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Infinitesimal generator

  1. Mar 12, 2008 #1
    Hi folks,

    I have a question concerning the infinitesimal generator of a stochastic ;process, more specificaly of Brownian motion.

    Let [itex]X_t[/itex] be a stochastic process, then the infinitesimal generator A acting on nice (e.g. bounded, twice differentiable) functions f is defined by
    [tex]
    (Af)(x)=\lim_{t\to 0}{\frac{1}{t}\left[E_x\left[X_t\right]-1\right]}
    [/tex]

    For (one-dimensional) Brownian motion this turns out to be just the second derivative operator.

    What happens however, if I were to consider reflected Brownian motion (reflected at zero). In distribution this process is equal to [itex]|B_t|[/itex] where [itex]B_t[/itex] is a (non-reflected) Brownian motion. My feeling is that for [itex]x \neq 0[/itex] the infintesimal generator should still be the second derivative, but what happens at x=0?

    Unfortunately I couldn't find this in any textbook.

    Any help appreciated:smile:

    -Pere
     
    Last edited: Mar 12, 2008
  2. jcsd
  3. Mar 13, 2008 #2
    According to http://www.sciencedirect.com/scienc...d=994540&md5=5e55acb14a85bd700328530ba37f2925

    the infinitesimal generator of reflected Brownian motion on a finite intervall [itex][0,\gamma][/itex] (reflected at both ends) is the "neumann laplacian" [itex]\mathfrak{L}_N[/itex], defined as "the closure
    of the operator [itex]L=\frac{1}{2}\frac{d^2}{dx^2}[/itex] in [itex][0,\gamma][/itex] on the domain [itex]\{u\in C^2([0,\gamma]):u'(0)=u'(\gamma)=0\}[/itex].

    I am not sure if I understand this. Assuming I have a function u in this set, what would [itex](\mathfrak{L}_Nu)0[/itex] be...? Just [itex]\frac{1}{2}u''(0)[/itex]...?

    Probably so, but what about functions f whose first derivative does not vanish at x=0 ...?

    Thanks
    -Pere
     
    Last edited: Mar 13, 2008
  4. Mar 13, 2008 #3

    gel

    User Avatar

    As well as the differential generator, you need to know its domain. The domain for Brownian motion contains all (*) twice continuously differentiable functions on R, and its generator is [itex](1/2)d^2/dx^2[/itex].

    Reflected Brownianian motion has all (*) twice continuously differentiable functions on[0,infinity) with zero derivative at zero in its domain. Then the generator is again [itex](1/2)d^2/dx^2[/itex].

    Essentially, any function on [0,infin) can be extended to a symmetric function on R by reflecting about 0, but in order for this to give a differentiable function, you need u'(0)=0.



    (*) satisfying nice boundary conditions, such as bounded support.
     
  5. Mar 13, 2008 #4

    gel

    User Avatar

    to be more specific...
    Yes.
    They're not in the domain of the generator.

    I think you could show that f is in the domain for RBM if and only if extending it to R by reflecting about 0 gives a function in the domain for BM.
     
    Last edited: Mar 13, 2008
  6. Mar 14, 2008 #5
    Hi gel, thanks for your answer.

    OK that makes sense to me. I have to think about it, there is still a question in my mind, but I cannot yet really pin it down.
     
  7. Mar 18, 2008 #6
    Ok I now know what I want to ask:smile:

    Assume I want to use the infinitesimal generator to calculate the stationary distribution of reflected Brownian Motion with negative drift [itex]-\mu,\quad\mu>0[/itex].
    Then the infinitesimal generator is given by

    [tex]
    (\mathfrak{L}f)(x)=\frac{1}{2}\frac{d^2}{dx^2}f(x)-\mu \frac{d}{dx}f(x)
    [/tex]
    with domain as described above:

    [tex]
    D_{\mathfrak{L}}=\{f\in C_b^2(\mathbb{R}^+):\lim_{x\to 0^+}{f'(x)}=0\}
    [/tex]

    The adjoint of the generator is given by
    [tex]
    (\mathfrak{L}^*f)(x)=\frac{1}{2}\frac{d^2}{dx^2}f(x)+\mu \frac{d}{dx}f(x)
    [/tex]

    and the stationary measure [itex]d\pi(x)=\rho_\infty(x)dx[/itex] satisfies [itex]A^*\rho_\infty=0[/itex].

    However this is the very same equation as for the stationary density of non-reflecting Brownian motion with constant drift, which is zero. So do I have to impose some boundary conditions on the stationary density? My feeling is the solution should be
    [tex]
    \rho_\infty(x)=\frac{1}{2\mu}e^{-2\mu x}\mathbb{I}_{\{x\geq 0\}}
    [/tex]

    Moreover, shouldn't
    [tex]\int_{\mathbb{R}^+}{d\pi(x)(Af)(x)}=\int_{\mathbb{R}^+}{d\pi(x)f(x)}[/tex]
    hold for all f in the domain of the generator... it seems not to be the case for my suggested solution...I computed the left hand side for some easy function f of the form [itex]f(x)=e^{-\alpha x},\quad\alpha >0[/itex] and it gives zero while the right hand side does not ...

    Is there a textbook containing a treatment of the boundary behavior of one-dimensional diffusions.. there is something in Breiman and in Freedman as well but I don't find it particularly illuminating. Then there is of course the paper by Harrison on general diffusions with boundary conditions, but I'd like to first understand this easy case:smile:

    Thanks

    -Pere
     
    Last edited: Mar 18, 2008
  8. Mar 19, 2008 #7

    gel

    User Avatar

    You can calculate the adjoint (and the domain) with a bit of integration by parts.
    I'm not sure what a good reference is though.

    (got my post just before this one totally wrong, so I deleted it)
     
    Last edited: Mar 20, 2008
  9. Mar 20, 2008 #8

    gel

    User Avatar

    If f'(0)=0, and p is a twice continuously differentiable function on R+
    then you can calculate the adjoint [itex]\mathfrak{L}^*p[/itex]

    [tex]
    \int_0^\infty (\mathfrak{L}^*p)f\,dx = \int_0^\infty p\mathfrak{L}f\,dx
    = \int_0^\infty\left( -\frac{1}{2}p'(x)f'(x)-\mu p(x)f'(x)\right)\,dx
    [/tex][tex]
    = \frac{1}{2}p'(0)f(0)+\mu p(0)f(0)+\int_0^\infty\left(\frac{1}{2}p''(x)+\mu p'(x)\right)f(x)\,dx
    [/tex]

    So,

    [tex]
    \mathfrak{L}^*p(x)=\left(\frac{1}{2}p'(0)+\mu p(0)\right)\delta(x)+\frac{1}{2}p''(x)+\mu p'(x)
    [/tex]

    where [itex]\delta(x)[/itex] is the Dirac delta function.
    Restricting to the domain

    [tex]
    D_{\mathfrak{L}^*}=\{p\in C_b^2(\mathbb{R}^+):p'(0)=-2\mu p(0)\}
    [/tex]

    then the adjoint is given by [itex]\mathfrak{L}^*p=\frac{1}{2}p''+\mu p'[/itex].

    The solution to [itex]\mathfrak{L}^*p=0[/itex] is [itex]p(x)= c \exp(-2\mu x)[/itex] as you suggest.
     
    Last edited: Mar 20, 2008
  10. Mar 20, 2008 #9

    gel

    User Avatar

    There's no reason the right hand side should give zero (that would imply that [itex]\pi=0[/itex]).
    If X is a diffusion with the given generator, then you should have
    [tex]
    \frac{d}{dt}E(f(X_t)) = E(\mathfrak{L}f(X_t)).
    [/tex]
    Using the stationary distribution, this would give
    [tex]
    \int \mathfrak{L}f\,d\pi = 0
    [/tex]
    as you in fact found.

    More generally
    [tex]
    f(X_t)-\int_0^t\mathfrak{L}f(X_s)\,ds
    [/tex]
    is a martingale, which is a very useful alternative way of characterizing the generator.
     
    Last edited: Mar 20, 2008
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