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In the completed natural Brownian filtration, a bounded continuously differentiable $\varphi$ with bounded derivative satisfiesThe integrand is unique up to $d\mathbb{P}dt$-almost everywhere equality. Gaussian integration by parts first gives the duality between the payoff and every square-integrable Itô integral. The Brownian martingale representation theorem then identifies the integrand as the projection of ${\varphi }^{\prime }(W_T)1_{\{t\le T\}}$ onto the predictable processes in the product $L^2$ space. This explains the conditional expectation in the formula, rather than a nonadapted terminal derivative.

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 stock diffusion $d{S}_t=S_t(rdt+\sigma (t,S_t)d{W}_t)$ has spot-dependent local volatility. With positive volatility and the Brownian filtration, Brownian martingale representation theorem converts discounted payoff martingales into stock gains.

The natural Brownian filtration is the natural filtration generated by a Brownian motion. Its usual augmentation adds null events and makes it right-continuous; the Brownian martingale representation theorem holds for this augmented natural filtration. A larger Brownian filtration can preserve independence of future increments while containing additional randomness, so it need not have martingale representation with respect to the specified Brownian motion.

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 $k=\delta$.

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.

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.

Use ${\theta }_t$ for the dollar amount invested in the stock; the number of shares is ${\theta }_t/S_t$. The bank account contains $w_t-{\theta }_t$ dollars. With no intermediate consumption, the self-financing portfolio equation isIf instead $\theta$ denotes share holdings, replace each dollar ${\theta }_t$ here by $S_t{\theta }_t$. The equation says that trading only reallocates existing portfolio wealth; gains come from cash interest and stock returns.

For the standard unconstrained complete-diffusion formulation, assume the completed natural Brownian filtration, initial wealth $w_0>0$, and nonnegative admissible wealth. Define the market price of risk ${\kappa }_t=({\mu }_t-r_t)/{\sigma }_t$. The boundedness assumptions make $\kappa$ bounded. The state-price density normalized at zero isIts money-market-adjusted process $Z_t=e^{{\int }_0^t{r}_{s}ds}{\zeta }_t$ is a true martingale by the Novikov condition. Under $d\mathbb{Q}=Z_{T}d\mathbb{P}$, the Girsanov theorem gives $W_t^{\mathbb{Q}}=W_t+{\int }_0^t{\kappa }_{s}ds$ as a Brownian motion, and the discounted stock has zero drift.

The Itô product rule applied to the wealth equation givesFor nonnegative admissible wealth this is a nonnegative local martingale, hence a supermartingale. Therefore every affordable terminal payoff $X=w_T$ satisfies the state-price budget constraint $\mathbb{E}[{\zeta }_{T}X]\le w_0$. Nonzero volatility and the Brownian martingale representation theorem allow replication of nonnegative claims with finite state-price cost.

For a differentiable strictly concave utility function with the usual conditions ensuring an interior integrable optimizer, put $I=(U^{\prime }{)}^{-1}$, the inverse marginal utility. Pointwise maximization of $U(x)-y{\zeta }_{T}x$ givesThe second equation determines the positive constant $y$. To prove optimality, use the concave supporting-tangent inequality:Taking expectations and using the budget bound proves that the candidate dominates every admissible affordable payoff, whenever the utility expectations are well defined. This is the complete-market terminal utility optimizer.

The actual wealth and dollar portfolio are as explicit as general adapted coefficients permit. LetThe Brownian martingale representation theorem, with localization if only $L^1$ integrability is initially known, gives $h$. The dollar portfolio from a deflated wealth martingale isThe bank holds $w_t^{\ast }-{\theta }_t^{\ast }$ dollars. Equivalently, with $B_t=e^{{\int }_0^t{r}_{s}ds}$, $w_t^{\ast }=B_t{\mathbb{E}}^{\mathbb{Q}}[w_T^{\ast }/B_T\mid {\mathcal{F}}_t]$. These formulas need no unjustified differentiability of a value function with random coefficients.

The printed assumption that $U$ is only increasing and concave does not guarantee that the interior formula exists, or even that the optimal value is finite. In general replace the inverse formula by an affordable selectionwhere $\mathcal{D}$ is the wealth domain; boundary choices or supporting slopes handle nondifferentiability. Existence, finite cost and replication still have to hold. A concrete obstruction is the unbounded linear terminal-wealth utility: take $U(x)=x$, $r=0$, $\mu =\sigma =1$, and the Brownian filtration. Then ${\zeta }_T=e^{-W_T-T/2}$. For $A_n=\{{\zeta }_T<1/n\}$, the bounded nonnegative claimis replicable, costs $w_0$, and has $\mathbb{E}{X}_n>n{w}_0$. Thus there is no finite optimal expected utility under the literal utility hypotheses alone. Also, bounded coefficients do not ensure market completeness in an arbitrarily enlarged filtration containing untraded randomness; the full replication formula uses the Brownian market information assumption stated above.

Use the usual Black-Scholes model with $\sigma >0$ and the filtration generated by $W$. SetFor $u>t$, independence and the Gaussian distribution exponential moment formula give ${\mathbb{E}}_P[Z_u/Z_t\mid {\mathcal{F}}_t]=1$. Thus $Z$ is a strictly positive martingale of mean one, defining an equivalent measure $Q$ on every finite horizon by $dQ/dP{|}_{{\mathcal{F}}_t}=Z_t$.

To verify the measure change directly, put ${\hat{W}}_t=W_t+\lambda t$. For any real $a$ and $h=u-t$,The conditional characteristic function is that of $N(0,h)$ and does not depend on past information. Therefore $\hat{W}$ has independent increments with Gaussian distribution and continuous paths, and is a Brownian motion under $Q$. Substituting $dW=d\hat{W}-\lambda dt$ gives $d{S}_t=S_t(rdt+\sigma d{\hat{W}}_t)$. Hence $S_t/B_t$ is a geometric Brownian motion that is a martingale and $Q$ is an equivalent martingale measure.

The form is also forced. By the Brownian martingale representation theorem, any density process that is a martingale has $d{Z}_t=a_{t}d{W}_t$. The change of measure identity makes $Z_t{S}_t/B_t$ a local martingale under $P$. Its drift, calculated by Itô formula, is $(S_t/B_t)[Z_t(\mu -r)+\sigma a_t]dt$, which must vanish. Thus $a_t=-\lambda Z_t$. Solving $d{Z}_t=-\lambda Z_{t}d{W}_t$ with $Z_0=1$ gives exactly the displayed stochastic exponential. Equivalence here is on each finite horizon, as usual for Black-Scholes pricing.

The terminal variable $F=\varphi (W_T)$ is bounded and hence square-integrable. In the completed natural Brownian filtration, the Brownian martingale representation theorem says that any square-integrable ${\mathcal{F}}_T$-measurable variable admits a representationwith $\beta$ predictable and $\mathbb{E}{\int }_0^T{\beta }_u^{2}du<\infty$. Extend $\beta$ by zero after $T$. Thus the requested constant and integrability areExpectation determines $c$ uniquely. If two integrands give the same representation, the Itô isometry gives $\mathbb{E}{\int }_0^{\infty }({\beta }_u-{\tilde{\beta }}_u{)}^{2}du=0$. Consequently $\beta$ is unique up to $d\mathbb{P}du$-almost everywhere equality, rather than pointwise equality at every time. The corresponding integral martingales are indistinguishable.

Set $Z_t=\mathbb{E}[{\varphi }^{\prime }(W_T)\mid {\mathcal{F}}_t]$. The Brownian martingale representation theorem applied to the bounded terminal variable ${\varphi }^{\prime }(W_T)$ supplies a continuous adapted version of $Z$, so it is predictable. Henceis predictable and satisfies $\mathbb{E}\int {\gamma }_t^{2}dt\le T\Vert {\varphi }^{\prime }{\Vert }_{\infty }^2$. For every square-integrable predictable process $\alpha$, conditioning at each deterministic time and using Fubini theorem givesThus part (d) says $\mathbb{E}\int ({\beta }_t-{\gamma }_t){\alpha }_{t}dt=0$. Taking $\alpha =\beta -\gamma$, which is an allowed predictable square-integrable process, makes its squared norm zero. We concludeThis is the Clark-Ocone formula for a smooth Brownian terminal payoff, with exactly the uniqueness established in part (c).

Assume the usual positive initial asset values. The coefficients determine a unique normalized local martingale deflator. The state-price density and local deflator distinction matters here: calling it a state-price density uses the local convention, while true expectation pricing needs an additional qualification, addressed below.

Let $ZB$ and $ZS$ be local martingales, with $Z_0=1$. The Brownian martingale representation theorem says that every local martingale in the usual natural Brownian filtration is continuous and is a stochastic integral against $W$. Applying it to $ZB$, and dividing by the positive finite-variation account, forcesfor a locally square-integrable predictable $\eta$. The Itô product rule for $ZS$ has driftIt must vanish. Since $S,Z,\sigma$ are positive, this yieldsConversely this drift choice makes both $ZB$ and $ZS$ local martingales. Continuity and strict positivity of $\sigma$ make $\lambda$ bounded along each path on every finite time interval, so its pathwise square integral is finite. The unique linear SDE solution isUniqueness follows from the forced drift and diffusion coefficients and uniqueness of this linear stochastic differential equation.

Continuity alone does not make the density a true martingale. For a true equivalent martingale measure, the stochastic exponential $D_t=Z_t{B}_t/B_0$ must have expectation one, for example under the Novikov condition on each horizon. True martingale pricing of all desired deflated payoffs also requires the relevant integrability. These stronger conclusions do not follow just from pathwise continuity.

An explicit counterexample to the stronger reading uses a three-dimensional Bessel process $R$ with $R_0=1$ and $dR=dW+R^{-1}dt$, which is a positive strong solution in the Brownian filtration. Take $B=1$ and $S=R$. Then $r=0$, $\mu =R^{-2}$ and $\sigma =R^{-1}$ are continuous and $\sigma >0$. The unique local candidate is $Z=R^{-1}$, with $ZS=1$. The reciprocal three-dimensional Bessel strict local martingale is not a true martingale. To verify the loss of expectation, use the standard Bessel transition densityIntegrating $p_t(1,y)/y$ gives $\mathbb{E}[R_t^{-1}]=2\Phi (1/\sqrt{t})-1<1$ for $t>0$, where $\Phi$ is the standard normal cumulative distribution function. A true pricing density for the constant bank account would have expectation one. Hence the printed hypotheses establish the local deflator statement, while a true-density reading needs an additional condition and is false as stated.

Here the bank account is continuous and of finite variation, but for $t>0$ its rate is $r_t=t^{-1/2}$. This is integrable at zero, so the account itself is well defined. The stock satisfies $d{S}_t=S_{t}d{W}_t$. The previous continuity hypothesis on the rate no longer holds at the initial time.

Suppose a positive normalized state-price density existed; even a local martingale deflator would suffice for a contradiction. Then $ZB$ would be a continuous local martingale by the Brownian martingale representation theorem. Dividing by the positive continuous account gives continuous $Z$, with $Z_0=1$, andThe Itô product rule for $ZS$ forces its drift to vanish, so ${\eta }_t=Z_t{t}^{-1/2}$ for almost every positive time. Continuity and $Z_0=1$ imply that each path has a positive interval on which $Z_t\ge 1/2$. But thencontradicting the local square-integrability required for the stochastic integral. ThusThis is the singular initial market-price-of-risk obstruction: the necessary market price of risk is $-t^{-1/2}$, whose squared integral diverges at zero. On an interval starting at a strictly positive time this particular obstruction disappears; the initial-time normalization 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.