Brownian local time density

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

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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 $L$.

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@zhoraster: +1. And $y=z$? – Did Feb 26 '11 at 15:42
@Didier thanks, indeed. – zhoraster Feb 26 '11 at 16:44

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 Markov 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 much further than writing the density of $L(t,x)$ 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 a set of lecture notes on Brownian motion by Peter Mörters and Yuval Peres, available here (chapter 2).

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have you ever seen: math.stackexchange.com/questions/682825/… I found statistics.berkeley.edu/sites/default/files/tech-reports/… which is about the local time of a Brownian Bridge but I cannot see how local time can tell us anything about about time spent above 0 except the mean. – lost1 Jul 18 '14 at 16:08
@lost1 No, the local time at 0 does not say anything about the distribution of the time spent above 0. Is this your question? (And no I had not seen the MSE question, whose answer is entirely standard.) – Did Jul 18 '14 at 20:20
Please have look at this link: math.stackexchange.com/questions/682825/… it is not my question but I did not find answer on Google. I thought if you integrate $L^x_1$ against the indicator function $1_{x>0}$, you get the time spent above the 0 for a Brownian motion. Can probably do something like this Brownian bridge too, but I do not think the representation is particularly useful. – lost1 Jul 18 '14 at 22:10
This paper by Takacs gives an explicit density for $L(1,x)$ being $(2\Phi(x+u)-1)du$. – petrelharp Sep 9 '15 at 23:34