ? For example, for a Brownian motion, if $f$ is $C^2$ then $$A f(x)=\frac{1}{2}\Delta f(x).$$

But I don't really understand which information does $A$ give is. Is it a sort of derivative of $X_t$ ?

Q2) What is exactely the measure $\mathbb P^x$ ? I know it is $\mathbb P^x\{X_t\in A\}=\mathbb P(\{X_t\in A\}\mid \{X_0=x\}),$ But does it mean that on $(\Omega ,\mathcal F,\mathbb P^x)$ we have that $\mathbb P\{X_0=x\}=1$ ? (i.e. is deterministic).

user621345
  • 674
  • 1
    (1) Intuitively, if $X_t$ is a deterministic function of $t$, $Af(x)$ is exactly $\frac{\rm d}{{\rm d}t}|_{t=0}f(X_t)$. When $X_t$ is an Ito process, $Af(x)$ measures how "fast" $f(X_t)$ changes with respect to $t$ in the sense of expectation. (2) $\mathbb{P}^x$ is a conditional probability, which conditions on $X_0=x$. That is, $X_t$ has a fixed, deterministic starting point $x$. In this sense, yes, $\mathbb{P}\left{X_0=x\right}=1$, which, rigorously, should be $\mathbb{P}\left(\left{X_0=x\right}|\left{X_0=x\right}\right)=1$. This is trivially true. – hypernova Jan 03 '19 at 13:49
  • 1
    You might want to take a look at this question and this question – saz Jan 03 '19 at 13:59
  • @saz : Very nice links. Thank you for this contribution, it's very helpful. I'll probably comment them in the future. – user621345 Jan 03 '19 at 23:02

1 Answers1

2

Q1) It seems that this link could help.

Q2) Also see this link for a formal definition of conditional probability. Even, the existence of regular conditional probabilities is a non-trivial fact even for simple cases like $\mathbb{R}^2$ equipped with Borel sigma algebra.

Aragon
  • 46