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
 A: 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 $L$.
A: 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).
A: From Borodin and Salminen (2002, Eqn. (1.3.4), p. 15) the probability that the BLT (Brownian local time) stands close to a given level $a$, for a time $dy$ is given by the density:
$$g(y; t, a) = \sqrt{\frac{2}{\pi t}}e^\frac{-(y + |a|)^2}{2t}$$
