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Let $B_t$ be the standard Brownian motion and $L_t^a$ be the local time at level $a$. It is known that the joint-density of $(L_t^0,B_t)$ is

$$ P\left(B_t\in d y, L_t^0\in d v\right) = \frac{|y|+v}{\sqrt{2\pi t^3}}\exp\left(-\frac{(|y|+v)^2}{2 t}\right) 1_{[0,\infty)}(v)1_{(-\infty,\infty)}(y) d y d v, $$

see, e.g., p. 181 of Chung & Williams.

The problem is whether anyone knows the joint density of $(L_t^a,B_t)$ for $a\ne 0$?

Thank you very much for any hints.

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    $\begingroup$ Can't you decompose according to the first hitting time of $a$? $\endgroup$ Commented May 2, 2015 at 21:48
  • $\begingroup$ Thanks Professor Zeitouni. This question might have been asked and solved long ago. Do you have any references in mind? Otherwise, one may study the joint law of $(L_t^a,B_t,T_a)$, where $T_a$ is the first hitting time of $a$. $\endgroup$
    – Anand
    Commented May 3, 2015 at 15:13
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    $\begingroup$ I don't have a reference, only a suggestion that the result you want follows from the result you quote by using the Markov property - decompose at the first hitting time of a. $\endgroup$ Commented May 3, 2015 at 18:27

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In "Trivariate Density of Brownian Motion, Its Local and Occupation Times, with Application to Stochastic Control"

at section 4 "The trivariate density with nonzero initial condition", they obtain the joint density of $B_{t}^{x},L_{t}$ starting from $x\neq 0$ (after you integrate the marginal for occupation time $\Gamma_{t}$). So then you can use translate by $x$

$$L^0(t,B^{x}) =\lim_{\varepsilon\downarrow 0} \frac{1}{2\varepsilon} \int_0^t 1_{\{ 0- \varepsilon < B_s^{x} < 0+\varepsilon \}}=\lim_{\varepsilon\downarrow 0} \frac{1}{2\varepsilon} \int_0^t 1_{\{ -x- \varepsilon < B_s^{0} < -x+\varepsilon \}}=L^{-x}(t,B^{0}).$$

and study $(B^{0}_{t}+x\in dy ,L^{-x}_{t})$ in terms $(B^{x}_{t},L^0_{t})$.

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