Infinitesimal generator : what is it exactly?












1














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 $$Af(x)=lim_{tto 0^+}frac{mathbb E^x[f(X_t)]-f(x)}{t},$$
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 $mathbb P^x{X_tin A}=mathbb P({X_tin A}mid {X_0=x}),$ 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).






probability measure-theory







share|cite|improve this question


















  • 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, $mathbb{P}left{X_0=xright}=1$, which, rigorously, should be $mathbb{P}left(left{X_0=xright}|left{X_0=xright}right)=1$. This is trivially true.
    – hypernova
    yesterday








  • 1




    You might want to take a look at this question and this question
    – saz
    yesterday












  • @saz : Very nice links. Thank you for this contribution, it's very helpful. I'll probably comment them in the future.
    – NewMath
    yesterday


















1














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 $$Af(x)=lim_{tto 0^+}frac{mathbb E^x[f(X_t)]-f(x)}{t},$$
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 $mathbb P^x{X_tin A}=mathbb P({X_tin A}mid {X_0=x}),$ 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).






probability measure-theory







share|cite|improve this question


















  • 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, $mathbb{P}left{X_0=xright}=1$, which, rigorously, should be $mathbb{P}left(left{X_0=xright}|left{X_0=xright}right)=1$. This is trivially true.
    – hypernova
    yesterday








  • 1




    You might want to take a look at this question and this question
    – saz
    yesterday












  • @saz : Very nice links. Thank you for this contribution, it's very helpful. I'll probably comment them in the future.
    – NewMath
    yesterday
















1












1








1


1





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 $$Af(x)=lim_{tto 0^+}frac{mathbb E^x[f(X_t)]-f(x)}{t},$$
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 $mathbb P^x{X_tin A}=mathbb P({X_tin A}mid {X_0=x}),$ 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).






probability measure-theory







share|cite|improve this question













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 $$Af(x)=lim_{tto 0^+}frac{mathbb E^x[f(X_t)]-f(x)}{t},$$
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 $mathbb P^x{X_tin A}=mathbb P({X_tin A}mid {X_0=x}),$ 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).






probability measure-theory




probability measure-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked yesterday









NewMath

3338




3338








  • 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, $mathbb{P}left{X_0=xright}=1$, which, rigorously, should be $mathbb{P}left(left{X_0=xright}|left{X_0=xright}right)=1$. This is trivially true.
    – hypernova
    yesterday








  • 1




    You might want to take a look at this question and this question
    – saz
    yesterday












  • @saz : Very nice links. Thank you for this contribution, it's very helpful. I'll probably comment them in the future.
    – NewMath
    yesterday
















  • 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, $mathbb{P}left{X_0=xright}=1$, which, rigorously, should be $mathbb{P}left(left{X_0=xright}|left{X_0=xright}right)=1$. This is trivially true.
    – hypernova
    yesterday








  • 1




    You might want to take a look at this question and this question
    – saz
    yesterday












  • @saz : Very nice links. Thank you for this contribution, it's very helpful. I'll probably comment them in the future.
    – NewMath
    yesterday










1




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, $mathbb{P}left{X_0=xright}=1$, which, rigorously, should be $mathbb{P}left(left{X_0=xright}|left{X_0=xright}right)=1$. This is trivially true.
– hypernova
yesterday






(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, $mathbb{P}left{X_0=xright}=1$, which, rigorously, should be $mathbb{P}left(left{X_0=xright}|left{X_0=xright}right)=1$. This is trivially true.
– hypernova
yesterday






1




1




You might want to take a look at this question and this question
– saz
yesterday






You might want to take a look at this question and this question
– saz
yesterday














@saz : Very nice links. Thank you for this contribution, it's very helpful. I'll probably comment them in the future.
– NewMath
yesterday






@saz : Very nice links. Thank you for this contribution, it's very helpful. I'll probably comment them in the future.
– NewMath
yesterday












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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.






share|cite|improve this answer








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    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.






    share|cite|improve this answer








    New contributor




    Aragon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      1














      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.






      share|cite|improve this answer








      New contributor




      Aragon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





















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        1








        1






        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.






        share|cite|improve this answer








        New contributor




        Aragon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        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.







        share|cite|improve this answer








        New contributor




        Aragon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        share|cite|improve this answer



        share|cite|improve this answer






        New contributor




        Aragon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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