What is the infinitesimal generator of reflected Brownian motion?

[Pere Callahan]

Hi folks,

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

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

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 |B_t| where B_t is a (non-reflected) Brownian motion. My feeling is that for x \neq 0 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

-Pere

 

[Pere Callahan]

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 [0,\gamma] (reflected at both ends) is the "neumann laplacian" \mathfrak{L}_N, defined as "the closure
of the operator L=\frac{1}{2}\frac{d^2}{dx^2} in [0,\gamma] on the domain \{u\in C^2([0,\gamma]):u&#039;(0)=u&#039;(\gamma)=0\}.

I am not sure if I understand this. Assuming I have a function u in this set, what would (\mathfrak{L}_Nu)0 be...? Just \frac{1}{2}u&#039;&#039;(0)...?

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

Thanks
-Pere

 

[gel]

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 (1/2)d^2/dx^2.

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 (1/2)d^2/dx^2.

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.

 

[gel]

to be more specific...

what would (\mathfrak{L}_Nu)0 be...? Just \frac{1}{2}u&#039;&#039;(0)...?

Yes.

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.

 

[Pere Callahan]

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.

 

[Pere Callahan]

Ok I now know what I want to ask

Assume I want to use the infinitesimal generator to calculate the stationary distribution of reflected Brownian Motion with negative drift -\mu,\quad\mu&gt;0.
Then the infinitesimal generator is given by

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

<br /> D_{\mathfrak{L}}=\{f\in C_b^2(\mathbb{R}^+):\lim_{x\to 0^+}{f&#039;(x)}=0\}<br />

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

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

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
<br /> \rho_\infty(x)=\frac{1}{2\mu}e^{-2\mu x}\mathbb{I}_{\{x\geq 0\}}<br />

Moreover, shouldn't
\int_{\mathbb{R}^+}{d\pi(x)(Af)(x)}=\int_{\mathbb{R}^+}{d\pi(x)f(x)}
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 f(x)=e^{-\alpha x},\quad\alpha &gt;0 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

Thanks

-Pere

 

[gel]

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)

 

[gel]

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

<br /> \int_0^\infty (\mathfrak{L}^*p)f\,dx = \int_0^\infty p\mathfrak{L}f\,dx<br /> = \int_0^\infty\left( -\frac{1}{2}p&#039;(x)f&#039;(x)-\mu p(x)f&#039;(x)\right)\,dx<br /><br /> = \frac{1}{2}p&#039;(0)f(0)+\mu p(0)f(0)+\int_0^\infty\left(\frac{1}{2}p&#039;&#039;(x)+\mu p&#039;(x)\right)f(x)\,dx<br />

So,

<br /> \mathfrak{L}^*p(x)=\left(\frac{1}{2}p&#039;(0)+\mu p(0)\right)\delta(x)+\frac{1}{2}p&#039;&#039;(x)+\mu p&#039;(x)<br />

where \delta(x) is the Dirac delta function.
Restricting to the domain

<br /> D_{\mathfrak{L}^*}=\{p\in C_b^2(\mathbb{R}^+):p&#039;(0)=-2\mu p(0)\}<br />

then the adjoint is given by \mathfrak{L}^*p=\frac{1}{2}p&#039;&#039;+\mu p&#039;.

The solution to \mathfrak{L}^*p=0 is p(x)= c \exp(-2\mu x) as you suggest.

 

[gel]

Moreover, shouldn't
\int_{\mathbb{R}^+}{d\pi(x)(Af)(x)}=\int_{\mathbb{R}^+}{d\pi(x)f(x)}
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 f(x)=e^{-\alpha x},\quad\alpha &gt;0 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 \pi=0).
If X is a diffusion with the given generator, then you should have
<br /> \frac{d}{dt}E(f(X_t)) = E(\mathfrak{L}f(X_t)).<br />
Using the stationary distribution, this would give
<br /> \int \mathfrak{L}f\,d\pi = 0<br />
as you in fact found.

More generally
<br /> f(X_t)-\int_0^t\mathfrak{L}f(X_s)\,ds<br />
is a martingale, which is a very useful alternative way of characterizing the generator.