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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 $L(t,x)$ 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 $x>0$, thus one getsx>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, writing $g(t,\cdot)$ for the centered Gaussian density with variance every $t$ 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 $s$, one gets for every positive $x$, $$P(L(t,x)\in\mathrm{d}u)=p_{t,x}\delta_0(\mathrm{d}u)+h_{t,x}(u)\mathrm{d}u $$ where, for every positive $u$, $$ h_{t,x}(u)=4\int_0^tg(t-s,u)\mathrm{d}s\int_x^{+\infty}\partial_sg(s,y)\mathrm{d}y, $$ and $$ p_{t,x}=P(L(t,x)=0)=P(S_t < x)=1-2P(W_t > x)=1-2\int_x^{+\infty}g(t,y)\mathrm{d}y. $h_{t,x}(u)=4\int_0^tg(t-s,u)\mathrm{d}s\int_{|x|}^{+\infty}\partial_sg(s,y)\mathrm{d}y, $$

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).
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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 $L(t,x)$ is twice the density of $W_t$ restricted to $(0,+\infty)$.

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 $W$. W$ (this is the strong Mrkov property of $W$ at time $\tau_x$). The distribution of $\tau_x$ is known hence this givessince $[\tau_x\le t]=[S_t\ge x]$ for every $x>0$, thus one gets, in principle, the distribution of $L(t,x)$ as a barycenter of the distributions of the 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.

Rather

Unless I am mistaken, writing $g(t,\cdot)$ for the centered Gaussian density with variance $t$ and $\partial_s$ for the partial derivative with respect to $s$, one gets for every positive $x$, $$P(L(t,x)\in\mathrm{d}u)=p_{t,x}\delta_0(\mathrm{d}u)+h_{t,x}(u)\mathrm{d}u $$ where, for every positive $u$, $$ h_{t,x}(u)=4\int_0^tg(t-s,u)\mathrm{d}s\int_x^{+\infty}\partial_sg(s,y)\mathrm{d}y, $$ and $$ p_{t,x}=P(L(t,x)=0)=P(S_t < x)=1-2P(W_t > x)=1-2\int_x^{+\infty}g(t,y)\mathrm{d}y. $$ In fact, rather than considering a fixed time $t$, one often looks at the distribution of $L(U,x)$, L(Z,x)$, where $U$ 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).
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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 $P(L(t,0)\ge z)=P(S_t\ge z)=2P(W_t\ge z)$ by Désiré André's reflexion principle.

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 $W$. The distribution of $\tau_x$ is known hence this gives, in principle, the distribution of $L(t,x)$ as a barycenter of the distributions of the 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.

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