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For a nonexplosive continuous-time Markov chain on the nonnegative integers, suppose zero is absorbing and each positive state has positive rate to zero. On every finite positive set the conditional chance of absorption at the next jump has a positive minimum. By the Strong Markov property, the chance of surviving $r$ departures from that set is bounded by $(1-\delta {)}^r$. Without absorption only finitely many such departures occur, and finite positive holding rates prohibit staying forever at a positive state. Thus eventually the path stays outside each finite set: it escapes to infinity. This conclusion needs no upward skip-free assumption.

Let $g_D(x)={\mathbb{E}}_x{T}_D$, possibly infinite, and suppose the closed ball $\overline{B(x,r)}$ lies in the open set $D$. A Strong Markov property decomposition at an independent random-radius exit time, using uniform ball sampling by Brownian stopping, gives the displayed identity in the nonnegative extended reals. Indeed the expected exit time from a radius-$s$ ball started at its centre is $s^2/n$, and averaging this against $n{s}^{n-1}/r^n$ gives $r^2/(n+2)$. The identity gives local integrability whenever $g_D(x)$ is finite.

Successive exits from intervals of radius $a$ around the current position embed a simple symmetric random walk into Brownian motion. The Strong Markov property makes the durations independent with common law $a^2{\sigma }_1$; the displacements divided by $a$ are independent symmetric signs. The displacement sequence is independent across exits; no independence between a given displacement and its duration is needed for these conclusions.

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.

In slotted ALOHA, let old packets retry independently with fixed probability $0<p<1$, and let independent Poisson random variables of mean $\nu >0$ transmit once on arrival. Write $q=1-p$ and $r=e^{-\nu p}$. A hypothetical process which never removes a packet has backlog $b+A_t$, where $A_t$ is the cumulative arrival count over $t$ slots. Couple its attempts to those of the actual process until the first noncollision. Its one-slot noncollision probability at backlog $n$ is $h(n)=e^{-\nu }[(1+\nu )q^n+np{q}^{n-1}]$. The union bound and the Poisson distribution giveIndeed $\mathbb{E}[q^{A_t}]=r^t$ and $\mathbb{E}[A_t{q}^{A_t}]=\nu tq{r}^t$. Thus $\epsilon (b)\to 0$. From every sufficiently large backlog, permanent collisions have probability at least $1/2$. Independent arrival batches reach that threshold almost surely; after each failed attempt the Strong Markov property permits another attempt. The probability of $k$ consecutive failures is at most $2^{-k}$, proving almost-sure eventual permanent collisions without assuming in advance that the backlog escapes.

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.

For any real constant $c$, one-dimensional Brownian motion repeatedly returns to zero. Consider successive segments from zero until exiting $(-1,1)$, separated by returns to zero. Their capped durations are independent identically distributed positive bounded random variables by the Strong Markov property, so their sum diverges by the strong law of large numbers. On each such segment $e^{c{B}_s}\ge e^{-|c|}$. Thus the exponential clock has infinite lifetime, despite arbitrarily deep negative excursions.

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.

For a nonconstant entire function, its derivative is nonzero somewhere and therefore bounded away from zero on a small disc. Recurrence of planar Brownian motion and the Strong Markov property imply infinite total occupation time in that disc: repeat entrances into a smaller concentric disc after exits from the larger one. On each entrance there is a uniformly positive chance of remaining inside the larger disc for a fixed positive duration. Iterating the conditional failure bound gives infinitely many such sojourns almost surely.

The displayed integral is bounded below by a positive constant times this occupation time, so it diverges. The same conclusion follows by comparing with the divergence of a positive planar Brownian occupation integral. At every finite time the clock is finite, because the Brownian path has compact range and $f^{\prime }$ is continuous. Thus the conformal Brownian clock covers every nonnegative image time; conformal invariance of planar Brownian motion then yields a Brownian image with no missing finite or infinite lifetime.

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 an alive barrier-triggered put, the difference from the corresponding European put option is the displayed discounted conditional expectation, evaluated under the risk-neutral measure. To prove it, condition the European payoff on the first hitting time or expiry and use the Strong Markov property. At a hit, continuation of the European claim is worth $P(B,{\tau }_B;K)$, whereas the fixed exercise rule pays $K-B$ immediately. On survival to expiry the two payoffs coincide. The relevant stopped quantities are bounded, so the conditional-expectation argument is justified.

Suppose motion within each level ends on hitting one of two boundaries, and each boundary transition resets the process to a fixed entry state of an adjacent level. Observe only the successive level exits. The resulting level process is a Markov chain, with transition probabilities equal to the within-level boundary-hitting probabilities. The fixed-entry condition matters: without it, retaining only the level may discard information needed for the next transition law.

For a simple symmetric random walk on local rooms $0,\dots ,r$, starting each excursion at room $1$, the gambler's ruin probability of reaching $r$ first is $1/r$. Thus the level chain moves up with probability $1/r$ and down with probability $(r-1)/r$. Independent future decisions, or more generally the Strong Markov property, justify restarting this calculation after each exit.

For two active levels between absorbing levels $0$ and $3$, starting on level $1$, put $p=1/r$ and $q=1-p$. The success probabilities satisfy $u=pv$ and $v=p+qu$, henceFor $r=4$, the result is $1/13$. This compression illustrates how unbiased microscopic choices can produce a biased coarse random walk when entry points are close to one boundary.

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.

Start a Markov chain at a nonabsorbing state $i$. If each departure from $i$ has probability $p>0$ of absorption before the next return, the Strong Markov property makes the number of visits to $i$, including the initial one, a positive-integer geometric random variable of parameter $p$. For a continuous-time Markov chain whose holding times at $i$ are independent rate-${\lambda }_i$ exponentials, independently of the embedded jump chain, the total occupation time at $i$ is a sum of $K$ such holding times and has expected value $1/(p{\lambda }_i)$.