= Solution
Here is the Brownian change-of-drift form of the <Girsanov theorem>. Let $W$ be a standard <Brownian motion> on $[0,T]$ and $\theta$ predictable with $\int_0^T\theta_s^2ds<\infty$ almost surely. Suppose the <stochastic exponential>
$$
Z_t=\exp\left(-\int_0^t\theta_s\,dW_s-\frac12\int_0^t\theta_s^2ds\right)
$$
is a true <martingale> with $\mathbb EZ_T=1$. Define $dQ=Z_TdP$. Then $Q$ is equivalent to $P$ and
$$
\boxed{W_t^Q=W_t+\int_0^t\theta_sds\text{ is a standard Brownian motion under }Q.}
$$
The same statement holds in $d$ dimensions with scalar products, squared norms and a vector drift. <Novikov's condition> $\mathbb E\exp(\frac12\int_0^T\theta_s^2ds)<\infty$ is a standard sufficient condition for the true-martingale hypothesis. Bounded $\theta$ is enough for all applications below.
For the proof, <Itô formula> gives $dZ_t=-Z_t\theta_t\,dW_t$. The <Itô product rule> gives the cancellation
$$
d(Z_tW_t^Q)=W_t^QdZ_t+Z_t(dW_t+\theta_tdt)+d[Z,W]_t
=Z_t(1-\theta_tW_t^Q)dW_t.
$$
The two drift terms cancel because $d[Z,W]_t=-Z_t\theta_tdt$. Stop $W^Q$ when its absolute value reaches $n$, at a time $\tau_n$ capped by $T$. The same product calculation makes $Z_tW_{t\wedge\tau_n}^Q$ a <local martingale> under $P$. Its absolute value is at most $nZ_t$; the density <martingale> is uniformly integrable over this finite horizon, so the stopped product is a true <martingale>. Bayes' conditional identity then gives
$$
\mathbb E_Q[W_{t\wedge\tau_n}^Q\mid\mathcal F_s]
=Z_s^{-1}\mathbb E_P[Z_tW_{t\wedge\tau_n}^Q\mid\mathcal F_s]
=W_{s\wedge\tau_n}^Q.
$$
As $n$ increases, these stopping times exhaust the horizon by continuity. Thus $W^Q$ is a continuous <local martingale> under $Q$. Adding a finite-variation drift does not change <quadratic variation>, so $[W^Q]_t=t$.
To prove the Brownian conclusion rather than just assert it, for every real $u$ apply <Itô formula> to $F_t=\exp(iuW_t^Q+u^2t/2)$. Its differential is $iuF_t\,dW_t^Q$, and its modulus is bounded on the fixed horizon, so it is a true <martingale>. Consequently
$$
\mathbb E_Q[e^{iu(W_t^Q-W_s^Q)}\mid\mathcal F_s]=e^{-u^2(t-s)/2}.
$$
The conditional <characteristic function> identifies a centered normal increment with <variance> $t-s$, independent of the past. Together with continuity and $W_0^Q=0$, this proves the Brownian assertion. For vector $u$, the same argument uses $\|u\|^2$ and gives the multivariate version. If $|\theta|\le K$, stopping $Z$ and applying <Itô formula> to $Z^2$ gives a uniform second-moment bound $\mathbb EZ_{t\wedge\tau}^2\le e^{K^2t}$ by <Gronwall inequality>; this proves uniform integrability of the stopped densities and the needed true-martingale property in the bounded case.
In the <Black-Scholes model> with its usual augmented Brownian <filtration>, write $dS_t=S_t(\alpha dt+\sigma dW_t)$, $B_t=e^{\rho t}$, with $\sigma>0$. Take $\theta=(\alpha-\rho)/\sigma$, the <market price of risk>. Under $Q$, the preceding change of drift gives
$$
dS_t=\rho S_tdt+\sigma S_t dW_t^Q,
$$
so $S_t/B_t$ is a <martingale>. The <Martingale representation theorem> then makes integrable Brownian-market claims attainable and gives their <risk-neutral pricing> as discounted <conditional expectations>. The physical drift $\alpha$ has disappeared from the price dynamics.
For the joint terminal-value and maximum law, first use the <Brownian reflection principle> at level $a>0$. Reflecting a path after its first hit of $a$ preserves its law as a <Brownian motion> and sends a terminal value $y<a$ to $2a-y>a$; the symmetry and strong Markov property of the post-hit increments justify this map. Thus, with $\varphi_t(y)=(2\pi t)^{-1/2}e^{-y^2/(2t)}$,
$$
P(W_t\in dy,\ \sup_{s\le t}W_s<a)
=[\varphi_t(y)-\varphi_t(2a-y)]\,dy\quad(y<a).
$$
Take the constant-drift exponential density $e^{\mu W_t-\mu^2t/2}$. By <Girsanov theorem>, the coordinate process under this tilted measure has the law of $W_s+\mu s$ under $P$. Consequently, for $t>0$ and $x\le a$,
$$
P(W_t^\mu\le x,M_t^\mu<a)
=\int_{-\infty}^x e^{\mu y-\mu^2t/2}[\varphi_t(y)-\varphi_t(2a-y)]\,dy.
$$
Completing the squares gives
$$
e^{\mu y-\mu^2t/2}\varphi_t(y)=\varphi_t(y-\mu t),\qquad
e^{\mu y-\mu^2t/2}\varphi_t(2a-y)=e^{2a\mu}\varphi_t(y-2a-\mu t).
$$
Integrating proves
$$
\boxed{P(W_t^\mu\le x,M_t^\mu<a)=
\Phi\!\left(\frac{x-\mu t}{\sqrt t}\right)
-e^{2a\mu}\Phi\!\left(\frac{x-2a-\mu t}{\sqrt t}\right).}
$$
Here $\Phi$ is the standard normal distribution function. The first variable is the terminal Brownian value, distinct from its running maximum.
Set $T=t_0$, $a=\log(b/S_0)/\sigma>0$ and $\mu=(\rho-\sigma^2/2)/\sigma$. Under the <risk-neutral measure>, $\log(S_s/S_0)/\sigma=W_s^Q+\mu s$. Taking $x=a$ in the joint law gives the <probability> of not hitting the upper level. Complementing it and discounting the unit payment at maturity gives the <one-touch option> price
$$
\boxed{V_0=e^{-\rho T}\left[
\Phi\!\left(\frac{\mu T-a}{\sqrt T}\right)
+e^{2a\mu}\Phi\!\left(\frac{-a-\mu T}{\sqrt T}\right)\right].}
$$
The exponential weight is also $(b/S_0)^{2\rho/\sigma^2-1}$. The payment occurs at $T$, even if the barrier was hit earlier, so its discount factor is $e^{-\rho T}$, not a hitting-time discount factor.
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