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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). 3 added 950 characters in body; added 4 characters in body 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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