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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.

Consider a binomial market with dates $0,\dots ,n$, bank account $B_k=b^k$ for $b>0$, and stock price $S_{k+1}=S_{k}u$ or $S_{k}d$, where $0<d<u$. Both moves have positive physical probability, and the information at date $k$ is the history of the first $k$ moves. Absence of arbitrage requires and is implied by $d<b<u$. If $b\le d$, borrowing cash to buy stock has nonnegative payoff with a strict gain in one state; if $b\ge u$, the reverse trade does. In the interior range, the risk-neutral probability of an up move isIt makes ${\mathbb{E}}_Q[S_{k+1}\mid {\mathcal{F}}_k]=b{S}_k$, so $S_k/B_k$ is a martingale. This proves absence of arbitrage by pricing discounted gains under the equivalent measure.

At a node with stock price $s$, let the desired next-period portfolio values be $V_u,V_d$. Solve the two replication equations $\Delta su+\beta B_{k+1}=V_u$ and $\Delta sd+\beta B_{k+1}=V_d$. The replicating portfolio in a binomial market hasIts current value isRebalance at the next node using exactly the value supplied by the previous holdings, so the strategy is self-financing. Backward induction from any terminal contingent claim $H$ gives its hedge and the unique priceThus the finite binomial market is a complete market. The construction works for path-dependent claims on the full history tree, even when their values do not recombine merely as functions of the current stock price. For $H=f(S_n)$, independence under $Q$ gives the explicit time-zero priceThe stock hedge is a difference quotient of successor values, the discrete analogue of the delta hedge.

Let $I=(U^{\prime }{)}^{-1}$ be the inverse marginal utility. The derivative $U^{\prime }$ is continuous and strictly decreasing: concavity makes it nonincreasing, and equality at two distinct points would make $U$ affine between them, contradicting strict concavity. The Inada conditions therefore make $U^{\prime }$ a bijection from $(0,\infty )$ onto $(0,\infty )$. For each $y>0$, the derivative of $U(x)-xy$ is positive below $I(y)$ and negative above it. Consequently its unique global maximum is finite andThere is no need to differentiate $I$ to find the first derivative of this utility conjugate. Comparing the maxima at $y$ and $y+h$ gives, for $h>0$,Continuity of $I$, together with the corresponding inequalities for $h<0$, givesThus $V$ is strictly decreasing and strictly convex, since its derivative is strictly increasing. This proves all the asserted properties except the second differentiability, which does not follow from the printed hypotheses.

Here is an explicit Inada utility with vanishing curvature. PutThe positive integral defining $p$ makes $U$ strictly increasing. Also $p^{\prime }=-w$ is negative except at one point, so $p$ is strictly decreasing and $U$ is strictly concave. Near zero $w(s)$ is asymptotic to $s^{-2}$, and at infinity it is asymptotic to $s^{-2}$; thus $p(0+)=\infty$ and $p(\infty )=0$. This smooth utility function satisfies every printed positive-domain assumption, but $U^{\prime \prime }(1)=0$. At $y_0=p(1)=3-4\log 2$, differentiability of $I$ would imply, by differentiating $p(I(y))=y$,which is impossible. Hence $V^{\prime }=-I$ has no finite derivative there. More explicitly, $p(x)-p(1)=-(x-1{)}^3/12+o((x-1{)}^3)$, so its inverse function has a cube-root singularity. The twice-differentiable conclusion is false as printed. With the additional assumption $U^{\prime \prime }(x)<0$ everywhere, the inverse function theorem proves the intended dual differentiability with nonvanishing utility curvature:

For the financial assertions, use the positive-wealth admissible trading strategy class appropriate to the positive-domain utility function, and expectations for which the expressions are defined. A normalized terminal state-price density satisfies $Z_N>0$, $\mathbb{E}{Z}_N=1$, and defines an equivalent martingale measure $Q$ by $dQ=Z_{N}dP$. A self-financing portfolio obeyswhere the holdings are measurable before the next price increment. Under $Q$ the gains are a martingale when integrable; for the usual lower-bounded admissible class, localization gives a supermartingale. In particular the state-price budget constraint is $\mathbb{E}[Z_N{X}_N]\le x$. The positive-domain qualification matters: assumptions only on $U$ at positive arguments do not control its values at nonpositive wealth.

The definition of the utility conjugate gives the pointwise inequalityTaking expectations, using the state-price budget constraint, and then taking the supremum over strategies gives $u(x)\le \mathbb{E}V(y{Z}_N)+yx$ for every $y>0$ and every state-price density. Taking the two infima proves the utility duality with martingale deflators upper bound:If an expectation is infinite, use the same inequality in the extended-real sense whenever the negative parts make it well defined; no subtraction of two infinities is needed.

A complete market is one in which every terminal contingent claim in the specified claim class is attainable by a self-financing portfolio. Bounded terminal claims suffice for the uniqueness argument. In the standard finite-horizon discrete-time market, absence of arbitrage gives existence of an equivalent martingale measure by the fundamental theorem of asset pricing. To prove uniqueness, let $Z_N$ and ${\tilde{Z}}_N$ be two normalized state-price densities. Every bounded claim $H$ has a replicating price $p(H)$, and its replication has the same price under either equivalent martingale measure:Take $H=1_A$ for every terminal event $A$. The integrable difference $Z_N-{\tilde{Z}}_N$ has zero integral over every event, hence vanishes almost surely: its positive and negative sets would otherwise give a nonzero integral. The normalized terminal state-price density is unique up to null sets. Its density process is then uniquely $Z_n=\mathbb{E}[Z_N\mid {\mathcal{F}}_n]$. This uses the usual pricing replication class, where bounded claims have bounded replicating value processes; arbitrary doubling strategies are not an admissible definition of completeness.

For logarithmic utility, direct maximization gives $V(y)=-\log y-1$. In the complete market, write $Z=Z_N$. The unique-density utility conjugate bound is minimized at $y=1/x$, givingThe candidate terminal portfolio wealth is $H^{\ast }=x/Z$, whose cost is exactly $\mathbb{E}[Z{H}^{\ast }]=x$. If completeness includes nonnegative $Q$-integrable claims, it is replicable and attains the bound. There is also an approximation proof using only bounded-claim completeness. DefineThen $0<b_n\le 1$, $b_n\uparrow 1$, $H_n$ is positive and bounded, and $\mathbb{E}[Z{H}_n]=x$. Replicating $H_n$ therefore gives an admissible terminal claim of initial cost $x$. Since $\mathbb{E}(\log Z{)}^{+}\le \mathbb{E}Z=1$, the negative parts of $-\log (Z\vee n^{-1})$ have an integrable common bound. Monotone convergence after adding this lower bound givesConsequently the logarithmic terminal wealth in a complete market isThe value may be $+\infty$ if $\mathbb{E}(\log Z_N{)}^{-}=\infty$; the displayed formula remains meaningful, while attainment by $H^{\ast }$ depends on the precise admissible claim class.

A complete market is one in which every contingent claim in the pricing class can be replicated: for each finite maturity $T$ and terminal ${\mathcal{F}}_T$-measurable payoff $H$, there are an initial value and an admissible self-financing strategy whose terminal value is $H$ almost surely. It is enough for the complete market characterization to include all claims with bounded discounted payoffs. Thus every such payoff is attainable by trading the existing assets, rather than needing an additional independent security.

A one-period self-financing strategy is a deterministic holdings vector $h=(h^0,\dots ,h^d)$, with initial cost $h\cdot S_0$ and terminal wealth $h\cdot S_1$. In the zero-initial-cost convention, an arbitrage satisfiesThe convention allowing a nonpositive initial cost is equivalent here: invest any initial proceeds in the strictly positive numéraire to obtain a zero-cost portfolio with nonnegative terminal wealth.

Suppose a strictly positive state-price density $\rho$ satisfies the asset-pricing equations, with the indicated products integrable. By linearity,For an arbitrage, the right-hand side is zero while the left-hand side is strictly positive, since the integrand is nonnegative and is positive with positive probability. This contradiction proves absence of arbitrage.

An attainable European contingent claim $X$ is a terminal random payoff for which some holdings vector $h$ satisfies $h\cdot S_1=X$ almost surely; its replication cost is $h\cdot S_0$. To prove uniqueness in a complete market, suppose ${\rho }^{\prime }$ is another positive state-price density. For every event $A$, attainability gives a portfolio replicating $S_1^{(0)}{1}_A$. Its initial cost has both representations, soThe signed random variable here is integrable, because $\mathbb{E}[\rho S_1^{(0)}]=\mathbb{E}[{\rho }^{\prime }{S}_1^{(0)}]=S_0^{(0)}<\infty$. Take $A$ to be the event where it is positive and then where it is negative. Both its positive and negative parts have expectation zero, and hence vanish almost surely. Since $S_1^{(0)}>0$,This proves that complete markets have unique state-price densities without requiring an extra lower bound on the random terminal numéraire.

For the finite-state example, order the states from top to bottom in the diagram and write the density values as ${\rho }_1,{\rho }_2,{\rho }_3$. The two pricing equations areAdding gives ${\rho }_1+{\rho }_2+{\rho }_3=3$. Put $t={\rho }_3$; substitution then givesThese are all the strictly positive solutions, and each satisfies both equations. The parameter is a state-price-density value, not a probability.

For a concrete unattainable contingent claim, take the payoff vector $(0,1,0)$. A replicating portfolio $(a,b)$ would have to satisfy $3a+6b=0$ and $6a+3b=0$ in the first and third states. These imply $a=b=0$, contradicting the required payoff one in the middle state. Thus the market is an incomplete market, despite having strictly positive state-price densities.

In the one-period model, initial holdings are a deterministic vector $H\in {\mathbb{R}}^n$, chosen with trivial initial information, and the terminal payoff of the portfolio is $H^T{P}_1$. A contingent claim ${\xi }_1$ is replicable if some such vector has $H^T{P}_1={\xi }_1$ almost surely; its initial claim replication cost is $H^T{P}_0$. A complete market replicates every claim in the specified terminal-information class; for the arguments below, it suffices that every bounded measurable claim is replicable. The $n$ assets count all assets of the model, including a cash asset if one is traded.

These are one-period definitions. If arbitrary terminal information were already available initially and holdings could depend on it, the finite-atomic conclusion in the next part would not follow; the deterministic-holdings convention is essential.

Pricing kernel and replication. In the nondegenerate Black-Scholes model, put $\kappa =(\mu -r)/\sigma$ and normalize the state-price density by ${\zeta }_0=1$. The process isThe density $e^{rt}{\zeta }_t$ changes probability to the risk-neutral measure. Under that measure $W_t+\kappa t$ is a Brownian motion and the stock drift is $r$. Thus an integrable contingent claim $H$ has time-$t$ priceIn the usual augmented natural Brownian filtration, the Brownian martingale representation theorem supplies a replicating strategy; this is the complete market assumption. For a nonnegative admissible trading strategy without intermediate consumption, the state-price budget constraint is $\mathbb{E}[{\zeta }_T{w}_T]\le w_0$, with equality for a fully invested replicated claim. In particular

Feasibility and the largest slope. First take the intended regime $r>0$, $T>0$, $w_0,S_0>0$, and the usual nonnegative portfolio wealth constraint. The terminal wealth floor isFor any feasible claim the state-price budget constraint impliesTherefore $\alpha \le w_0/S_0$. Conversely, when $0\le \alpha \le w_0/S_0$, hold $\alpha$ shares and put the remaining $w_0-\alpha S_0$ in the continuous-time bank account. Its terminal portfolio wealth isHenceAt equality, $\xi =\bar{\alpha }{S}_T$ has cost exactly $w_0$. Positivity of the state-price density forces $w_T=\bar{\alpha }{S}_T$ almost surely: any strict improvement would cost more. Invest all initial portfolio wealth in $w_0/S_0$ shares and hold them until $T$.

Optimal payoff below the feasibility limit. For $\alpha <\bar{\alpha }$ the floor is strictly positive, and its price is strictly less than $w_0$. Assume the utility function is increasing, differentiable and strictly concave, satisfies the Inada conditions, and has the integrability needed for the finite-budget optimization. These are the usual hypotheses implicit in using inverse marginal utility. For each positive multiplier $\lambda$, maximizeseparately in every state. Its derivative decreases through zero at $I(\lambda {\zeta }_T)$, so the floored marginal utility optimizer isThe multiplier is characterized byUnder the stated integrability hypotheses the left side is continuous and decreasing, tends to the floor cost as $\lambda \uparrow \infty$, and tends to infinity as $\lambda \downarrow 0$. It is strictly decreasing wherever it exceeds the floor cost: on the event where the inverse-marginal-utility payoff exceeds the floor, a larger multiplier strictly reduces that payoff. Therefore the budget determines a unique finite multiplier. For CRRA utility the inverse marginal utility is $I(y)=y^{-1/R}$; lognormal moments provide the needed integrability.

For completeness, pointwise maximality gives, for any feasible competing terminal portfolio wealth $Y$,Taking expectations and using $\mathbb{E}[{\zeta }_{T}Y]\le w_0=\mathbb{E}[{\zeta }_T{w}_T^{\ast }]$ proves optimality. Strict concavity gives uniqueness of the terminal claim. Its price processis nonnegative and starts from $w_0$; claim replication therefore turns the payoff optimizer into an admissible portfolio.

What the missing interest-rate hypothesis changes. The PDF does not explicitly assume $r>0$ or give the utility and admissibility hypotheses above. These omissions matter. At $r=0$, under nonnegative admissibility, the largest feasible slope remains $w_0/S_0$: for larger slopes the positive-part floor costs strictly more than $w_0$, since $S_T$ has support $(0,\infty )$. But every $\alpha \le w_0/S_0$ already gives floor cost exactly $w_0$, so the only feasible terminal claim is $\xi$. There is no spare budget for an inverse-marginal-utility improvement. For example $\alpha =0$, $\kappa \ne 0$ and CRRA utility make $\mathbb{E}[{\zeta }_T\max \{w_0,I(\lambda {\zeta }_T)\}]>w_0$ for every finite $\lambda$. The prescribed positive finite multiplier then does not exist.

For negative $r$, nonnegative admissibility requires the effective floor ${\xi }_{+}=\max \{\xi ,0\}$. Define its costThis is continuous and convex. For $\alpha \le w_0/S_0$ it equals $w_0{e}^{-rT}+\alpha S_0(1-e^{-rT})$, exceeding $w_0$ below the endpoint. At ${\alpha }_0=w_0/S_0$, $G({\alpha }_0)=w_0$ and $G^{\prime }({\alpha }_0)=S_0(1-e^{-rT})<0$. Above that endpoint the lognormal stock gives strict convexity, and $G(\alpha )\to \infty$. Thus there is a unique second root ${\alpha }_1>{\alpha }_0$ of $G({\alpha }_1)=w_0$, the feasible slopes are $[{\alpha }_0,{\alpha }_1]$, and the largest is ${\alpha }_1$. Equivalently, for $\alpha >{\alpha }_0$ the floor price is $\alpha$ times the European call option price with strike $S_0-w_0/\alpha$. At the upper endpoint replicate ${\xi }_{+}$; in the interval with strict budget slack the same floored marginal utility optimizer applies with ${\xi }_{+}$.

If instead portfolio wealth may be negative and utility is defined on all real portfolio wealth, the floor itself has its affine replication cost: at $r=0$ every slope is feasible, and at negative $r$ every $\alpha \ge w_0/S_0$ is feasible. There is then no largest finite slope. The intended stock-only endpoint and strict-slack optimizer use positive interest and standard nonnegative admissibility.

For any finite contingent claim $C$ measurable at the final date in the natural binomial tree, backward induction givesAt a node with stock $S_j$, let $V_{j+1}^u,V_{j+1}^d$ be its two successor values. Hold, during the next period,The stock and bank holdings reproduce both successor values and cost ${\Delta }_j{S}_j+{\psi }_j{B}_j=R^{-1}[q{V}_{j+1}^u+(1-q)V_{j+1}^d]=V_j$. Rebalancing at every node is self-financing, and terminal value is $C$. Thus every such claim is replicated and the no-arbitrage price is unique.

The model has a unique equivalent martingale measure $\mathbb{Q}$ under whichand the discounted stock is a martingale. The time-zero value of an integrable time-$T$ contingent claim $C$ is therefore

A contingent claim whose payoff depends on the underlying price history, rather than only its terminal price. Asian options use averages, barrier options use whether a threshold has been crossed, and lookback options use running extrema. In the Black-Scholes model, the Brownian Martingale representation theorem still gives a replicating strategy; its portfolio wealth need not be a function of the current stock price alone. Adding a running integral or extremum can give a finite-dimensional state for pricing.

A replicating strategy is a self-financing portfolio whose terminal wealth equals the prescribed payoff of a contingent claim. Admissibility specifies the permitted wealth bounds and integrability of its holdings.

A finite signed measure with total mass one under which discounted asset gains have zero integrals. Its density can be negative, so it need not be a probability measure or give arbitrage-free prices to every nonnegative contingent claim. A square-integrable signed density is useful in one-period least-squares hedging, even when it is not an equivalent martingale measure.

An admissible self-financing portfolio superhedges a contingent claim when its terminal wealth is at least the claim payoff almost surely. The least permitted initial cost is the superhedging price. Exact claim replication requires equality.