What is the infinitesimal generator of reflected Brownian motion?

In summary, the conversation discusses the infinitesimal generator of a stochastic process, specifically Brownian motion and reflected Brownian motion. The infinitesimal generator is defined as the limit of the expected value of the process divided by time, and for Brownian motion it is equal to the second derivative operator. For reflected Brownian motion, the generator is the "Neumann Laplacian" with a specific domain. The conversation also discusses the stationary distribution and its relation to the adjoint of the generator. The solution for the stationary density of reflected Brownian motion with negative drift is given as \rho_\infty(x)=\frac{1}{2\mu}e^{-2\mu x}\mathbb{I}_{\{x\geq
  • #1
Pere Callahan
586
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
 
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  • #2
According to http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V1B-4FV9J4V-1&_user=994540&_rdoc=1&_fmt=&_orig=search&_sort=d&view=c&_acct=C000050024&_version=1&_urlVersion=0&_userid=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
 
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  • #3
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.
 
  • #4
to be more specific...
Pere Callahan said:
what would [itex](\mathfrak{L}_Nu)0[/itex] be...? Just [itex]\frac{1}{2}u''(0)[/itex]...?
Yes.
Pere Callahan said:
Probably so, but what about functions f whose first derivative does not vanish at x=0 ...?
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.
 
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  • #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.
 
  • #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
 
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  • #7
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)
 
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  • #8
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.
 
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  • #9
Pere Callahan said:
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 ...

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.
 
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FAQ: What is the infinitesimal generator of reflected Brownian motion?

What is an infinitesimal generator?

An infinitesimal generator is a mathematical concept used in the study of stochastic processes. It represents the operator that controls the evolution of a stochastic process over time.

How does an infinitesimal generator work?

An infinitesimal generator is typically represented by a matrix, and its action on a function is equivalent to taking the derivative of that function. It governs the transition rates between different states in a stochastic process.

What is the importance of infinitesimal generators in mathematics?

Infinitesimal generators play a crucial role in the study of stochastic processes and their applications in various fields such as physics, engineering, and finance. They provide a mathematical framework for understanding the behavior of systems that involve randomness.

What are some real-world applications of infinitesimal generators?

Infinitesimal generators are used in various fields, including physics, biology, economics, and finance. They are used to model the behavior of complex systems, such as stock prices, population dynamics, and chemical reactions.

What are some common misconceptions about infinitesimal generators?

One common misconception is that infinitesimal generators only apply to continuous-time processes. However, they can also be used for discrete-time processes. Another misconception is that infinitesimal generators are limited to Markov processes, but they can also be applied to non-Markov processes. Lastly, it is important to note that infinitesimal generators are not physical generators, but rather mathematical operators.

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