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For standard Brownian motion started at zero, $H_a=\inf \{t\ge 0:B_t\ge a\}$ is finite almost surely for $a\ge 0$ andThe Brownian reflection principle proves finiteness. For $u\ge 0$, the Exponential martingale for Brownian motion stopped at $H_a\wedge t$ is bounded by $e^{ua}$. The dominated convergence theorem gives $\mathbb{E}{e}^{ua-u^2{H}_a/2}=1$, and taking $u=\sqrt{2\lambda }$ gives the transform.

Stop the Exponential martingale for Brownian motion $e^{\theta B_s-{\theta }^{2}s/2}$ when $B$ first reaches $\delta$ or at $t$. On the hitting event its value is at least $e^{\theta \delta -{\theta }^{2}t/2}$, so that event has probability at most $e^{-\theta \delta +{\theta }^{2}t/2}$. Minimize over $\theta >0$ and apply the same bound to $-B$. The union bound gives the displayed inequality for $t>0$; at zero the probability is zero.

For standard Brownian motion started at zero and first exit ${\tau }_x$ from $(-x,x)$, the Laplace transform is $1/\cosh (x\sqrt{2\lambda })$ for $\lambda >0$. The stochastic process $e^{-\lambda t}\cosh (\sqrt{2\lambda }{B}_t)$ is the average of two copies of the Exponential martingale for Brownian motion. Its values stopped at $t\wedge {\tau }_x$ are bounded, so bounded-time application of the optional stopping theorem followed by dominated convergence evaluates the transform without assuming uniform integrability of an unstopped exponential martingale on the infinite horizon.

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.

For $\theta >0$, $E_s=\exp (\theta B_s-{\theta }^{2}s/2)$ is an Exponential martingale for Brownian motion, as follows directly from independent Gaussian increments. Stop it at ${\tau }_{\delta }\wedge t$, where ${\tau }_{\delta }$ is the first time $B$ reaches $\delta$. On $\{{\tau }_{\delta }\le t\}$,The stopped expectation is one and the other contribution is nonnegative, soFor $t>0$, minimize the exponent at $\theta =\delta /t$. Apply the same argument to $-B$ and use the union bound to obtainAt $t=0$ the probability is zero; the right side can be interpreted by its limit. This proves the Gaussian maximal bound for Brownian motion without losing an extra factor from a two-sided reflection argument.

Let $L$ be a global Lipschitz constant of $b$. Realize the diffusion with the stated generator by $d{X}_s^{\epsilon }=b(X_s^{\epsilon })ds+\epsilon d{W}_s$, where $W$ is a standard $d$-dimensional Brownian motion. Global Lipschitz continuity gives existence and uniqueness; its generator is exactly the printed operator. Couple it with the ordinary differential equation having the same initial value $x(0)=x_0$. Subtraction gives, with $D_t={\sup }_{s\le t}|X_s^{\epsilon }-x_s|$,The Gronwall inequality implies $D_t\le \epsilon e^{Lt}{\sup }_{s\le t}|W_s|$. If a Euclidean norm exceeds $u$, some coordinate has absolute value exceeding $u/\sqrt{d}$. The scalar bound and a union over the coordinates therefore giveTaking the logarithm and multiplying by ${\epsilon }^2$ yields the explicit answerFor $t=0$ the probability is zero and the limit is $-\infty$, with $\log 0=-\infty$. This is exponential small-noise concentration for a Lipschitz diffusion.

Matching the deterministic initial value is necessary. The printed last line specifies the differential equation but does not separately write this initial condition. Without it, take $b=0$ and a deterministic solution displaced from $x_0$ by $2\delta$ in one fixed direction. The deviation event already holds at time zero, giving logarithmic rate zero instead of a strictly negative rate.

Fix $\lambda >0$ and choose the positive rootUse the Exponential martingale for Brownian motion $M_t=\exp (\theta B_t-{\theta }^{2}t/2)$. Before the linear-boundary stopping time $\tau$, path continuity gives $B_s<a+bs$. At the boundary there is equality. ConsequentlyThe optional stopping theorem at the bounded stopping time $t\wedge \tau$ yields $\mathbb{E}{M}_{t\wedge \tau }=1$. Separate the two events:The last term is at most $e^{\theta a-\lambda t}$ and tends to zero. The first term converges by the monotone convergence theorem. Taking $e^{-\lambda \infty }=0$ therefore provesThis bounded-stopping argument does not assume in advance that $\tau$ is finite. For $b\le 0$, letting $\lambda \downarrow 0$ makes the right side tend to one, so it also proves $\mathbb{P}(\tau <\infty )=1$. Setting $b=0$ gives the Brownian first-passage Laplace transformThe zero-parameter value follows from the just-proved almost-sure finiteness.

For the Brownian first-passage time to a line, set $\lambda =\theta >0$ and choose $h=b+\sqrt{b^2+2\lambda }>0$, so $h^2/2-bh=\lambda$. The Exponential martingale for Brownian motion $M_s=e^{h{W}_s-h^{2}s/2}$, stopped at $T_{a,b}$, obeysThe optional stopping theorem at the bounded time $s\wedge T_{a,b}$ gives expectation one. The contribution from $\{T_{a,b}>s\}$ is at most $e^{ha-\lambda s}$, and tends to zero. Taking $e^{-\lambda \infty }=0$ therefore yieldsLetting $\theta \downarrow 0$ shows that the hitting probability is one for $b\le 0$ and $e^{-2ab}$ for $b>0$; in the latter case the hitting-time law has mass at infinity.

To derive the finite-time linear-boundary Brownian first-passage distribution, put $Z_s=W_s-bs$. This is Brownian motion with drift $\mu =-b$, and the event is $\{{\sup }_{s\le t}{Z}_s\ge a\}$. For driftless Brownian motion, the Brownian reflection principle says that the joint endpoint density on this event equals the normal density $p_t(x)$ when $x\ge a$ and the reflected density $p_t(2a-x)$ when $x<a$, where $p_t(x)=(2\pi t{)}^{-1/2}{e}^{-x^2/(2t)}$. The constant-drift change of measure weights these densities by $e^{\mu x-{\mu }^{2}t/2}$. Thus the crossing probability isCompleting the square gives respectively $\Phi ((\mu t-a)/\sqrt{t})$ and $e^{2\mu a}\Phi ((-\mu t-a)/\sqrt{t})$. Substituting $\mu =-b$ provesThe density weighting follows directly from the Exponential martingale for Brownian motion, or the Girsanov theorem, on the fixed finite horizon; it does not require eventual hitting.

For the barrier digital call, under the risk-neutral measure in the Black-Scholes model,Assume $\sigma >0$, put $a=\log (c/S_0)/\sigma >0$ and $b=-(\rho -{\sigma }^2/2)/\sigma$. The upper stock barrier is reached exactly when $W_s^Q$ reaches $a+bs$. The payment is at the fixed maturity $t_0$, not at the random hitting time. Hence its price isIf $\sigma =0$, use the deterministic path $S_s=S_0{e}^{\rho s}$ instead: the value is $e^{-\rho t_0}$ if that path reaches $c$ by $t_0$, and zero otherwise. The nondegenerate formula concerns the usual positive-volatility model.

A useful finite-horizon form of the Girsanov theorem is as follows. Let $W$ be a Brownian motion and $\eta$ an adapted, progressively measurable process with ${\int }_0^T{\eta }_s^{2}ds<\infty$ almost surely. Suppose its stochastic exponentialis a true martingale with mean one. The Novikov condition $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T{\eta }_s^{2}ds)<\infty$ is a sufficient condition. Under the equivalent probability measure $dQ=Z_{T}dP$, the process $W_t^Q=W_t+{\int }_0^t{\eta }_{s}ds$ is a Brownian motion.

For the proof sketch, the Itô formula gives $dZ=-Z\eta dW$. Integration by parts cancels the drift in $d(Z{W}^Q)$, since $Z\eta dt$ from $d{W}^Q$ cancels $d[Z,W^Q]=-Z\eta dt$. After localization, the Bayes formula for conditional expectation makes $W^Q$ a continuous local martingale under $Q$. Its quadratic variation is still $t$, because adding a finite-variation process does not change quadratic variation. One can see the Brownian conclusion directly: for real $z$, the Itô formula makes $\exp (iz{W}_t^Q+z^{2}t/2)$ a local martingale. Its modulus is bounded on the fixed horizon, so it is a true martingale. ThusThis is the conditional characteristic function of a centered normal distribution with variance $t-s$, independent of the past. Continuity and these independent normal increments give the defining properties of Brownian motion.

Now put ${\varphi }_t(x)=(2\pi t{)}^{-1/2}{e}^{-x^2/(2t)}$ and $\Phi (z)={\int }_{-\infty }^z{\varphi }_1(x)dx$, the standard normal distribution function. At zero drift the endpoint density of paths staying below $a$ isIndeed, reflect the portion of a path after its first hitting time of $a$. By continuity and the independent symmetric Brownian motion increments after this stopping time, reflection preserves its probability law and maps a crossing path ending at $x<a$ to a path ending at $2a-x>a$. This subtracts exactly ${\varphi }_t(2a-x)$ from the unrestricted endpoint density; it is the Brownian reflection principle.

Taking the constant drift tilt $dQ/dP=e^{\mu W_t-{\mu }^{2}t/2}$ in the Girsanov theorem makes $W$ under $Q$ have the same law as $W+\mu s$ under $P$. Therefore the desired finite-horizon maximum of Brownian motion with drift has probabilityCompleting squares gives $e^{\mu x-{\mu }^{2}t/2}{\varphi }_t(x)={\varphi }_t(x-\mu t)$ and $e^{\mu x-{\mu }^{2}t/2}{\varphi }_t(2a-x)=e^{2\mu a}{\varphi }_t(x-2a-\mu t)$. HenceThe maximum has no atom at a positive level, so strict and weak barrier inequalities agree. This also follows from continuity of the displayed distribution function. At $\mu =0$ the formula reduces to $2\Phi (a/\sqrt{t})-1$.

For the up-and-out power claim, write $T=t_0>0$ and use the risk-neutral measure for the Black-Scholes model, under whichwith $\sigma >0$. The barrier survives exactly when ${\sup }_{t\le T}(W_t+m_{0}t)<A$. Its risk-neutral pricing value isThe factor inside the expectation is a mean-one Exponential martingale for Brownian motion. Under the probability tilted by it, $W_t-2\sigma t$ is a Brownian motion, so the barrier process has drift $m=m_0+2\sigma =(\rho +3{\sigma }^2/2)/\sigma$. Applying the preceding survival formula yieldsAs $c\to \infty$, the bracket tends to one and the price becomes the discounted second moment $S_0^2{e}^{(\rho +{\sigma }^2)T}$. As $c\downarrow S_0$, it tends to zero. If $\sigma =0$, use the deterministic path instead: the price is $S_0^2{e}^{\rho T}$ when $S_0\max (1,e^{\rho T})<c$, and zero otherwise. At $T=0$, with $c>S_0$, the price is simply $S_0^2$.

Independence and the normal exponential moment give $\mathbb{E}[S_t\mid {\mathcal{F}}_s]=S_s\exp [(\mu +{\sigma }^2/2)(t-s)]$. ThusFor $\sigma \ne 0$, $B_t/t\to 0$ almost surely, so $\log S_t/t\to -{\sigma }^2/2$ and $S_t\to 0$ almost surely. Nevertheless $\mathbb{E}{S}_t=e^{x_0}$ for every finite $t$, whereas $\mathbb{E}{S}_{\infty }=0$. This Exponential martingale for Brownian motion is not uniformly integrable, so it does not converge in $L^1$. If $\sigma =0$, the process is instead constantly $e^{x_0}$.

Put $L_r={\theta }_{T_r}$. The hitting times $T_r$ are finite almost surely by recurrence of planar Brownian motion, and increase with $r$. At $T_r$, the position is $e^{-r}{e}^{i{L}_r}$. By the Strong Markov property, Brownian scaling, and rotational invariance of planar Brownian motion, the future processis a standard planar Brownian motion started at $1$, independent of ${\mathcal{F}}_{T_r}$. The random rotation does not change its conditional law, since that law is rotationally invariant.

The first time this rescaled process reaches radius $e^{-h}$ corresponds to the original time $T_{r+h}$. Its continuous argument starts at zero and equals the increment of the original winding angle. HenceFor any ordered finite collection of levels, apply this assertion successively. Earlier winding values are measurable at the current hitting time, so all increments are independent and their laws depend only on the level differences. This proves the Lévy process property of winding at logarithmic radial passage levels.

One can also verify stochastic continuity and identify this process. The logarithmic-radius and argument martingales have the common clockThe same Cauchy-Riemann equations calculation as in part (a), applied locally to the logarithm and joined along the continuous argument, gives independent Brownian motions $U,V$ such that $\log R_t=U_{H_t}$ and ${\theta }_t=V_{H_t}$. This clock diverges: if it had a finite limit, convergence of martingales with finite total quadratic variation would make $\log R_t$ converge to a finite value, forcing the clock integral itself to diverge.

Thus, with ${\eta }_r=\inf \{u:U_u=-r\}$, we have $L_r=V_{{\eta }_r}$. The Brownian first-passage Laplace transform is $\mathbb{E}{e}^{-q{\eta }_r}=e^{-r\sqrt{2q}}$. It follows either from the Exponential martingale for Brownian motion stopped at ${\eta }_r\wedge n$, or from the one-dimensional exit equation. Conditioning on the independent process $U$ givesSo this is a symmetric Cauchy process, with level parameter $r$. The formula implies stochastic continuity at zero and, by stationary increments, everywhere. Under the usual definition requiring càdlàg paths, use the right-continuous passage clock ${\eta }_r^{+}=\inf \{u:U_u<-r\}$. It agrees with ${\eta }_r$ for each fixed level almost surely, and $V_{{\eta }_r^{+}}$ is the càdlàg modification. This distinction concerns simultaneous passage levels and does not alter any increment law.

Write $\tau ={\tau }_x$ and $T_t=t\wedge \tau$. Path continuity gives $|B_{T_t}|\le x$. Stop the martingale $B_s^2-s$ at the bounded stopping time $T_t$ and apply the optional stopping theorem:By the monotone convergence theorem, $\mathbb{E}\tau \le x^2$, so $\tau <\infty$ almost surely. Path continuity then gives $B_{\tau }\in \{-x,x\}$. The dominated convergence theorem for the bounded variables $B_{T_t}^2$ shows

Next stop the quartic Hermite polynomial martingale from part (a), again only at $T_t$. Its expectation is zero, soMonotone convergence proves $\mathbb{E}{\tau }^2\le 2x^4$, establishing the needed second-moment integrability before the final passage to the limit. Since $T_t{B}_{T_t}^2\le x^2\tau$ and $B_{T_t}^4\le x^4$, dominated convergence givesConsequently the Brownian symmetric interval-exit moments are

To obtain the Laplace transform of symmetric Brownian interval-exit time, put $a=\sqrt{2\lambda }$. The Exponential martingale for Brownian motion shows thatis a martingale with $H_0=1$. Bounded-time stopping gives $\mathbb{E}{H}_{T_t}=1$. Its stopped values are bounded by $\cosh (ax)$, so dominated convergence applies as $t\to \infty$. Since $\cosh (a{B}_{\tau })=\cosh (ax)$, it yieldsEvery use of stopping at $\tau$ has thus been justified through bounded stopping and an explicit integrability or domination argument.

For $a>0$, the Brownian reflection principle gives, for $t>0$,where $\Phi$ is the standard normal distribution function. Hence $H_a<\infty$ almost surely; for $a=0$, $H_0=0$.

For $u\ge 0$, use the Exponential martingale for Brownian motion $Z_t=\exp (u{B}_t-u^{2}t/2)$. At $H_a\wedge t$, it is bounded by $e^{ua}$ because $B_s<a$ before the first hit and $B_{H_a}=a$ by continuity. The bounded optional stopping theorem gives $\mathbb{E}{Z}_{H_a\wedge t}=1$. Its limit is $e^{ua-u^2{H}_a/2}$, and the dominated convergence theorem yields $\mathbb{E}{e}^{ua-u^2{H}_a/2}=1$. Equivalently, the Brownian first-passage Laplace transform isTaking $\lambda =u^2/2$ proves the displayed exponential identity. The cases $u=0$ and $a=0$ are included. The restriction $u\ge 0$ is what gives the upper bound on the stopped exponential.

The Exponential martingale for Brownian motion is a true martingale: independent increments and the normal distribution identity $\mathbb{E}{e}^{aN}=e^{a^{2}v/2}$ for $N\sim N(0,v)$ give $\mathbb{E}[Z_t\mid {\mathcal{F}}_s]=Z_s$. Its second moment isThe uniform integrability from bounded second moments criterion now applies: for $K>0$,Thus $\{Z_t:0\le t\le T\}$ has uniform integrability. In fact, all values stopped at stopping times bounded by $T$ have uniform integrability, since $Z_{\tau }=\mathbb{E}[Z_T\mid {\mathcal{F}}_{\tau }]$ and uniform integrability of conditional expectations applies. This is a finite-horizon assertion, not an assertion of uniform integrability over all $t\ge 0$.

For $\theta >0$, the Exponential martingale for Brownian motion and the Doob maximal inequality for a nonnegative submartingale giveTaking $\theta =a/s$ gives $e^{-a^2/(2s)}$. Apply the same argument to $-B$ and use the union bound:

The stationary law is the law of ${\sup }_{t\ge 0}(\sigma B(t)-dt)$. Let $\kappa =2d/{\sigma }^2$. The Exponential martingale for Brownian motion $e^{\kappa (\sigma B_t-dt)}$, stopped on first hitting $b$, is bounded by $e^{\kappa b}$. On no hitting, the driving path tends to minus infinity. Dominated convergence and optional stopping therefore give $1=e^{\kappa b}\mathbb{P}(\text{hit }b)$. The reflected process started at zero converges to this supremum law: its value at time $t$ has the law of the driving path's maximum up to $t$, by reversing the increments. The resulting density is $(2d/{\sigma }^2)e^{-2dx/{\sigma }^2}$ for $x\ge 0$.