Brownian local time density - MathOverflow

Brownian local time density

2011-02-26T05:26:38Z

Dear All,

I am not a mathematican, please be patient if I ask something in a not appropriate way!

Let we suppose a Brownian motion with inital value of W(0)=0, and we look its possible realizations on time interval [0,T).

MY QUESTION: What is the probability density function (with closed form) of local time related to a given time (t) and value (x) (let we note it as L(x,t)) ?

The intuitive meaning of L(x,t): the probability density that the realization is "staying" at point x for time lasting t.

I marked "staying" because Brownian motion continuous but non differentiating function, it does not stay but cross the points. There is no probability of a given point but there is a probability density of crossings the given point on interval [0,T). Which can be computed as integral of Brownian motion with dirac delta (if I am not wrong).

I simulated the process by Matlab and based on the results it seems for me, that pdf consist of two parts.

If we look L(t,x) as a slice at a given x (I mean we slice the 2 dimensional "joint" pdf at a given x) we will get the "time profil" of L(x,t). First part is a profound density for t=0 (meaning the possibility that it never cross x on interval [0,t)). second part is similar to a half gaussian pdf (decreasing slope).

I have read articles related local time, but I did not find closed form representation of pdf of L(x,t) but I am sure that there should be, because this is relative simple problem in SDE problem. Maybe the first article of Levy in 1939 can have the formula (but I do not have read the article and I do not read French).

Thanks in advance. Tomi

Answer by Didier Piau for Brownian local time density

2011-02-26T08:12:16Z

The case $x=0$ can be deduced from the identity in law due to Paul Lévy between $(S_t-W_t,S_t)$ and $(|W_t|,L(t,0))$, where $S_t$ is the maximum of $W$ on $[0,t)$. Hence $P(L(t,0)\ge z)=P(S_t\ge z)$ and, since $P(S_t\ge z)=2P(W_t\ge z)$ by Désiré André's reflexion principle, the density of the distribution of $L(0,x)$ is twice the density of $W_t$ restricted to $(0,+\infty)$. That is, writing $g(t,\cdot)$ for the centered Gaussian density with variance $t$, $$ P(L(t,0)\in\mathrm{d}u)=2g(t,u)\mathbf{1}_{u > 0}\mathrm{d}u. $$ If $x\ne0$, $L(t,x)$ is distributed like $L((t-\tau_x)^+,0)$ where $\tau_x$ is independent of $W$ and distributed like the first hitting time of $x$ by $W$ (this is the strong Mrkov property of $W$ at time $\tau_x$). The distributions of $\tau_x$ and $\tau_{-x}$ coincide and the distribution of $\tau_x$ is known since $[\tau_x\le t]=[S_t\ge x]$ for every $x>0$. This yields, in principle, the distribution of $L(t,x)$ as a barycenter of distributions of random variables $L(s,0)$ for $s$ in $[0,t)$ but I doubt one can go further than writing the density of $L(t,x)$ more nicely than as an integral.

Unless I am mistaken, for every $x\ne0$, one gets $$ P(L(t,x)\in\mathrm{d}u)=p_{t,x}\delta_0(\mathrm{d}u)+h_{t,x}(u)\mathbf{1}_{u > 0}\mathrm{d}u, $$

where $$p_{t,x}=P(L(t,x)=0)=P(S_t < |x|)=1-2P(W_t > |x|)=2\int_0^{|x|}g(t,y)\mathrm{d}y. $$ and, writing $\partial_s$ for the partial derivative with respect to $s$, for every positive $u$, $$h_{t,x}(u)=4\int_0^tg(t-s,u)\mathrm{d}s\int_{|x|}^{+\infty}\partial_sg(s,y)\mathrm{d}y, $$

In fact, rather than considering a fixed time $t$, one often looks at the distribution of $L(Z,x)$, where $Z$ is a random time independent on $W$ and exponentially distributed. In other words, one considers the Laplace transform (with respect to the time argument) of the random function $L(\cdot,x)$.

All this is done in many places, see for example the set of lecture notes on Brownian motion by Peter Mörters and Yuval Peres available here (chapter 2).

Answer by zhoraster for Brownian local time density

2011-02-26T14:59:31Z

Another useful way to study local time is related to the very useful occupation formula (its meaning is obvious if you think a little bit): $\int_0^t g(W_s) ds = \int_{\mathbb R} g(x) L(x,t) dx $.

Putting $g(x) = e^{izx}$: $$ \int_{\mathbb R} e^{izx} L(x,t) dx = \int_0^t e^{iz W_s}ds. $$ Now the lhs is the Fourier transform of $L$; inverting it, we get $$ L(x,t) = \frac{1}{2\pi} \int_{\mathbb R} e^{-izx} \int_0^t e^{iz W_s} ds\, dz = \frac{1}{2\pi} \int_{\mathbb R} \int_0^t e^{iz (W_s-x)} ds\, dz. $$ From here one can e.g. more or less easily find the moments of $L$.