= Solution
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 $\int_0^T\eta_s^2ds<\infty$ almost surely. Suppose its <stochastic exponential>
$$
Z_t=\exp\left(-\int_0^t\eta_s\,dW_s-\frac12\int_0^t\eta_s^2ds\right)
$$
is a true <martingale> with mean one. The <Novikov condition> $\mathbb E\exp(\frac12\int_0^T\eta_s^2ds)<\infty$ is a sufficient condition. Under the <equivalent probability measure> $dQ=Z_TdP$, the process $W_t^Q=W_t+\int_0^t\eta_sds$ is a <Brownian motion>.
For the proof sketch, the <Itô formula> gives $dZ=-Z\eta\,dW$. Integration by parts cancels the drift in $d(ZW^Q)$, since $Z\eta\,dt$ from $dW^Q$ 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 $\exp(izW_t^Q+z^2t/2)$ a local martingale. Its modulus is bounded on the fixed horizon, so it is a true martingale. Thus
$$
\mathbb E_Q[e^{iz(W_t^Q-W_s^Q)}\mid\mathcal F_s]=e^{-z^2(t-s)/2}.
$$
This 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 $\phi_t(x)=(2\pi t)^{-1/2}e^{-x^2/(2t)}$ and $\Phi(z)=\int_{-\infty}^z\phi_1(x)dx$, the <standard normal distribution function>. At zero drift the endpoint density of paths staying below $a$ is
$$
P(W_t\in dx,\ \sup_{s\leq t}W_s<a)=\{\phi_t(x)-\phi_t(2a-x)\}\,dx,\qquad x<a.
$$
Indeed, 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 $\phi_t(2a-x)$ from the unrestricted endpoint density; it is the <Brownian reflection principle>.
Taking the constant drift tilt $dQ/dP=e^{\mu W_t-\mu^2t/2}$ 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 probability
$$
\int_{-\infty}^a e^{\mu x-\mu^2t/2}\{\phi_t(x)-\phi_t(2a-x)\}\,dx.
$$
Completing squares gives $e^{\mu x-\mu^2t/2}\phi_t(x)=\phi_t(x-\mu t)$ and $e^{\mu x-\mu^2t/2}\phi_t(2a-x)=e^{2\mu a}\phi_t(x-2a-\mu t)$. Hence
$$
\boxed{P\left(\sup_{s\leq t}(W_s+\mu s)\leq a\right)=\Phi\left(\frac{a-\mu t}{\sqrt t}\right)-e^{2\mu a}\Phi\left(\frac{-a-\mu t}{\sqrt t}\right).}
$$
The 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 which
$$
S_t=S_0\exp\left(\sigma W_t+(\rho-\tfrac12\sigma^2)t\right),\qquad
A=\frac{\log(c/S_0)}{\sigma},\qquad m_0=\frac{\rho-\sigma^2/2}{\sigma},
$$
with $\sigma>0$. The barrier survives exactly when $\sup_{t\leq T}(W_t+m_0t)<A$. Its <risk-neutral pricing> value is
$$
V_0=S_0^2e^{(\rho+\sigma^2)T}\mathbb E_Q\left[e^{2\sigma W_T-2\sigma^2T}\mathbf1_{\{\sup_{t\leq T}(W_t+m_0t)<A\}}\right].
$$
The 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 yields
$$
\boxed{V_0=S_0^2e^{(\rho+\sigma^2)T}\left[\Phi\left(\frac{A-mT}{\sqrt T}\right)-e^{2mA}\Phi\left(\frac{-A-mT}{\sqrt T}\right)\right].}
$$
As $c\to\infty$, the bracket tends to one and the price becomes the discounted second moment $S_0^2e^{(\rho+\sigma^2)T}$. As $c\downarrow S_0$, it tends to zero. If $\sigma=0$, use the deterministic path instead: the price is $S_0^2e^{\rho T}$ when $S_0\max(1,e^{\rho T})<c$, and zero otherwise. At $T=0$, with $c>S_0$, the price is simply $S_0^2$.
Back to article page
