# Infinitesimal generator

1. Mar 12, 2008

### 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
$$(Af)(x)=\lim_{t\to 0}{\frac{1}{t}\left[E_x\left[X_t\right]-1\right]}$$

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

Last edited: Mar 12, 2008
2. Mar 13, 2008

### Pere Callahan

According to http://www.sciencedirect.com/scienc...d=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 $\{u\in C^2([0,\gamma]):u'(0)=u'(\gamma)=0\}$.

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 $\frac{1}{2}u''(0)$...?

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

Thanks
-Pere

Last edited: Mar 13, 2008
3. Mar 13, 2008

### 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.

4. Mar 13, 2008

### gel

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
5. Mar 14, 2008

### 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.

6. Mar 18, 2008

### 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 $-\mu,\quad\mu>0$.
Then the infinitesimal generator is given by

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

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

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

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

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 $f(x)=e^{-\alpha x},\quad\alpha >0$ 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

Last edited: Mar 18, 2008
7. Mar 19, 2008

### 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)

Last edited: Mar 20, 2008
8. Mar 20, 2008

### gel

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

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

So,

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

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

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

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

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

Last edited: Mar 20, 2008
9. Mar 20, 2008

### gel

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
$$\frac{d}{dt}E(f(X_t)) = E(\mathfrak{L}f(X_t)).$$
Using the stationary distribution, this would give
$$\int \mathfrak{L}f\,d\pi = 0$$
as you in fact found.

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

Last edited: Mar 20, 2008
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