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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 ${\int }_0^T{\theta }_s^{2}ds<\infty$ almost surely. Suppose the stochastic exponentialis a true martingale with $\mathbb{E}{Z}_T=1$. Define $dQ=Z_{T}dP$. Then $Q$ is equivalent to $P$ andThe same statement holds in $d$ dimensions with scalar products, squared norms and a vector drift. Novikov's condition $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T{\theta }_s^{2}ds)<\infty$ is a standard sufficient condition for the true-martingale hypothesis. Bounded $\theta$ is enough for all applications below.

For the proof, Itô formula gives $d{Z}_t=-Z_t{\theta }_{t}d{W}_t$. The Itô product rule gives the cancellationThe two drift terms cancel because $d[Z,W{]}_t=-Z_t{\theta }_{t}dt$. Stop $W^Q$ when its absolute value reaches $n$, at a time ${\tau }_n$ capped by $T$. The same product calculation makes $Z_t{W}_{t\wedge {\tau }_n}^Q$ a local martingale under $P$. Its absolute value is at most $n{Z}_t$; the density martingale is uniformly integrable over this finite horizon, so the stopped product is a true martingale. Bayes' conditional identity then givesAs $n$ 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 $F_t=\exp (iu{W}_t^Q+u^{2}t/2)$. Its differential is $iu{F}_{t}d{W}_t^Q$, and its modulus is bounded on the fixed horizon, so it is a true martingale. ConsequentlyThe 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 $\Vert u{\Vert }^2$ and gives the multivariate version. If $|\theta |\le K$, stopping $Z$ and applying Itô formula to $Z^2$ gives a uniform second-moment bound $\mathbb{E}{Z}_{t\wedge \tau }^2\le e^{K^{2}t}$ 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 $d{S}_t=S_t(\alpha dt+\sigma d{W}_t)$, $B_t=e^{\rho t}$, with $\sigma >0$. Take $\theta =(\alpha -\rho )/\sigma$, the market price of risk. Under $Q$, the preceding change of drift givesso $S_t/B_t$ 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 ${\phi }_t(y)=(2\pi t{)}^{-1/2}{e}^{-y^2/(2t)}$,Take the constant-drift exponential density $e^{\mu W_t-{\mu }^{2}t/2}$. 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 $x\le a$,Completing the squares givesIntegrating provesHere $\Phi$ 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 priceThe 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.

Expected utility of terminal wealth. Let initial wealth be $x_0>0$, maturity $T=t_0$, and stock-dollar investment ${\pi }_t$. A self-financing strategy in the Black-Scholes model has wealthChoose admissible strategies with nonnegative wealth, and an increasing strictly concave differentiable utility function $U$ satisfying the Inada conditions. Assume the required expectations are finite and the optimization is well posed; the resulting budget equation below must have a solution. Write $\lambda =(\alpha -\rho )/\sigma$ andThis state-price density is the discounted density of the unique equivalent martingale measure. Nonnegative discounted wealth is a supermartingale under that measure, so every admissible terminal wealth $Y$ obeys the state-price budget constraint $\mathbb{E}[H_{T}Y]\le x_0$.

Conversely a nonnegative terminal claim with finite such cost is financed by its conditional martingale price, which remains nonnegative; Brownian Martingale representation theorem constructs its holdings. Thus market completeness turns dynamic optimization into a terminal-payoff problem. Let $I=(U^{\prime }{)}^{-1}$ be the inverse marginal utility, and choose $y>0$ from $\mathbb{E}[H_{T}I(y{H}_T)]=x_0$. Then the complete-market terminal utility optimizer isFor any feasible $Y$, concavity gives the pointwise tangent inequalityTaking expectations and using the budget proves optimality; strict concavity gives uniqueness up to null events. The optimal wealth process isIf its discounted martingale has representation $d(e^{-\rho t}{X}_t^{\ast })={\zeta }_{t}d{W}_t^Q$, hold ${\delta }_t=e^{\rho t}{\zeta }_t/(\sigma S_t)$ stock units and invest the remainder in the bank account. This derives the strategy, not only the terminal first-order condition.

The equivalent dynamic approach is the Bellman equation for terminal-wealth utility. For a smooth concave value $v(t,x)$, dynamic programming and Itô formula giveFor $v_{xx}<0$, maximizing the concave quadratic gives ${\pi }^{\ast }=-(\alpha -\rho )v_x/({\sigma }^2{v}_{xx})$. For an arbitrary admissible control the Itô drift of $v(t,X_t)$ is nonpositive; localization and integrability therefore bound its expected terminal utility by $v(0,x_0)$. The maximizing control makes the drift zero, giving equality when the verification expectations are valid.

For CRRA utility $U(x)=x^{1-\gamma }/(1-\gamma )$, $\gamma >0$, $\gamma \ne 1$, substitution givesThe optimal stock fraction is constant. Its geometric wealth dynamics stay positive, so it is feasible. For logarithmic utility the fraction is $(\alpha -\rho )/{\sigma }^2$ and $v(t,x)=\log x+(\rho +{\lambda }^2/2)(T-t)$. Risk aversion controls the risky exposure, while the market price of risk controls its reward.

Pricing claims depending on the path. For a path-dependent contingent claim $C=F((S_u{)}_{0\le u\le T})$, the risk-neutral pricing process isFor square-integrable discounted $C$, the Brownian Martingale representation theorem gives $d(e^{-\rho t}{V}_t)={\phi }_{t}d{W}_t^Q$. Since $d(e^{-\rho t}{S}_t)=\sigma e^{-\rho t}{S}_{t}d{W}_t^Q$, the replicating holding is ${\delta }_t={\phi }_t/(\sigma e^{-\rho t}{S}_t)$, with bank units $(V_t-{\delta }_t{S}_t)/B_t$. Thus path dependence does not destroy market completeness; it changes the information needed to determine the price and hedge.

For an arithmetic Asian option, introduce $A_t={\int }_0^t{S}_{u}du$ and seek $V_t=v(t,S_t,A_t)$. Because $d{A}_t=S_{t}dt$, Itô formula givesThe stock holding is $v_s$. The state $a$ records the already observed average; current stock price alone cannot recover it.

A Geometric Asian option admits an explicit conditional calculation. Let $I_t={\int }_0^t\log S_{u}du$ and $G_T=e^{I_T/T}$. Conditional on time $t$, the normal law of $\log G_T$ has mean and varianceIndeed its random part is $\sigma T^{-1}{\int }_t^T(T-u)d{W}_u^Q$, by stochastic Fubini; the Itô isometry gives $q_t$. Completing the square in a normal exponential integral proves the conditional geometric-average Asian option formulafor $K>0,t<T$. This explicitly includes the known past log-average. Holding that past integral fixed when differentiating the price gives the stock holdingThe normal-density derivative terms cancel because $e^{m_t+q_t/2}\phi (d_1)=K\phi (d_2)$.

For a lookback option use $M_t={\max }_{u\le t}{S}_u$. A floating-strike put pays $M_T-S_T$. Its price $v(t,s,m)$ obeys the ordinary Black-Scholes equation in $0<s<m$, with terminal value $m-s$. The extra Itô term is $v_{m}d{M}_t$; since $d{M}_t$ is carried by $S_t=M_t$, self-financing requires the running-maximum boundary for a lookback option $v_m(t,m,m)=0$ before maturity. Equivalently its price follows from the upper-hitting probabilities in Question 3:This uses $\mathbb{E}\max (m,Z)=m+{\int }_m^{\infty }P(Z\ge z)dz$ and ${\mathbb{E}}_Q{S}_T=s{e}^{\rho (T-t)}$. For a barrier option, the state must additionally record whether the barrier has already been hit; a no-rebate knockout price has an absorbing zero boundary. These augmentations give tractable PDEs or conditional-expectation computations while retaining the same martingale pricing and claim replication principle.

Let ${\varphi }_T(y)=(2\pi T{)}^{-1/2}{e}^{-y^2/(2T)}$ be the normal distribution density. The Brownian reflection principle gives the killed endpoint densityIndeed, reflection after the first hit of $a$ maps paths ending below $a$ that have hit the barrier to paths ending at $2a-y$; symmetry gives ${\varphi }_T(2a-y)={\varphi }_T(y-2a)$.

By the Girsanov theorem, weighting the path law by $\exp (\nu W_T-{\nu }^{2}T/2)$ makes the coordinate process a Brownian motion with drift $\nu$. Completing the two squares yieldsIntegration up to $x\le a$ proves the joint endpoint and maximum law for drifted Brownian motion:Here $\Phi$ is the standard normal distribution function. Setting $x=a$ also gives a continuous barrier distribution, so equality at the barrier has zero probability.

For the up-and-in option, use the risk-neutral measure of the Black-Scholes model with interest rate $\rho$ and spot volatility $\sigma >0$. ThenAbove $a$, the barrier has certainly been reached. Below $a$, subtracting the killed density from the unrestricted endpoint density leaves $e^{2a\nu }{\varphi }_T(y-2a-\nu T)$. Thus the discounted expected value of its payoff isIn the second integral put $z=y-2a$, and defineThen $S_0{e}^{\sigma y}=S_0{e}^{\sigma z}/\kappa$, $z\le -a$ means $S_0{e}^{\sigma z}\le \kappa b$, and $e^{2a\nu }={\kappa }^{-\nu /\sigma }$. Both integrals now use the unrestricted terminal stock distribution. This proves that the reflected European payoff for an up-and-in option isTheir initial risk-neutral pricing values agree for payoffs with finite absolute expectations in these integrals. The identity concerns prices; the two terminal payoffs need not agree along individual paths.

Let $T=t_0$, $\tau =T-t$, and let $N$ have the standard normal distribution. For a twice differentiable payoff with enough growth control for the following expectations and derivatives, risk-neutral pricing in the dividend-free Black-Scholes model givesThe conditional discounted payoff is a martingale under the risk-neutral measure. Applying the Itô formula to $e^{-\rho s}V(S_s,s)$ and setting its drift to zero therefore gives the Black-Scholes equationThe delta hedge holds $\Delta =V_x$ shares; the remaining bond value is $\beta =V-x\Delta$, or $h=\beta /e^{-\rho (T-t)}$ units of the zero-coupon bond. Substitution in the Itô formula and the displayed equation gives $dV=\Delta dS+hdB$, proving replication and self-financing.

Differentiating the pricing expectation in the current stock price gives its option delta and option gamma:Thus a nondecreasing payoff has a nondecreasing price; a nonincreasing payoff has a nonincreasing price. A convex payoff has $\Gamma \ge 0$, and a concave payoff has $\Gamma \le 0$. This is convexity preservation in Black-Scholes pricing.

The Black-Scholes parameter sensitivities follow directly from the same expectation. Differentiation in the interest rate gives the option rhoFor spot volatility, differentiation first givesThe Gaussian integration by parts identity $\mathbb{E}[N\psi (N)]=\mathbb{E}[{\psi }^{\prime }(N)]$, applied to $\psi (N)=f^{\prime }(xM)xM$, cancels the term containing $f^{\prime }$ and gives the option vegaConsequently increasing spot volatility increases a convex payoff's value and decreases a concave payoff's value, with weak inequalities in affine or degenerate cases. A short bond position gives positive option rho, while a long bond position gives negative option rho. The physical stock drift $\mu$ does not enter these prices: the risk-neutral measure replaces it by $\rho$.

The option theta is the calendar-time derivative at fixed current stock price. The Black-Scholes equation givesFor $\rho \ge 0$, a convex payoff with $\beta <0$ therefore has $V_t\le 0$, while a concave payoff with $\beta >0$ has $V_t\ge 0$. The signs are strict when $\rho >0$. At $\rho =0$, an affine payoff can give equality. Remaining-maturity sensitivity is $V_T=-V_t$ when the payoff is independent of $T$. All these time comparisons hold at fixed current stock price.

The stated time-monotonicity conclusion needs the nonnegative-rate condition. For example, the convex affine payoff $f(x)=x-K$ hasThus with negative interest rates it provides an explicit counterexample even though the replica is short in bonds.

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.

Use the physical dynamics of the Black-Scholes model, $d{S}_t/S_t=\mu dt+\sigma d{W}_t$, with bank rate $\rho$ and $\sigma >0$. Write $\lambda =(\mu -\rho )/\sigma$ for the market price of risk. If $w_t$ is portfolio wealth, ${\pi }_t$ the dollar amount in stock, and $c_t\ge 0$ the consumption rate, the budget dynamics areAdmissible controls are progressively measurable, locally integrable in the required drift and diffusion norms, and keep $w_t\ge 0$. Such a solvency restriction excludes doubling-type strategies. In a complete Brownian market, the investment-consumption problem is to maximizewhere the utility functions are increasing and concave. A terminal-only problem omits the integral. Strict concavity gives uniqueness of optimal consumption and terminal wealth when an optimum exists.

The state-price density isThe Itô formula givesThus $H_t{w}_t+{\int }_0^t{H}_s{c}_{s}ds$ is a nonnegative local martingale, hence a supermartingale, yielding the state-price budget constraintConversely, a nonnegative consumption plan and terminal payoff $\xi$ satisfying budget equality and the necessary integrability can be financed in the Brownian complete market. DefineThe Brownian martingale representation theorem for the total future budget supplies a stochastic-integral coefficient; comparing it with $H_t(\sigma {\pi }_t-\lambda w_t)$ defines ${\pi }_t$. This reconstructs the wealth equation with $w_t\ge 0$. Localization gives the representation for integrable positive budgets when an $L^2$ assumption is not available.

For smooth strictly concave utilities with the usual marginal-utility range, let $I_u=(u^{\prime }{)}^{-1}$ and $I_U=(U^{\prime }{)}^{-1}$ be the inverse marginal utility functions. The Lagrange-multiplier solution iswhere $y>0$ makes the budget an equality. This is a global argument, not merely formal differentiation: the concave supporting-tangent inequality giveswith the analogous terminal inequality. Taking expectations, integrating, and applying the budget bound makes the difference in objectives nonpositive. Replication of the candidate budget proves attainability and optimality whenever these expectations are well-defined and finite. If the marginal-utility ranges or integrability conditions fail, existence must be established separately rather than inferred from the first-order equations.

A detailed example is constant relative risk aversion utility $u(c)=c^{1-\gamma }/(1-\gamma )$ and $U(w)=\eta w^{1-\gamma }/(1-\gamma )$, with $\gamma >0$, $\gamma \ne 1$, $\eta >0$, and $\delta \ge 0$. PutThe optimal plans from the marginal-utility equations are $c_t^{\ast }=(y{e}^{\delta t}{H}_t{)}^{-1/\gamma }$ and ${\xi }^{\ast }={\eta }^{1/\gamma }(y{e}^{\delta T}{H}_T{)}^{-1/\gamma }$. The normal distribution of $W_t$ gives, for $a=1-1/\gamma$,The budget therefore becomes $y^{-1/\gamma }b(T)=w_0$, determining the multiplier explicitly. Independent Brownian increments in the conditional budget formula give $w_t^{\ast }=c_t^{\ast }b(T-t)$. The diffusion coefficient of $c_t^{\ast }$ is $(\lambda /\gamma )c_t^{\ast }$, so matching the wealth diffusion determines the dollar stock holding. Consequently the finite-horizon power-utility investment and consumption optimum isAll candidate moments are finite on a finite horizon, since powers of a lognormal state-price density have finite moments. The budget and tangent argument above therefore verify optimality in this example, including admissibility and positive wealth.

For another verification of the formulas, the Hamilton-Jacobi-Bellman equation for a value function with discount measured from the current time isSubstitute $V(w,t)=b(T-t{)}^{\gamma }{w}^{1-\gamma }/(1-\gamma )$. The two concave maximizations give the displayed controls, and the remaining equation is $b^{\prime }(\tau )=1-kb(\tau )$ with $b(0)={\eta }^{1/\gamma }$, exactly solved by the formula above. ThusFor $k=0$, interpret the integral as $\tau$, so $b(\tau )=\tau +{\eta }^{1/\gamma }$. Unlike the infinite-horizon Merton consumption-investment problem, finite horizons do not require $k>0$. Consumption vanishes in the terminal-only formulation; then the integral term in $b$ is absent and the optimal stock fraction remains $(\mu -\rho )/(\gamma {\sigma }^2)$. Logarithmic utility is handled directly with inverse marginal utility $1/z$ and the same wealth-to-consumption formula at $\gamma =1$, with $k=\delta$.

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 $S_0^2$.