Let $(L\_t)\_{0\leq t \leq 1}$ be the local time at $0$ of a brownian bridge. Let $(T\_l)\_{0\leq l \leq L\_1}$ be its generalized inverse (as in $T\_l := \inf\{t\geq 0 : L\_t \geq l\}$). What is the joint distribution of
$$(T\_{L\_1/4}, T\_{L\_1/2}, T\_{3L\_1/4}) ?$$


In particular, I am interested in the joint distribution of $(T\_{L\_1/2}, T\_{L\_1/4} - T\_{L\_1/2}+T\_{3L\_1/4})$, which I conjecture to have both marginals uniform in $[0,1]$.



Quartiles of local time of brownian bridge at origin
*Joint Law of Brownian Motion and Its Local Time*.


In what follows, we will use the joint law of Brownian motion $B\_t$ with $B\_0=x \in \mathbb{R}$ and its local time accumulated at zero $\ell\_t$, which can be found, e.g., in (1.3.8) on pg. 140 of:


A. Borodin and P. Salminen (1996). **Handbook of Brownian Motion**, Springer. 


This formula states that: 
$$
\mathbb{P}(\ell\_t \in dy , B\_t \in dz) = p(x,y,z,t) dy dz \tag{1}
$$ where we have introduced the joint density function:
$$
p(x,y,z,t) = \frac{1}{t \sqrt{2 \pi t}} (|x| + y + |z|) \exp\left(-\frac{(|x| + y + |z|)^2}{2t} \right) 
$$ 


*Law of $T\_{L\_1/2}$.*


Let $X\_t$ denote a Brownian bridge between $X\_0=a$ and $X\_1=b$, and let $L\_t$ be its local time accumulated at zero over $[0,t]$. Given $ t\_1 \in (0,1)$, note that the event $\{ T\_{L\_1/2}>t\_1 \}$ is the same as the event $\{ L\_{t\_1} < L\_1/2 \}$. To determine this latter probability, we use the law of the local time of a Brownian bridge. This is obtained from (1) by conditioning on the endpoints of the Brownian motion. Doing this we get:
\begin{align\*}
\mathbb{P}^{a,b}\left( T\_{L\_1/2} > t\_1 \right)&=
\mathbb{P}^{a,b}\left( L\_{t\_1} < \frac{L\_1}{2} \right) \\
&=Z^{-1} \int\_{R} p(a,y\_1,z\_1,t\_1) p(z\_1,y\_2,b,1-t\_1) dy\_1 dy\_2 dz\_1
\end{align\*} where $R$ is the region:
$$
R=\left\{ (y\_1,y\_2,z\_1) \in \mathbb{R}^3: \begin{array}{c} 0 \le y\_1 \le (y\_1+y\_2)/2 \\ 0 \le y\_2 \le \infty \end{array} \right\}
$$ and $Z$ is a normalization constant:
$$
Z=\int\_{-\infty}^{\infty} \int\_0^{\infty} \int\_0^{\infty} p(a,y\_1,z\_1,t\_1) p(z\_1,y\_2,b,t\_2-t\_1) dy\_1 dy\_2 dz\_1
$$ One can check from this formula that when $a=b=0$, the law of $T\_{L\_1/2}$ is indeed uniform over $(0,1)$, as conjectured by the OP.


*Joint Law of $(T\_{L\_1/4}, T\_{L\_1/2}, T\_{3L\_1/4})$.*


Similar to the previous calculation, but this time given $0 < t\_1 < t\_2 < t\_3 < 1$, 
\begin{align\*}
 \mathbb{P}^{a,b}&\left(T\_{L\_1/4} > t\_1 , T\_{L\_1/2}>t\_2, T\_{3L\_1/4} > t\_3 \right) \\ &= \mathbb{P}^{a,b}\left(L\_{t\_1} < \frac{L\_1}{4} , L\_{t\_2} < \frac{L\_1}{2} , L\_{t\_3} < \frac{3 L\_1}{4} \right) \\
 &= 
Z^{-1} \int\_{R} p(a,y\_1,z\_1,t\_1) p(z\_1,y\_2,z\_2,t\_2-t\_1) \\
& \times p(z\_2,y\_3,z\_3,t\_3-t\_2) p(z\_3,y\_4,b,1-t\_3) dz\_1 dz\_2 dz\_3 dy\_1 dy\_2 dy\_3 dy\_4 
\end{align\*} where $R$ is the region:
$$
R=\left\{ (y\_1,y\_2,y\_3,y\_4,z\_1,z\_2,z\_3) \in \mathbb{R}^7: 
\begin{array}{c}
0 \le y\_1 \le S/4 \\ 0 \le y\_2 \le S/2 \\ 0 \le y\_3 \le 3 S/4 \\ 0 \le y\_4 \le \infty \\ S= y\_1+y\_2+y\_3+y\_4 \end{array} \right\}
$$ and $Z$ is a normalization constant.


*Law of $(T\_{L\_1},T\_{L\_1/4}-T\_{L\_1/2}+ T\_{3L\_1/4})$.*


One can use the joint CDF given above to compute this law. For $a=b=0$, the OP's conjecture that the marginals are uniform seems to be correct.


