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For three-dimensional Brownian motion, every radius $r>0$ and every initial point satisfyIf the initial point is on one of the spheres, it is already a hit at time zero. If it is inside a lattice-centered open ball, its eventual exit hits that sphere. Otherwise choose at each integer time a nearest lattice center. The current displacement from it has norm at most $\sqrt{3}/2$, so the multivariate normal distribution of the next unit increment gives a uniform positive chance of entering its radius-$r/2$ open ball. The geometric tail bound from a uniform escape probability makes eventual entry certain, and continuity forces a boundary crossing. This periodic-family recurrence does not contradict the Brownian sphere-hitting probability in dimension three for one fixed sphere.

For Brownian motion with drift $dX=dB+\mu dt$ in $(-b,b)$, the displayed probability of exiting at $-b$ follows by stopping the harmonic function $e^{-2\mu X}$. A unit-time increment larger than $2b$ has a uniform positive probability and forces exit, so the exit time has a geometric tail bound from a uniform escape probability and all polynomial moments. At zero drift the limiting probability is $(b-x)/(2b)$.

Let $a<i<b$ and $a<j<b$ be integers. A simple symmetric random walk started at $i$ exits the integer interval at $\tau =\inf \{n\ge 0:S_n\in \{a,b\}\}$. Its expected occupation of $j$ before exit, including its initial state, isA block of $b-a$ consecutive positive steps forces exit, so the geometric tail bound from a uniform escape probability proves $\tau <\infty$ almost surely. Optional stopping at the bounded stopping time $n\wedge \tau$, followed by the bounded convergence theorem for $S_{n\wedge \tau }\in [a,b]$, gives the gambler's ruin probabilitiesTranslate the absolute value martingale of a simple symmetric random walk from zero to $j$. Bounded-time optional stopping yieldsUse monotone convergence on the increasing visit count and the bounded convergence theorem on the right. Substituting the exit probabilities giveswhich simplifies to the displayed expression by considering $i\le j$ and $i\ge j$. In particular, $G_{a,b}(0,0)=-2ab/(b-a)$ when $a<0<b$.

The proof establishes finite expected value of the visit count. Therefore each translated stopped absolute-value martingale is dominated in absolute value by a constant plus that total count, an integrable random variable; its family of values is uniformly integrable. This supplies the limit justification without assuming uniform integrability in advance.

For real $\lambda$, the normal moment-generating formula gives $\mathbb{E}{e}^{\lambda (B_t-B_s)}=e^{{\lambda }^2(t-s)/2}$. Using independent increments and conditioning on the Brownian filtration,All terms are integrable by the same normal formula. Thus the displayed process is an Exponential martingale for Brownian motion.

For the interval exit time, we first show both almost-sure finiteness and finite expectation. On $\{T>n\}$, the position at time $n$ lies in $(-b,a)$. If $X_{n+1}-X_n\ge a+b$, the continuous path must exit by time $n+1$. This increment is independent of ${\mathcal{F}}_n$ and has law $N(\mu ,1)$, so the conditional probability of this event is a fixed $c>0$. ThereforeThis geometric tail bound from a uniform escape probability implies $T<\infty$ almost surely and $\mathbb{E}T<\infty$.

Take $\lambda =-2\mu$. Then $e^{-2\mu X_t}=e^{-2\mu B_t-2{\mu }^{2}t}$ is the exponential martingale just proved. Its stopped values lie between $e^{-2\mu a}$ and $e^{2\mu b}$. The optional stopping theorem at $T\wedge t$, followed by bounded convergence, gives $\mathbb{E}{e}^{-2\mu X_T}=1$. With $v=\mathbb{P}(X_T=a)$ and the two possible boundary values,Solving yieldsNext apply bounded-time optional stopping to $B_t$. Since $B_{T\wedge t}=X_{T\wedge t}-\mu (T\wedge t)$, we obtain $\mathbb{E}{X}_{T\wedge t}=\mu \mathbb{E}(T\wedge t)$. The stopped position is bounded, and the stopped time increases to an integrable $T$. Passing to the limit gives $\mu \mathbb{E}T=av-b(1-v)$. ThereforeAs a check, letting $\mu \downarrow 0$ gives the driftless formulas $v\to b/(a+b)$ and $\mathbb{E}T\to ab$. The expectation calculation uses a justified stopping limit, rather than assuming that optional stopping holds at every unbounded time.

Write $X_t=x+W_t+\mu t$ with $W$ a zero-starting Brownian motion. A unit-time increment exceeds $2b$ with probabilityIf the process has not exited before an integer time $k$, it is in $(-b,b)$; an increment exceeding $2b$ forces it outside at time $k+1$ and hence forces an exit by then. Independence of Brownian increments gives $\mathbb{P}(T>k)\le (1-p{)}^k$. Thus the geometric tail bound from a uniform escape probability yieldsIn particular $T$ is finite almost surely.

The diffusion generator is $\mathcal{L}=\frac{1}{2}{\partial }_{xx}+\mu {\partial }_x$. The function $f(z)=e^{-2\mu z}$ satisfies $\mathcal{L}f=0$, so the Itô formula gives $df(X_t)=-2\mu f(X_t)d{W}_t$. It is a true martingale, since it is $e^{-2\mu x}$ times the Gaussian exponential martingale $e^{-2\mu W_t-2{\mu }^{2}t}$. The stopped values $f(X_{t\wedge T})$ are uniformly bounded because the stopped path stays in $[-b,b]$. Optional sampling and bounded convergence givewhere continuity ensures that $X_T$ is exactly one of the two endpoints. Solving,This drifted Brownian interval-exit probability tends to $(b-x)/(2b)$ as $\mu \to 0$, as expected.

Write $c=1/(a-1)$. If $X_0=x>c$, then for either possible sign,as long as $X_n>c$. Thus the process stays above $c$, is strictly increasing, and induction gives $X_n\ge c+a^n(x-c)\to \infty$. If $x<-c$, the symmetric argument makes the process strictly decreasing to $-\infty$. This proves the first assertion pathwise, with the strict sufficient threshold $|x|>c$.

For an arbitrary starting point, a proof is needed that the process actually reaches such a region. Fix $L>c$ and choose an integer $m$ soAt a block starting at time $km$, if $X_{km}\ge 0$, the next $m$ signs all equal to $+1$ give $X_{(k+1)m}>L$. If $X_{km}<0$, all $-1$ signs instead give $X_{(k+1)m}<-L$. Conditional on the current natural filtration, the selected block has probability $\epsilon =2^{-m}$, by independence of future signs. If $\tau =\inf \{n:|X_n|>L\}$, iterated conditional expectation givesThis is a geometric tail bound from a uniform escape probability, so $\tau <\infty$ almost surely. Once outside, the sign is preserved and, for every future noise sequence,Consequently the explosive affine recursion with symmetric bounded noise hasThe block argument closes the gap between escape from a large initial value and escape from an arbitrary one; no assertion about the absence of atoms in an infinite random series is necessary.

Let $\tau =T_{-a}\wedge T_b$ for the simple symmetric random walk. From any interior point, $a+b$ successive $+1$ increments force exit through $b$. The conditional probability of this block is $2^{-(a+b)}$, so a blockwise geometric tail bound from a uniform escape probability gives $\tau <\infty$ almost surely.

Apply the bounded optional stopping theorem to $\tau \wedge n$. Since $S_n$ is a martingale, ${\mathbb{E}}_0{S}_{\tau \wedge n}=0$. The stopped positions lie in $[-a,b]$, so the bounded convergence theorem gives ${\mathbb{E}}_0{S}_{\tau }=0$. If $h={\mathbb{P}}_0(T_{-a}<T_b)$, nearest-neighbor jumps ensure $S_{\tau }=-a$ on that event and $S_{\tau }=b$ otherwise. ThusNext apply the bounded optional stopping theorem to the square-minus-time martingale of a simple symmetric random walk:Again the right side converges by the bounded convergence theorem, while the left side increases to ${\mathbb{E}}_0\tau$ by the monotone convergence theorem. ThereforeThis argument justifies passage to the unbounded stopping time rather than applying optional stopping without an integrability check.

Use the natural filtration ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$. The first exit time is a stopping time. If $r\le x$, it is zero, so its expectation is already finite. Now assume $0<x<r$.

The positive-increment hypothesis gives a number $h>0$ with $\mathbb{P}(X_1>h)>0$. Choose an integer $m$ so large that $r/m<h$, and put $p=\mathbb{P}(X_1>r/m)>0$. A block of $m$ increments all exceeding $r/m$ has probability $p^m$. From any point still in $(0,r)$, that block forces an upper exit before the block ends.

By independence, conditional on ${\mathcal{F}}_{km}$ and survival to time $km$, the next block has this same probability. Thus the geometric tail bound from a uniform escape probability givesUsing the tail-sum formula for the expected value of a nonnegative integer-valued random variable,This is the random-walk exit bound from a positive-increment block. In particular, the exit occurs with probability one. The mean-zero assumption is not needed for this first bound.

For $m\ge 1$, the increments have exponential momentThus $Z_n=2^{S_n}$ is an exponential martingale of a random walk relative to the natural filtration. We justify its stopping limit. From any surviving state in $(-m,m)$, a block of $2m$ successive increments all equal to $-1$ forces an exit. The block has probability $q=(6/7{)}^{2m}>0$, independently of the preceding increments. The geometric tail bound from a uniform escape probability givesso $T$ is finite almost surely and has finite expectation.

At the lower exit $S_T=-m$ exactly; at the upper exit $S_T$ is either $m$ or $m+1$. Before exit the same overall bound $-m\le S_{n\wedge T}\le m+1$ holds. Hence $Z_{n\wedge T}$ is bounded by $2^{m+1}$, uniformly in $n$. The bounded optional stopping theorem gives $\mathbb{E}{Z}_{n\wedge T}=Z_0=1$, and the dominated convergence theorem now yieldsAllowing the upper overshoot by one is essential; the terminal state need not equal $m$ on upper exit.

Let ${\tau }_r=\inf \{t\ge 0:|B_t|=r\}$, ${\tau }_R=\inf \{t\ge 0:|B_t|=R\}$, and $\tau ={\tau }_r\wedge {\tau }_R$. The function $u(y)=|y{|}^{-1}$ is harmonic away from the origin: its radial Laplacian in dimension three isThe Itô formula therefore makes $u(B_{t\wedge \tau })$ a local martingale, bounded between $1/R$ and $1/r$, hence a true martingale. The annulus exit time is finite almost surely. For instance, exit from the containing open ball is finite because from every point in the open ball a sufficiently large Gaussian vector increment has a uniform positive probability of leaving it in unit time; iterate the geometric tail bound from a uniform escape probability.

The optional stopping theorem at $t\wedge \tau$, followed by the dominated convergence theorem, gives $\mathbb{E}u(B_{\tau })=u(x)$. By continuity, there is no radial overshoot and the two exit spheres cannot be hit simultaneously. Writing $p={\mathbb{P}}_x({\tau }_r<{\tau }_R)$ givesAll harmonicity and boundedness statements apply only in the annulus, which stays away from the singularity at zero.

For each $n$,since $S_k$ is adapted. This verifies directly that $\tau$ is a stopping time.

Set $d=b-a$. On $\{\tau >md\}$, the random walk is at some integer strictly between $a$ and $b$. If its next $d$ steps are all $+1$, it must hit $b$ within that block. These future steps are independent of ${\mathcal{F}}_{md}$ and have probability $2^{-d}$ of being all $+1$. HenceThe law of total expectation and mathematical induction give the geometric tail bound from a uniform escape probability:Therefore $\mathbb{P}(\tau <\infty )=1$. In fact this also proves $\mathbb{E}\tau \le d{2}^d<\infty$ by grouping the tail-sum formula into blocks of length $d$. No assertion of independence between successive survival events is needed.

Set $D(s)=4-2s-s^2$, so $D(s)G(s)=s^2$. Differentiate once and evaluate at $s=1$, using $D(1)=1$, $D^{\prime }(1)=-4$, $G(1)=1$:A second differentiation gives $D^{\prime \prime }G+2D^{\prime }{G}^{\prime }+D{G}^{\prime \prime }=2$, henceThe derivatives of a probability generating function give $G^{\prime }(1)=\mathbb{E}[F]$ and $G^{\prime \prime }(1)=\mathbb{E}[F(F-1)]$. Consequently $\mathrm{Var}(F)=52+6-6^2=22$. The earlier geometric tail bound from a uniform escape probability ensures the needed moments are finite.

Let $E_0$ be the expected additional flips when there is no immediately preceding tail, and $E_1$ the expectation when the preceding flip is a tail. A one-step first-step analysis givesSolving gives $E_1=4$ and $\mathbb{E}[F]=E_0=6$. To justify finiteness before manipulating these expectations, split the flips into independent two-flip blocks. Each block has probability $1/4$ of being $TT$; hence $\mathbb{P}(F>2m)\le (3/4{)}^m$. This geometric tail bound from a uniform escape probability makes the expectation finite. The state $E_1$ is essential because successive possible pairs overlap: the waiting time is not itself a geometric distribution with parameter $1/4$.

Let $t_i={\mathbb{E}}_i[X]$. Time is accumulated in each transient chamber before the next move; arrival at either terminal state ends the trial immediately. Thus $t_0=t_k=0$. A block of $k$ identical rightward choices forces absorption, so the geometric tail bound from a uniform escape probability gives finite expected value of $X$.

First-step analysis now givesThe quadratic $t_i=i(k-i)$ has the correct boundary values and second difference $t_{i+1}-2t_i+t_{i-1}=-2$, so it solves this system. The difference of two solutions would be linear in $i$ with zero values at both endpoints, hence zero. ThereforeThis is the expected duration of symmetric gambler's ruin. It includes the first minute in the starting transient chamber and does not add a minute after absorption.

For a random walk with independent and identically distributed increments and an exit time from $(0,r)$, suppose $p=\mathbb{P}(X_1>r/m)>0$ for some integer $m\ge 1$. A block of $m$ such positive increments forces an exit. Conditional on survival, every successive block has success probability at least $p^m$, so $\mathbb{P}(T>km)\le (1-p^m{)}^k$ and $\mathbb{E}T\le m/p^m$. This is an instance of the geometric tail bound from a uniform escape probability; centering of the increments is not required.

Here $w$ is a finite word over an alphabet, every letter of the word has positive probability, and $B(w)$ is the set of lengths of its borders of a word, including its full length. Start a unit-capital gambler before every draw, betting successively on the word's letters and dividing surviving capital by the required letter probability. Total capital minus the number of entrants is a martingale. At the first full match, the surviving gamblers correspond exactly to the borders, giving the displayed formula by optional stopping. Disjoint blocks of full word length give a geometric tail bound from a uniform escape probability, so the waiting time is integrable; bounded word length and positive letter probabilities give bounded martingale increments and justify stopping.