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For standard Brownian motion started at zero, the strict first-passage times form a subordinator in the level parameter $a$. The Strong Markov property gives independent increments and stationary increments; the strict inverse of the continuous running maximum has càdlàg paths. The Brownian first-passage Laplace transform gives Laplace exponent $\Phi (\lambda )=\sqrt{2\lambda }$, so the process is strictly stable of index $1/2$. Choosing the non-strict hitting times preserves each fixed-level law but generally loses right continuity at random levels, as in the fixed-level versus simultaneous Brownian passage-time equality.

For standard Brownian motion and $a>0$, $T_a=\inf \{t\ge 0:B_t\ge a\}$ is finite almost surely andThe Brownian reflection principle gives this distribution function. The Brownian scaling identity $T_{ca}\stackrel{d}{=}{c}^2{T}_a$ and the Strong Markov property show its probability distribution is a strictly stable distribution with index $1/2$.

Let $\tau$ be the first attainment of the maximum of standard Brownian motion on $[0,T]$, with $T>0$. Time reversal shows $\tau <T$ with probability one. Then $Y_t\le 0$ for $0\le t\le T-\tau$, an initial interval of positive length. Standard Brownian motion has probability zero of this property by the Brownian reflection principle, so $Y$ is not Brownian. Moreover $\tau$ is not a stopping time for the natural Brownian filtration: for $0<t<T$, $\mathbb{P}(\tau \le t\mid {\mathcal{F}}_t)=2\Phi ((M_t-B_t)/\sqrt{T-t})-1$, which is strictly between zero and one on $\{B_t<0\}$. The Strong Markov property is therefore inapplicable at this random time.

The occupation time of a measurable set $A$ through time $t$ is ${\int }_0^t{1}_{\{B_s\in A\}}ds$. The occupation-times formula expresses such integrals through local time of a semimartingale. Recurrence and the Strong Markov property imply that one-dimensional Brownian motion spends an unbounded total time in every nonempty open interval.

Reflecting a Brownian motion $B$ after a stopping time $T$ by ${\hat{B}}_t=2B_T-B_t$ for $t>T$, and leaving it unchanged for $t\le T$, preserves its path probability distribution. For finite $T$, the Strong Markov property and symmetry of the future Brownian motion prove this. When $T=\infty$, the path is left unchanged.

Let simple random walk travel from $a\ne z$ to its first visit to $z$, then stop on its subsequent first hit of $a$ in a finite connected unweighted loopless graph. Every directed edge $u\to v$ has expected traversal countThe Strong Markov property splits the count into the two killed legs. By killed-walk occupation voltage, their contributions are $v_{az}(u)$ and $v_{za}(u)$. Their Laplacians cancel, so the harmonic maximum principle on a finite graph makes the sum constant, with value $v_{az}(a)+v_{za}(a)=R_{\mathrm{eff}}(a,z)$. Summing over all directed edges recovers the commute time identity.

If $f$ is a continuous probability density on ${\mathbb{R}}^2$ and $B$ is planar Brownian motion, thenalmost surely. The density is bounded below on some disc, recurrence supplies infinitely many visits to a smaller concentric disc, and the Strong Markov property gives infinitely many visits of a fixed positive duration.

Conditionally on the entire past up to $t$, the future domains mapped and centered by $g_t-U_t$ have the original chain law and are independent of that past. For Schramm–Loewner evolution this follows from independent stationary Brownian motion increments and the composition rule for chordal Loewner driving functions. The Strong Markov property gives the version at almost surely finite stopping times.

For a simple symmetric random walk on the integers, let $T_k$ be the first time that $k$ distinct vertices have been visited. The visited set is an interval, and at $T_k$ the walk is at an endpoint. The Strong Markov property turns the time to add one vertex into gambler's ruin between the two immediately exterior vertices, starting one step from a boundary. The expected duration of symmetric gambler's ruin is therefore $k$. This also holds at $k=1$.

A planar Brownian motion exits a conformal image of a bounded planar domain in finite time almost surely. If the image lifetime were infinite, the inverse derivative is bounded away from zero on a small interior disc. Recurrence of planar Brownian motion and the Strong Markov property give infinite Brownian occupation time there, forcing the inverse conformal Brownian clock to diverge. It cannot exceed the finite original exit time. The expected image exit time can nevertheless be infinite.

For standard Brownian motion, set $H_a=\inf \{t:B_t\ge a\}$ and $T_a=\inf \{t:B_t>a\}$. The Strong Markov property and immediate positive excursions give $T_a=H_a$ almost surely for each fixed $a\ge 0$. However, these level-indexed processes are not indistinguishable stochastic processes. Almost surely the random level $A={\sup }_{t\le 1}{B}_t$ is positive and exceeds $B_1$. It is first attained before time one but is not exceeded until after time one, so $H_A<1<T_A$. The simultaneous-equality event therefore has probability zero. The strict passage process $T_a$ is the right-continuous choice needed for a subordinator.

A one-dimensional Brownian motion spends an infinite total time in every nonempty open interval almost surely. To prove this for $(-1,1)$ after first hitting zero, alternate exits from that interval with returns to zero. By recurrence of one-dimensional Brownian motion all these stopping times are finite, and the Strong Markov property makes the successive exit durations independent and identically distributed random variables that are strictly positive. Some fixed positive duration is exceeded with positive probability, so the Borel-Cantelli lemma makes the sum of the durations infinite. Translate and scale for other intervals. In particular, if a nonnegative function is bounded below by a positive constant on such an interval, its infinite-horizon Brownian occupation time integral diverges. Recurrent visits alone, without this duration argument, would not prove divergence of an integral.

For a bounded domain whose boundary points are all regular boundary points, continuous Dirichlet boundary data $f$ have a unique solutionHere ${\tau }_D$ is the Brownian exit time. The Strong Markov property gives the mean value property for harmonic functions; a barrier for the Dirichlet problem gives continuity at the boundary. This is an existence theorem as well as the Brownian representation of the Dirichlet problem for an already known solution.