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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.

Let the bank account be $B_r=R^r$, where $R>0$ is the one-period riskless gross return. In the binomial market, $S_{r+1}$ is either $u{S}_r$ or $d{S}_r$, with $0<d<1<u$ and both successors having positive physical probability. The information at time $r$ consists of the first $r$ moves. A trading strategy chooses its holdings using only this information. A self-financing portfolio funds rebalancing from its current value; there is no external contribution.

Absence of arbitrage requires $d<R<u$. If $R\le d$, borrowing to buy the stock yields a nonnegative gain, strictly positive on the up branch; if $R\ge u$, shorting the stock and investing its price yields the reverse arbitrage. When $d<R<u$, define the risk-neutral probability in a binomial marketGiving each successive up move conditional probability $q$ defines an equivalent martingale measure $Q$, since every finite path has positive probability and ${\mathbb{E}}_Q[S_{r+1}\mid {\mathcal{F}}_r]=R{S}_r$.

For a contingent claim with successor values $V_u,V_d$ at a node of current stock value $s$, solve the two replication equations. The number of shares and current cash value areThen $\Delta us+RC=V_u$ and $\Delta ds+RC=V_d$, so the replicating strategy costsHold $C/B_r$ bank-account units over that period. Starting with the terminal payoff and applying these equations backwards supplies an adapted self-financing strategy replicating every path-dependent contingent claim. This proves market completeness. A different claim price would permit an arbitrage by buying the cheaper claim or replica and selling the dearer one. Iterating the backward pricing relation gives the unique risk-neutral pricing formulaThe physical up probability affects real-world frequencies but does not enter this replication price.

For a terminal payoff $f(S_n)$, write $k=n-r$. Its current price is the restriction to the available nodes of the functionIf $f$ is convex, each summand is convex in $s$ and each weight is positive; summing their defining convexity inequalities proves that $g_r$ is convex. This provides an explicit convex extension between the actual stock nodes, so “convex on the nodes” is unambiguous. If $f$ is strictly convex, each positive scaling preserves strict convexity, so $g_r$ is strictly convex as well.

For the replicating portfolio in a binomial market, the stock holding isFor $r+1<n$, let $F=g_{r+2}$. The next down and up holdings are the secant slopes of $F$ on the adjacent intervals with endpoints $d^{2}s,uds,u^{2}s$:For any convex function and $a<b<c$, its defining inequality at $b=(1-t)a+tc$ rearranges to $(F(b)-F(a))/(b-a)\le (F(c)-F(b))/(c-b)$. The inequality is strict for a strictly convex function. Applying this to the three stock values orders the two next holdings. Meanwhile the backward pricing relation gives, by subtracting its values at $us$ and $ds$,The coefficients are strictly positive and sum to one. Therefore the current share holding lies between the next two holdings:Both inequalities are strict for a strictly convex payoff. For an affine payoff the share holding is constant, explaining the required weak interpretation. These are numbers of shares, not the fraction of portfolio wealth invested in the stock.

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 $S_0^2$.

Write $T=t_0$, and specify the physical Black-Scholes model as $d{S}_t/S_t=\mu dt+\sigma d{W}_t$, with constant $\sigma >0$ and bank rate $\rho$. The physical drift $\mu$ is needed for an expected utility calculation; a pricing model specified only under a risk-neutral measure does not determine the investor's preferences over risky positions. PutHere $\vartheta$ is the market price of risk, $Z_T$ is the density of the risk-neutral measure $Q$, and $\xi$ is the state-price density. The Girsanov theorem makes $W_t^Q=W_t+\vartheta t$ a Brownian motion under $Q$.

For a dollar holding ${\pi }_t$ in the stock and consumption rate $c_t$, the portfolio wealth equation isSince $d{\xi }_t=-\rho {\xi }_{t}dt-\vartheta {\xi }_{t}d{W}_t$, integration by parts gives the useful deflated wealth equation with consumptionFor strategies with a true deflated gain martingale, terminal wealth $X=V_T$ and consumption obey the state-price budget constraintFor nonnegative wealth and consumption, the admissibility condition can instead give a budget inequality; increasing utilities use the full budget at an optimum. Conversely, suitable integrable payoff and consumption plans with this budget are financeable in the Brownian complete market. To construct the strategy, setBy the Brownian martingale representation theorem, write $d{M}_t=H_{t}d{W}_t$. The required dollar stock holding is ${\pi }_t=(H_t/{\xi }_t+\vartheta V_t)/\sigma$; substitution in the deflated gain equation verifies financing. Remaining wealth goes into the bank account. For square-integrable data the representation holds directly, and localization handles the customary integrable admissible data.

First take no consumption. For an increasing strictly concave differentiable utility function whose marginal derivative can be inverted, let $I_u=(u^{\prime }{)}^{-1}$. Choose $\lambda >0$ so thatThis candidate is not merely a first-order guess. The concave supporting-tangent inequality gives, pointwise,Taking expectations and using the budget proves $\mathbb{E}u(X)\le \mathbb{E}u(X^{\ast })$ for every feasible competitor. Finance the optimizer by its risk-neutral pricing value $V_t=e^{-\rho (T-t)}{\mathbb{E}}_Q[X^{\ast }\mid {\mathcal{F}}_t]$, using the preceding representation to find the holdings.

For a merely concave differentiable utility function, the general formulation is to select $X^{\ast }(\omega )$ from the maximizers of $u(x)-\lambda {\xi }_T(\omega )x$ over its wealth domain, and choose the multiplier to satisfy the budget. If increasing utility is not assumed, the equality-budget multiplier may have either sign; the positive-multiplier inverse formula above is for increasing utility. Flat portions allow multiple maximizers; boundaries require one-sided inequalities rather than equality of marginal derivatives. Existence of a multiplier, integrability and an admissible maximizer must be checked. Concavity alone does not guarantee a well-posed problem: with unrestricted dollar positions, linear utility $u(x)=x$ and $\mu >\rho$ permits unbounded expected terminal wealth. The explicit utilities below satisfy the needed conditions in their stated domains.

For exponential utility $u(x)=(1-e^{-ax})/a$ on $\mathbb{R}$, $u^{\prime }(x)=e^{-ax}$ and $I_u(y)=-\log y/a$. ThereforeThe budget is equivalently ${\mathbb{E}}_Q{X}^{\ast }=w_0{e}^{\rho T}$, since ${\xi }_T=e^{-\rho T}dQ/dP$. Under $Q$, ${\mathbb{E}}_Q{W}_T=-\vartheta T$, giving $\log \lambda =\rho T-{\vartheta }^{2}T/2-a{w}_0{e}^{\rho T}$. HenceMatching the Brownian coefficient of $dV$ with $\sigma {\pi }_t$ gives the finite-horizon exponential-utility portfolioHold ${\pi }_t^{\ast }/S_t$ shares and invest $V_t-{\pi }_t^{\ast }$ in the bank account. This is a dollar amount, not a fixed wealth fraction. The Gaussian exponential formula also gives the optimal expected utilityThis calculation uses unrestricted wealth and an admissibility class permitting the displayed Gaussian terminal wealth, with integrable deflated gains and the relevant exponential moment. A nonnegative-wealth constraint would change the optimizer; no such constraint is imposed on this exponential-utility example.

For both terminal and running utility function, choose $X$ and a nonnegative adapted consumption rate $c$ to maximize ${\mathbb{E}}_P[u(X)+{\int }_0^{T}v(c_t)dt]$ subject to the combined budget. Pointwise maximization of $u(x)-\lambda {\xi }_{T}x$ and $v(c)-\lambda {\xi }_{t}c$ gives, for interior strictly concave cases,Apply the concave supporting-tangent inequality to $u$ and to $v$ at every time, then integrate and take expectations. The difference in objectives is at most $\lambda$ times the difference in budget expenditures, which is nonpositive. This proves optimality, not just the marginal conditions. The financing value isThe question's consumption rate $R_t$ is denoted $c_t$ here to keep it distinct from interest-rate notation.

For logarithmic utility with $u(x)=a\log x$, $v(x)=b\log x$, assume $a,b,w_0>0$ and positive wealth and consumption. The marginal conditions and budget giveEach state-price-weighted payment is deterministic, so conditional valuation immediately gives $V_t=[a+b(T-t)]/(\lambda {\xi }_t)$. Put $A_t=a+b(T-t)$. Since the Itô formula gives $d(1/{\xi }_t)=(1/{\xi }_t)[(\rho +{\vartheta }^2)dt+\vartheta d{W}_t]$, this value satisfiesComparison with the wealth equation verifies the complete finite-horizon logarithmic investment and consumption strategy:Thus the optimal share holding is ${\pi }_t^{\ast }/S_t$, and the remaining value $V_t-{\pi }_t^{\ast }$ is in the bank account; borrowing or shorting is allowed if that value or the stock holding is negative. Terminal wealth is positive and consumption is positive throughout. Finally $\mathbb{E}\log {\xi }_t=-(\rho +{\vartheta }^2/2)t$, so the optimal objective, if desired explicitly, isFinite-horizon lognormal moments ensure the budget and expected utilities are finite, completing the feasibility and optimality checks.

The usual replication argument needs two hypotheses implicit in the pricing conclusion: the strategy must replicate the payoff, $V_N=X$, and its discounted gains must be integrable. Bounded predictable risky holdings suffice for the latter. Under these conditions, the equivalent martingale measure $\mathbb{Q}$ and part (c) giveThus discounted wealth is a martingale, and repeated conditioning givesFor a normalized initial bank account, $B_0=1$, deterministic terminal bank value, and trivial initial information, this reduces to the intended risk-neutral pricing formulaIf $B_N$ is random it must remain inside the expectation. If a claim is not replicable, the existence of an equivalent martingale measure alone does not make its arbitrage-free price unique.

The discounted wealth martingale integrability condition cannot simply be dropped. For a concrete counterexample, set $B_n=1$, $S_0=3/2$, $S_1=1+U$, and $S_2=1+U+\epsilon /2$, where $U$ is uniform on $(0,1)$ and $\epsilon$ an independent fair sign. These positive bounded prices form a martingale in the natural filtration. Start with zero holdings; at time one buy $1/U$ shares financed by borrowing $S_1/U$. This is previsible and self-financing, but its terminal wealth is $\epsilon /(2U)$, whose absolute expectation is infinite. It is therefore not a martingale. The stated pricing proof applies to the standard integrable replication class, rather than all unconstrained previsible strategies.

Since $d(B_t^{-1})=-r_t{B}_t^{-1}dt$, multiplying the given bond dynamics by $B_t^{-1}$ cancels the drift and gives $d{Z}_t=Z_t\Sigma (t,T)d{W}_t^Q$. For an integrable discounted terminal payoff, the risk-neutral value process is the discounted-payoff martingaleFor an attainable claim this also follows directly from its self-financing discounted wealth being a true martingale, with terminal value $B_T^{-1}X$. In a Brownian market, the Brownian martingale representation theorem provides $d{M}_t={\psi }_{t}d{W}_t^Q$; a traded bond with nonzero diffusion exposure can hedge it by holding ${\psi }_t/(Z_t{\Sigma }_t)$ bond units, with the remainder in the bank account. Thus, under the usual attainability and integrability assumptions, this construction gives the replicated price as well as the risk-neutral pricing formula.

Because $B_t$ is ${\mathcal{F}}_t$-measurable,In a general Heath-Jarrow-Morton model, writing this value as $V(t,r_t)$ requires an additional state-reduction assumption. The conditional expectation is valid without asserting that the short rate alone is the full state.

Let the one-period riskless gross return be $R>0$, with $0<d<R<u$, and let the physical up probability be $\pi \in (0,1)$. The binomial market has risk-neutral probabilityUnder its equivalent martingale measure $Q$, successive multipliers $J_k$ are independent, taking $u,d$ with probabilities $q,1-q$. The risk-neutral pricing function has the extensionFor every fixed positive multiplier $A$, the function $s\mapsto f(As)$ is convex when $f$ is convex. Taking a positive weighted sum preserves convexity. More explicitly, for $0\le \lambda \le 1$,Thus the pricing extension is convex, and so are its values on the attainable stock-price grid.

The replicating portfolio in a binomial market holdsshares over the next interval. If $r+1<n$, put $A={\Delta }_{r+1}(us)$ and $D={\Delta }_{r+1}(ds)$. These are the secant slopes of $g_{r+2}$ on $[uds,u^{2}s]$ and $[d^{2}s,uds]$, respectively. For a convex function, the slope of a secant on a later adjacent interval is no smaller than that on the earlier interval: apply the defining convexity inequality to the middle point $uds$, then rearrange. Hence $A\ge D$. Using backward option pricing twice givesThe two coefficients are positive and sum to one. Therefore convex-payoff delta monotonicity in a binomial market yieldsIn particular an up move cannot reduce the stock holding. “Increases” here means nondecreases: an affine payoff gives equality, so strict increase cannot be claimed for every convex payoff.

For the expected utility maximization problem, let $P$ be the physical measure and $Z=dQ/dP$ on terminal paths. If a path has $j$ up moves, its binomial-market probability density isEvery terminal payoff is replicable in this complete market. Buying a positive payoff $H$ with initial wealth $w_0$ therefore imposes the state-price budget constraint ${\mathbb{E}}_{Q}H=R^n{w}_0$. Since the utility function is increasing, the full budget is used. Its derivative and second derivative arePut $a=\gamma /(\gamma -1)$. The first-order condition $v^{\prime }(H_{\ast })=\lambda Z$ for a positive multiplier $\lambda$ gives $H_{\ast }=C{Z}^{-a}$. The budget determines $C$ uniquely. Independence of the successive moves givesThus the optimal claim, as a function of the number of up moves, isBecause $S_n=S_0{u}^j{d}^{n-j}$ uniquely determines $j$, this is also a terminal-stock-price payoff, attainable by backward option pricing and claim replication. It rewards states relatively more probable under the investor's physical measure than under the pricing measure. If $\pi =q$, the optimum is the riskless payoff $R^n{w}_0$.

To prove global optimality rather than merely assert a stationary point, use the tangent inequality for the strictly concave function $v$:Taking physical expectations makes the last term zero for any fully financed competing claim, and nonpositive for any claim costing less. Therefore ${\mathbb{E}}_{P}v(H)\le {\mathbb{E}}_{P}v(H_{\ast })$, with equality only when $H=H_{\ast }$ almost surely. Finiteness causes no difficulty on the finite binomial tree, and $H_{\ast }>0$ in every terminal state.

Subtract the boundary slope from the Brownian motion: $X_t=B_t-bt$. Then $T_{a,b}$ is the first-passage time of $X$ to the fixed level $a$. The zero-drift Brownian reflection principle givesHere $\Phi$ is the standard normal distribution function. To change the drift to $-b$, use the Girsanov theorem on each finite horizon, with density $L_t=\exp (-b{B}_t-b^{2}t/2)$. On $\{T_a\le t\}$, conditional expectation at the stopped time gives $\mathbb{E}[L_t\mid {\mathcal{F}}_{T_a}]=L_{T_a}$, because the density is a true martingale. Since $B_{T_a}=a$, the drifted Brownian first-passage density is consequentlyThere may be additional mass at infinite hitting time; the displayed density only describes finite passage. With $e^{-\theta \infty }=0$, apply the given zero-drift Brownian first-passage Laplace transform:The zero-discount limit is $1$ for $b\le 0$ and $e^{-2ab}$ for $b>0$, so the hitting probability is less than one precisely when the line moves away with positive slope.

For a direct verification of the linear-boundary Brownian first-passage distribution, defineBoth terms tend to zero as $t\downarrow 0$. Let $\varphi$ be the standard normal density. The identityfollows by expanding the two squares. Differentiating the two normal probabilities then givesSince this is the previously derived hitting density and both distribution functions start at zero,This proof applies to every real $b$, including the defective distribution when $b>0$.

For the barrier digital put, use the dividend-free Black-Scholes model with constant interest rate $\rho$ and spot volatility $\sigma >0$. Under the risk-neutral measure,Set $T=t_0$ andReaching the barrier $c$ is equivalent to $W_t^Q$ reaching $a+bt$. The contingent claim pays one only on the survival event $\{T_{a,b}>T\}$. Risk-neutral pricing therefore givesThis is a path-survival price, not the price of a terminal digital put option. Continuity of the stock paths and the absence of an atom in the finite-horizon maximum imply that alternative touching conventions change no price. As $c\to \infty$, the value tends to $e^{-\rho T}$; as $c\downarrow S_0$, it tends to zero. Both checks agree with the survival interpretation.