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A contingent claim is replicated when an admissible self-financing portfolio has exactly its terminal payoff. In the absence of arbitrage, two replicating strategies for the same payoff have the same initial cost.

In a one-factor market whose filtration is the usual augmentation of the natural Brownian filtration, with nonzero spot volatility and local martingale deflator $Y$, the Brownian martingale representation theorem constructs a nonnegative replicating strategy for a bounded nonnegative contingent claim. The minimal initial cost among nonnegative self-financing portfolios is $\mathbb{E}[Y_T{\xi }_T]$.

The claim $\max (K,S_T)$ for $K>0$ combines a guaranteed cash floor with a European call option, equivalently a stock with a put. In the Black-Scholes model, for $\tau =T-t>0$, its price is $V=S_t\Phi (d_1)+K{e}^{-r\tau }\Phi (-d_2)$, where $d_1=[\log (S_t/K)+(r+{\sigma }^2/2)\tau ]/(\sigma \sqrt{\tau })$ and $d_2=d_1-\sigma \sqrt{\tau }$. Under the risk-neutral lognormal law, the truncated stock moment is $S_t{e}^{r\tau }\Phi (d_1)$ and the probability of ending below the floor is $\Phi (-d_2)$, proving the price. Its option delta is $\Phi (d_1)$, since $S_t\varphi (d_1)=K{e}^{-r\tau }\varphi (d_2)$ cancels the derivative terms. The replicating strategy holds that many stock units and $K{e}^{-r\tau }\Phi (-d_2)/B_t$ bank-account units.

Determining the value of a European option or American option from a specified financial market model and its permitted exercise rules. Risk-neutral pricing, replicating strategies and optimal stopping address different parts of this problem.

The derivative of a claim price with respect to the interest rate, holding the payoff and other model parameters fixed. For a dividend-free Black-Scholes model with remaining time $\tau$, it is $\tau (x{p}_x-p)$, so its sign is opposite the bank-account value in the replicating strategy.

A replicating strategy holds $\Delta$ shares and a current cash amount $b$ in the bank account. Matching the two next values givesSolving this linear system,By no-arbitrage, the claim costs $V_t=\Delta S_t+b$. Collecting the coefficients of its two future values yieldsThe physical probability $p$ does not appear: replication fixes the price of this contingent claim from traded prices.

Take bank-account values $B_k=R^k$ with $R>0$, and let the stock multiply independently by $u$ or $d$ at each step, where $0<d<R<u$. Under the physical probability measure $P$, the up probability is $p\in (0,1)$; the filtration records the successive moves. These assumptions specify the standard binomial market. If $R$ lies outside the strict interval $(d,u)$, a long or short stock position financed through the bank account gives an arbitrage.

The risk-neutral probability in a binomial market is fixed by $qu+(1-q)d=R$, givingFor successor claim values $V_u,V_d$ at a node with current stock price $S$, solve the two replication equations. The current stock holding and current cash value areThe cost of this replicating portfolio in a binomial market isApply this backward option pricing recursion from any terminal contingent claim $H$, including claims depending on the whole sequence of moves. It constructs a self-financing portfolio and proves market completeness by induction on the remaining number of periods. Taking successive conditional expectations under the measure $Q$ with independent up probability $q$ gives the unique risk-neutral pricing valueAny other price would give an arbitrage by buying the cheaper of the claim and its replica and selling the dearer.

If $K$ is the number of up moves, a particular path has probabilities $p^K(1-p{)}^{n-K}$ under $P$ and $q^K(1-q{)}^{n-K}$ under $Q$. Therefore the binomial-market probability density, namely the Radon-Nikodym derivative, isThis last expression makes $Z$ a function of the terminal stock price. Summing over paths gives ${\mathbb{E}}_{P}Z=1$, and strict positivity proves equivalence of the two probability measures.

For unrestricted terminal portfolio wealth $H$, its initial replication cost is $R^{-n}{\mathbb{E}}_{Q}H$. Hence feasible wealth satisfies the state-price budget constraint ${\mathbb{E}}_{Q}H=R^n{w}_0$. For the stated exponential utility, $v^{\prime }(x)=e^{-ax}$ and $v^{\prime \prime }(x)=-a{e}^{-ax}<0$. The binomial exponential-utility terminal wealth iswhereIt satisfies the budget, and its marginal utility is $v^{\prime }(H^{\ast })=\lambda Z$ with $\lambda =\exp (-a{R}^n{w}_0-{\mathbb{E}}_Q\log Z)>0$. The concave supporting-tangent inequality givesTaking expectations under $P$ makes the last term zero by the budget. Strict concavity makes equality possible only when $H=H^{\ast }$ in every positive-probability state. The already constructed complete market supplies its replicating strategy, so this proves global optimality and uniqueness. The formula allows negative final wealth, as permitted by the unrestricted trading problem.

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.

Let $T=t_0$, $\tau =T-t$, $x=S_t>0$, and $\sigma >0$. Use a dividend-free Black-Scholes model with constant interest rate $\rho$. The terminal payoff is assumed independent of the parameters being varied. In addition to twice differentiability, impose enough integrability to differentiate the pricing formula; for example, $f,f^{\prime },f^{\prime \prime }$ of polynomial growth suffice. Twice differentiability alone is not enough: the smooth payoff $f(x)=e^{x^2}$ has infinite expectation under every nondegenerate lognormal distribution.

Under the risk-neutral measure $Q$, future stock values conditional on $S_t=x$ have the formClaim replication, or equivalently the martingale property of discounted value in the complete market, givesThe replicating strategy holds the option delta $p_x$ shares and bank-account value $\beta =p-x{p}_x$. No physical drift parameter appears: the Girsanov theorem replaces the physical drift by $\rho$ under the pricing measure, and the replication argument cancels it directly. The risk-neutral pricing operator is linear in the payoff and preserves pointwise payoff inequalities.

Here is a verification of the Black-Scholes equation using the integral itself. Differentiation in spot givesDifferentiating with respect to $\tau$ givesFor $G(Z)=S_T{f}^{\prime }(S_T)$, the Gaussian integration by parts identity $\mathbb{E}[ZG]=\mathbb{E}[G^{\prime }]$ yieldsSubstitution cancels the $-{\sigma }^2/2$ term and provesSince $p_t=-p_{\tau }$, this is precisely the Black-Scholes equation. As $\tau \downarrow 0$, $M\to 1$ and growth control permits passage to the limit, giving $p(x,T)=f(x)$. The delta hedge then supplies the admissible replicating gains under the same regularity assumptions.

The Black-Scholes parameter sensitivities describe the dependence on the model parameters without assuming a particular payoff. If $f$ is increasing, then $p_x\ge 0$. If $f$ is convex, then $p_{xx}\ge 0$, so its price is convex in spot. The option gamma $p_{xx}$ measures the response of the option delta to spot changes.

For the interest rate, the distribution of the future stock also changes with $\rho$, so looking only at discounting gives the wrong general answer. Differentiation gives the option rhoThus a short bank-account position gives positive rate sensitivity, while a positive bank-account position gives negative rate sensitivity. For a cash payoff the result is negative; for the payoff $f(x)=x$ it is zero.

For spot volatility, differentiate the lognormal multiplier and then use Gaussian integration by parts:Consequently a convex payoff has nonnegative option vega. This monotonicity can also be understood by a mean-preserving increase of dispersion in the discounted stock. Strict positivity requires nonzero curvature seen with positive probability; an affine payoff has zero option vega.

Time enters through the remaining maturity $\tau$. Holding spot fixed, the Black-Scholes equation givesUnder the conventional assumption $\rho \ge 0$, a convex payoff and a short bond position $\beta <0$ imply $p_t\le 0$. The inequality is strict if $\rho >0$, or if $p_{xx}>0$ with positive volatility. Thus the price at fixed stock value is nonincreasing in calendar time, and nondecreasing in maturity, under these hypotheses. It is not a claim that the realized stochastic price path decreases: that path still has Brownian exposure $\sigma S_t{p}_{x}d{W}_t^Q$.

The printed conclusion needs the nonnegative-rate qualification; it is false for arbitrary negative rates. For $K>0$, the twice-differentiable convex payoff $f(x)=x-K$ hasThis is an exact counterexample, not a failure of the pricing formula. Also, with $\rho =0$ and an affine payoff, “decreasing” can only mean nonincreasing. Without payoff monotonicity or the specified bond-position sign, neither the spot nor rate nor time sensitivity has a universal sign. Any additional payoff parameter, such as a strike, is differentiated through $f$; pointwise payoff ordering, rather than a generic parameter rule, determines its price effect.

Let $\varphi$ denote the standard Gaussian distribution probability density function. Directly from the definitions, $S\varphi (d_1)=K{e}^{-r\tau }\varphi (d_2)$. Differentiating the price in part (ii), the terms involving derivatives of $d_1,d_2$ cancel, yielding $V_S=\Phi (d_1)$. Therefore, in units of the two traded assets, the replicating strategy isIts value is $h_t^B{B}_t+h_t^S{S}_t=V(t,S_t)$. To check self-financing explicitly, differentiation also givesHence $V_t+rS{V}_S+\frac{1}{2}{\sigma }^2{S}^2{V}_{SS}=rV$. Applying Itô formula under the physical measure givesThus the holdings form a self-financing strategy and their terminal value is the required payout. Holdings at the single maturity instant can be assigned by their limiting values; the continuous-time trading gains do not depend on that assignment. If ${\pi }_t$ denotes wealth fractions instead, the stock fraction is $S_t\Phi (d_1)/V(t,S_t)$ and the bank fraction is $K{e}^{-r\tau }\Phi (-d_2)/V(t,S_t)$. For $K\le 0$, hold one stock and no bank-account units.

The diffusion coefficient becomes $S\sigma (S)=\sqrt{S}$, so the stock is a driftless square-root diffusion, with zero taken as an absorbing boundary. Substitution of $V(t,S)=e^{A(t)S+B(t)}$ into the pricing equation for a local volatility model givesEquating the constant and linear terms and imposing the terminal payoff givesThus $1/A(t)=(1+t)/2$ and $B(t)=0$, and the price isThis is the exponential claim in a driftless square-root model. At the absorbing boundary its value is $1$, matching the payoff at zero.

The payoff in this part is unbounded, unlike the payoff assumed in part (i), so it is useful to verify that the proposed price is a finite conditional expectation. For $h>0$, starting from stock value $s$, defineThen $F_v+\frac{1}{2}x{F}_{xx}=0$, $F(h,x)=e^{-ux}$, and $0\le F\le 1$. By Itô formula, $F(v,S_v)$ is a bounded local martingale, hence a true martingale. It follows thatThis is also the Laplace transform of a sum of $N$ independent variables with exponential distribution of rate $2/h$, where $N$ has Poisson distribution of mean $2s/h$ and is independent of the summands: its transform is $\exp [(2s/h)((2/h)/(2/h+u)-1)]$. Uniqueness of the Laplace transform therefore gives the compound Poisson transition law of a driftless square-root diffusion. The exponential distribution jump has exponential moment $\mathbb{E}[e^Y]=(2/h)/(2/h-1)$ whenever $h<2$. Summing over the Poisson distribution yieldsFor the remaining horizon $h=1-t\le 1$, this is finite and, by the Markov property, is precisely ${\mathbb{E}}_Q[e^{S_1}\mid {\mathcal{F}}_t]=e^{2S_t/(1+t)}$. Thus the displayed process is indeed the risk-neutral pricing value and a true martingale. The replicating strategy from part (ii) has $h_t^S=[2/(1+t)]V(t,S_t)$ and $h_t^B=[1-2S_t/(1+t)]V(t,S_t)/B_t$.

The delta replication from a local-volatility pricing equation uses the asset-unit holdingsTheir wealth is exactly $V(t,S_t)$. Moreover,where the final equality is the calculation in part (i). Thus the holdings are a self-financing strategy with terminal wealth $g(S_T)$, proving replication. Since $V\ge 0$, this replicating strategy has nonnegative wealth and is an admissible trading strategy. The equality of trading gains is unchanged by replacing the drift with its physical-measure value: the stock drift in Itô formula and in the stock holding changes by the same amount. If ${\pi }_t$ denotes wealth fractions, then, wherever $V>0$, its stock fraction is $S_t{V}_S/V$ and its bank fraction is $1-S_t{V}_S/V$.

For positive $S_0,K,T,\sigma$, the Black-Scholes formula usesTo hedge the sold European call option, buy $\Phi (d_1)$ shares and borrow $K{e}^{-rT}\Phi (d_2)$ in cash. This long replicating strategy has initial value $S_0\Phi (d_1)-K{e}^{-rT}\Phi (d_2)$, financed by the option premium. The share holding must subsequently be rebalanced according to the option's option delta in a self-financing strategy.

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.

A trading strategy chooses a vector $H_t$ of holdings for the period $(t-1,t]$, with $H_t$ measurable with respect to ${\mathcal{F}}_{t-1}$. Its end-of-period wealth is $X_t=H_t\cdot P_t$. Rebalancing at date $t$ is self-financing whennew holdings cost exactly the value released by the old holdings. With fixed initial capital $x$, the equivalent gains identity isA European contingent claim is a maturity-$T$ payoff $\xi$ measurable with respect to ${\mathcal{F}}_T$. It is attainable if there exists a predictable self-financing strategy and an initial capital $x$ for which $X_T=\xi$ almost surely. Such a strategy is a replicating strategy, and $x$ is its initial replication cost. Holdings are understood only up to the maturity being replicated.

Let $b$ and $h$ be the numbers of units of the numéraire and the stock in a replicating strategy. The two terminal payoffs of the European call option are $2$ and $0$, soSolving these simultaneous equations gives $b=-6/35$ and $h=8/35$. The initial replication cost isAs an independent pricing check, let $q$ be the risk-neutral probability of the state $(15,20)$ under the numéraire measure. The discounted stock must have initial value one and terminal mean one:Risk-neutral valuation then gives $10(3/7)(2/15)=4/7$. The physical probabilities $1/2,1/2$ are not the numéraire-measure probabilities.

The positive-part function is convex, so pathwise Jensen inequality givesTo compare the Asian option price with European call option prices at different dates, carry each earlier payoff forward using the numéraire. Purchase $1/T$ of each replicating strategy for maturity $t$, and, when its payoff is received, reinvest it in $N$ until $T$. This is a self-financing portfolio, with terminal wealthThe first inequality uses $N_T\ge N_t$ and nonnegative payoffs. Since the Asian option is replicable in the complete market, absence of arbitrage makes its replication cost no larger than this superhedging cost. ThereforeEquivalently, divide the first pathwise bound by $N_T$, use $(S_t-K{)}^{+}/N_T\le (S_t-K{)}^{+}/N_t$, and take expectations under the numéraire equivalent martingale measure. Merely averaging earlier payoffs without reinvestment would miss the difference in payment dates.

Take the original measure as the risk-neutral measure for this zero-interest model, or interpret $S$ as a discounted stock price. ${\xi }_t$ is the arbitrage-free value of the European contingent claim with payoff $g(S_T)$, and ${\pi }_t$ is its option delta and replicating stock holding. Part (b) supplies the self-financing portfolio:With the cash asset identically one, its value is ${\xi }_t$ and its gains satisfy $d{\xi }_t={\pi }_{t}d{S}_t$, so it exactly replicates $g(S_T)$. This delta hedge is admissible since the assumed nonnegative $V$ gives nonnegative wealth. In this continuous-path setting the holdings can be taken predictable.

The auxiliary change of measure in part (c) expresses the option delta as an expectation of terminal payoff sensitivity. It changes the stock drift to $a{a}^{\prime }$ and is generally not the original risk-neutral measure for the zero-interest stock market. Its role is to value sensitivity under the tilted state dynamics, while the actual replicating strategy trades under the original dynamics. No general claim of market completeness is needed, including when $a$ can vanish.

Write $q_{15},q_{12},q_9$ for the state probabilities under an equivalent martingale measure. Since cash is constant and the European put option pays only in the lowest state, the pricing equations areTheir unique solution is $q_9={\xi }_0/2$, $q_{15}={\xi }_0/2-2/3$, $q_{12}=5/3-{\xi }_0$. Every physical state has positive probability, so equivalence requires all three values to be strictly positive. ThusFor each price in this open interval the pricing kernel takes values $3q_{15},3q_{12},3q_9$ in the three equally likely states, proving absence of arbitrage.

The necessity, including the exclusion of endpoints, can also be checked directly. The three terminal payoff vectors of cash, the stock and the European put option form the invertible matrixEach unit state payoff therefore has a replicating strategy, whose initial cost is the corresponding $q$. A zero or negative $q$ supplies an arbitrage, so the endpoints are genuinely excluded.

Fix the maturity horizon. Since $YB$ is a nonnegative local martingale and $r$ is bounded, $\mathbb{E}{Y}_T<\infty$: $B_T\ge B_0{e}^{-\Vert r{\Vert }_{\infty }T}$ and $\mathbb{E}(Y_T{B}_T)\le B_0$. The bounded nonnegative payoff thus makes $Y_T{\xi }_T$ integrable. SetThe Brownian martingale representation theorem yields $N_t=N_0+{\int }_0^t{\zeta }_{s}d{W}_s$, using its locally square-integrable version for an integrable terminal random variable. The Itô formula for $N/Y$ givesChoose the replicating strategyIts self-financing portfolio equation has exactly the displayed drift and diffusion, because $\mu -r=\sigma \lambda$. All coefficients are locally integrable after stopping; the holdings are predictable in the augmented natural Brownian filtration. This construction has $X_t\ge 0$ and $X_T={\xi }_T$, so it is an admissible replicating strategy under the question's nonnegative-wealth convention.

For any other nonnegative self-financing portfolio $\hat{X}$ replicating the same payoff, part (b) implies ${\hat{X}}_0\ge \mathbb{E}[Y_T{\xi }_T]$. The constructed strategy attains equality, since $Y_0=1$ and $N_0=\mathbb{E}[Y_T{\xi }_T]$ in the augmented natural Brownian filtration. HenceThis is deflator-based claim replication; it does not require upgrading the local martingale deflator to a true martingale density.