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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 $S_0^2$.

Let $M_t={\sup }_{s\le t}{W}_s$ and write ${\varphi }_t(y)=(2\pi t{)}^{-1/2}{e}^{-y^2/(2t)}$ for the Brownian terminal density. By the Brownian reflection principle, for $y<a$ the reflected paths that hit $a$ and end at $y$ have terminal density ${\varphi }_t(2a-y)$. HenceThe Girsanov theorem changes zero drift into drift $\nu$ using the finite-horizon density $e^{\nu W_t-{\nu }^{2}t/2}$. The finite-horizon maximum of Brownian motion with drift therefore has distribution functionIn the first integral term, complete the square to get a normal density with mean $\nu t$. In the second, substitute $z=2a-y$ and complete the square with mean $-\nu t$; the prefactor is $e^{2a\nu }$. Here $\Phi$ is the standard normal distribution function, $a>0$ and $t>0$.

Let $T_{a,b}=\inf \{s\ge 0:W_s=a+bs\}$, with $b$ real. Survival means that $W_s-bs$ stays below $a$. Tilt by $e^{\theta W_t-{\theta }^{2}t/2}$; under the tilted probability, $W_s$ has drift $\theta$. Applying the maximum formula with drift $\theta -b$ gives the exponential terminal moment below a linear Brownian boundaryThis holds for every real $\theta$. The maximum distribution has no atom at the positive barrier, so strict or weak survival inequalities give the same probability here.

For the hitting-time moment, first use drift $-b$ in the maximum formula and take the complementary probability:If $z_1=(-a-bt)/\sqrt{t}$ and $z_2=(-a+bt)/\sqrt{t}$, their standard normal densities satisfy $\varphi (z_1)=e^{-2ab}\varphi (z_2)$. Differentiating the two terms, with $z_1^{\prime }=(a-bt)/(2t^{3/2})$ and $z_2^{\prime }=(a+bt)/(2t^{3/2})$, yields the drifted Brownian first-passage densityIts total mass need not be one, because the upward-sloping boundary may never be reached.

For $2\theta \le b^2$, put $\kappa =\sqrt{b^2-2\theta }\ge 0$. Multiplication by $e^{\theta s}$ gives $e^{\theta s}{f}_b(s)=e^{a(\kappa -b)}{f}_{\kappa }(s)$. Integrating to $t$ produces truncated discounted Brownian first passage, with negative discount rate allowed when the square root remains real:This includes the equality case $\kappa =0$, where it becomes $2e^{-ab}\Phi (-a/\sqrt{t})$. The expression integrates only finite hitting times up to $t$; no value is assigned to an exponential at an infinite hitting time.

For the cash-at-hit digital call, use the risk-neutral measure in the Black-Scholes model and denote expiry by $T=t_0$. ThenHitting $c>S_0$ corresponds to $W_s^Q=a+bs$, where $a=\log (c/S_0)/\sigma >0$ and $b=({\sigma }^2/2-\rho )/\sigma$. The unit-cash payment is made at the hitting time, so risk-neutral valuation discounts at that time, not at expiry. Set $\theta =-\rho$ andThe identity on the right verifies the square-root condition even for negative interest rates. The time-zero price isThis is in pounds for the unit-pound contract. It differs from a maturity-paid digital call option, whose discounting and exercise event are different.

An up-and-out power claim pays a power of the terminal stock price provided an upper barrier has never been reached. In the Black-Scholes model, set $A=\log (c/S_0)/\sigma$ and $m_p=(\rho +(p-\frac{1}{2}){\sigma }^2)/\sigma$, where $\sigma >0$, $c>S_0$ and $T>0$. The risk-neutral pricing value isHere $\Phi$ is the standard normal distribution function. Weighting the Brownian motion endpoint by $e^{p\sigma W_T-p^2{\sigma }^{2}T/2}$ changes the logarithmic drift to $m_p$; the finite-horizon maximum of Brownian motion with drift then supplies the survival factor.