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Use the usual normalization $Z_0=1$, finite deterministic initial wealth $x=X_0\ge 0$, positive state-price density, and nonnegative consumption. The deflated asset prices $ZP$ are local martingales, so predictable stochastic integrability makes $M=\int H\cdot d(ZP)$ a zero-starting local martingale. Integrating part (b) givesA nonnegative local martingale is a supermartingale: localize to martingales, apply Conditional Fatou lemma, and obtain the conditional expectation inequality and integrability. Therefore $N$, and hence $M=N-x$, are supermartingales. In particular,This is supermartingale control of deflated consumption gains. Nonnegative consumption is essential to the lower bound; it is the ordinary meaning of a consumption rate here. Wealth nonnegativity alone would not supply that bound if arbitrary signed cash flows were allowed.

Construct the strictly positive rolling one-period bond account using successive one-period zero-coupon bonds:At time $t-1$, invest the entire account value in the bond maturing at $t$. This gives a positive self-financing portfolio usable as a numéraire.

We use the finite-discrete-time fundamental theorem of asset pricing: in a frictionless market with finitely many adapted assets and trading dates, no arbitrage is equivalent to existence of an equivalent martingale measure for asset prices, including dividends, expressed in a positive traded numéraire. Maturing bond payoffs are reinvested in that numéraire. Let $\mathbb{Q}$ be such a measure and $L_t=\mathbb{E}[d\mathbb{Q}/d\mathbb{P}\mid {\mathcal{F}}_t]$ its positive density process. The martingale pricing relation for a unit zero-coupon bond isDefine $Z_t=L_t/{\beta }_t$. The Bayes formula for conditional expectation then givesThe positive expectations are finite because these are the traded finite bond prices; in particular $\mathbb{E}{Z}_T=Z_0{P}_0(T)$ when initial information is trivial. Normalize $Z_0=1$ without changing any ratio. With nontrivial initial information the same identities are conditional on that information. The state-price density need not be unique when the bond market is incomplete; existence is sufficient.

A unit zero-coupon bond pays one at its maturity, so $P_t(t)=1$. Monotonicity in maturity gives $P_t(t+1)\le 1$. Using the state-price density representation with $T=t+1$,The process is positive and integrable, as noted in part (a), and adapted. Thus it is a supermartingale. Strictly decreasing maturity prices yield a strict one-step conditional inequality; weak decrease is already enough for the conclusion.

The spot interest rate $r_T$ is known at $T-1$. The state-price density price of its floating payment is, by the law of total expectation,Here $P_0(0)=1$ if $T=1$. All these terms are integrable: $Z_T{r}_T$ is bounded in absolute value by $Z_T/P_{T-1}(T)+Z_T$, whose expectation is finite from the one-step pricing relation. Subtracting the fixed payment $f$ givesThis also follows directly from a floating-rate payment bond replication. At time zero buy one unit of the bond maturing at $T-1$ and short $1+f$ units of the bond maturing at $T$. Their initial cost is the displayed ${\xi }_0$. Hold them until $T-1$. The first bond then pays one; spend that one to buy $1/P_{T-1}(T)$ units of the maturity-$T$ bond, leaving the earlier short position in place. This rebalance is self-financing. At maturity the net payment isFor $T=1$, the first unit is time-zero cash, and the same immediate rebalance gives the deterministic payoff. Thus the replication establishes the zero no-arbitrage price without requiring completeness of other claims.

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.

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.

Use the state-price density as a positive local martingale deflator, so each component of $YP$ is a local martingale. The Itô product rule and the wealth equation giveThe consumption term has finite variation, and the quadratic covariation of a stochastic integral satisfies $d[Y,X]=H\cdot d[Y,P]$. Also $X=H\cdot P$. Regrouping the terms therefore givesThe differential before $YP$ is necessary: the first term is a stochastic gain, not the level of the deflated portfolio. It is missing in the printed display. This is the deflated wealth equation with consumption.

Concavity gives the supporting-tangent inequalityMultiply by the discount factor $e^{-bt}$ and use the first-order condition $e^{-bt}{U}^{\prime }(c_t)=Y_t$. This gives the pointwise marginal-utility verification of optimal consumption inequalityApply the budget inequality from part (d) to the competing admissible strategy, and use equality for the proposed one:The two weighted consumption integrals are finite, so their difference is integrable. Under the usual positive discount-rate assumption $b>0$, the utility integrals are also integrable: $U(0)\le U(x)\le L$ for a finite upper bound $L$, and ${\int }_0^{\infty }{e}^{-bt}dt=1/b$. Integrating the tangent inequality and taking expectations therefore givesEconomically, both consumers face the same state-price budget, and the candidate spends it exactly where its discounted marginal utility equals the state price. This is utility duality with martingale deflators in its consumption form.

A positive $b$, or another hypothesis making the infinite-horizon objectives well defined and permitting this integration, is needed. The printed question does not specify the sign of $b$. Bounded utility alone does not ensure that an undiscounted infinite time integral exists: a bounded integrand can have both infinite positive and negative parts. Thus the conclusion is established under the standard discount convention $b>0$, and also whenever the displayed objectives satisfy the stated integrability conditions; without either convention the literal infinite-horizon comparison need not be a defined mathematical expression.

This can occur within an admissible financial model, not just for an abstract bounded integrand. Take $U(x)=1-2e^{-x}$ and $b=0$. Choose a deterministic smooth nonnegative $c$ with successive unit-length plateaus alternating between zero and the integer $n$, and connect them over intervals whose lengths have finite sum. Then ${\int }_0^{\infty }2c_t{e}^{-c_t}dt<\infty$: the high-plateau contributions sum to ${\sum }_{n\ge 1}2n{e}^{-n}$ and the transition integrand is bounded by $2/e$. SetThe bank account is a positive deterministic Itô process and $Y_t{B}_t=Y_0$, so $Y$ is a state-price density. Holding $X_t/B_t$ bank units finances consumption with nonnegative wealth and exact budget equality. Also $U^{\prime }(c_t)=Y_t$. Nevertheless every zero plateau contributes $-1$ to the utility integral, while each sufficiently high plateau contributes a fixed positive amount. Both its negative and positive parts are infinite, so the undiscounted objective is undefined. This establishes why the missing discount or objective-integrability hypothesis is substantive.

Let $F(y)$ be the manager's utility function as a function of nonnegative terminal fund wealth. WriteThenBoth pieces are increasing and strictly concave functions. However, the derivative jumps upward from $a{e}^{-a{w}_0}$ to $b{e}^{-a{w}_0}$ at $w_0$. The payoff utility is increasing but not concave. A concave function must have nonincreasing one-sided slopes.

For concavification of incentive utility, the common tangent for exponential incentive utility replaces the upward kink by its common tangent. There are contact points $\ell <w_0<h$ and a common slope $s>0$ satisfyingSince $F(\ell )=-s/a$ and $F(h)=-s/b$, the chord condition gives $h-\ell =1/a-1/b$. Solving the derivative matching yields the contact pointsThe assumption that $w_0$ is large enough means, explicitly, thatso $\ell >0$. The inequalities $\ell <w_0<h$ follow from $\log q<q-1$ and $\log q>1-1/q$ for $q>1$.

The least concave majorant on $[0,\infty )$ isIt is increasing and concave: the derivative is decreasing outside the interval and equals $s$ inside. The tangent line lies above each original branch. Any concave majorant must lie above the chord joining the two contacts, so this one is the least.

The state-price budget constraint now reduces the problem to pointwise maximization of $\bar{U}(y)-qy$, where $q=\eta {\zeta }_T$ and $\eta >0$ is chosen so that $\mathbb{E}[{\zeta }_T{Y}^{\ast }]=w_0$. Because ${\bar{U}}^{\prime }(0)=a$ is finite, the nonnegative wealth constraint must be included. The optimal terminal wealth isAt $q=s$, any point of $[\ell ,h]$ maximizes the concavified objective. If $\kappa \ne 0$, the state-price density has a continuous log-normal distribution, so this event has probability zero. The optimizer then avoids $(\ell ,h)$ almost surely and satisfies $\bar{U}(Y^{\ast })=F(Y^{\ast })$. The supporting-line proof from part (i), with supergradients at the contacts, proves optimality for the concavified problem; the pointwise equality proves the same optimizer and value solve the original manager's problem. Nonnegative replication is available in the complete market.

If $\kappa =0$, the state-price density is deterministic. If the required deterministic mean terminal wealth lies in the linear segment, randomize between $\ell$ and $h$ at the common slope instead of choosing an interior wealth. This maintains the budget and attains the same concavified value. The Brownian market can replicate that bounded lottery even when its market price of risk is zero.

In the special case $r=\mu =0$, ${\zeta }_T=1$ and the fund has a fair-game wealth process. Since $\ell <w_0<h$, the manager uses fair-game gambling induced by an incentive fee:For example choose a threshold event in $W_T$ with the displayed probability and replicate its payoff. Then $w_t=\mathbb{E}[Y^{\ast }\mid {\mathcal{F}}_t]$ remains between $\ell$ and $h$, so the strategy respects nonnegative wealth. It earns no risk premium but raises expected incentive utility function above the value at the unrandomized $w_0$, because the common tangent lies strictly above the kink.

Assume $\sigma \ne 0$ and the usual Brownian filtration, so the single-risky-asset market is a complete market. With market price of risk $\kappa =(\mu -r)/\sigma$, the normalized state-price density isThe Itô formula shows that ${\zeta }_t{w}_t$ is a local martingale for a self-financing portfolio. For nonnegative admissible wealth it is a supermartingale, yielding the state-price budget constraintEvery integrable nonnegative terminal claim with equality is attainable in the complete market, by the Brownian martingale representation theorem. Its wealth process is $w_t={\zeta }_t^{-1}\mathbb{E}[{\zeta }_{T}X\mid {\mathcal{F}}_t]\ge 0$.

The Inada conditions $u^{\prime }(0+)=\infty$ and $u^{\prime }(\infty )=0$, together with strict concavity, make $I=(u^{\prime }{)}^{-1}$ a decreasing map from $(0,\infty )$ onto $(0,\infty )$. Pointwise optimization of $u(x)-y{\zeta }_{T}x$ gives the optimal terminal wealthwith scalar multiplier $y>0$. The question assumes a multiplier giving the required budget; utility expectations must also be well defined.

For any admissible terminal wealth $X$, the supporting-line inequality for a concave function givesTaking expectations and using the state-price budget constraint proves optimality:Strict concavity makes the optimal terminal wealth unique up to almost-sure equality.

A pricing kernel is a strictly positive random variable $R$ such that $R{P}_1^i$ is integrable andIt is a one-period state-price density, converting a terminal payoff into its initial price. Strict positivity means $R>0$ almost surely, not merely $R\ge 0$. When cash has price $1$ at both dates, this identity forces $\mathbb{E}R=1$, so $d\mathbb{Q}=Rd\mathbb{P}$ defines an equivalent martingale measure.

Let $(h_k)$ be a bounded minimizing sequence. A subsequence converges to some $h_{\ast }$, and continuity gives $F(h_{\ast })=f$. Since $F$ is smooth and $h_{\ast }$ is an unconstrained minimizer,DefineThen $\rho >0$, $\mathbb{E}\rho =1$, andThus the normalized exponential tilt is the required state-price density.

For deterministic initial asset values $A_0$ and a finite-dimensional terminal random vector $A_1$, exactly one possibility holds: a portfolio has nonpositive initial cost and nonnegative payoff with strict gain at at least one date, or there is a strictly positive integrable state-price density pricing $A_1$ as displayed. Apply the strictly positive barycentre cone lemma. Outside the closed support cone, strict separation gives a negative-cost portfolio. On its relative boundary, a supporting functional gives a zero-cost portfolio with positive payoff on a set of positive probability. With a traded positive bank account, normalize its discounted density to obtain an equivalent martingale measure.

Suppose a productive asset pays total consumption ${\delta }_t$, while price-taking agents value their consumption shares $p_t^i=c_t^i/{\delta }_t$ through discounted utility functions. Marginal pricing by a common state-price density ${\xi }_t$ gives $e^{-{\rho }_{i}t}{U}_i^{\prime }(p_t^i)/{\delta }_t=y_i{\xi }_t$. Writing $b(t)={\xi }_t{\delta }_t$, market clearing reduces to the scalar equation ${\sum }_i{I}_i(y_i{e}^{{\rho }_{i}t}b(t))=1$, where $I_i$ is the inverse marginal utility. Its unique positive solution is deterministic. Therefore the consumption shares are deterministic even when dividend levels are random.

With finite fundamental prices, the asset's price is $S_t={\delta }_{t}b(t{)}^{-1}{\int }_t^{\infty }b(s)ds$. Initial budget equations determine the multipliers. Identical preferences and discount rates make the shares constant and equal to initial ownership fractions. The aggregate benchmark is taken as given in the price-taking first-order conditions.

A state-price density converts terminal wealth and consumption into initial cost. For a nonnegative admissible self-financing portfolio with consumption, the deflated wealth plus cumulative deflated consumption is a supermartingale, giving $\mathbb{E}[{\zeta }_{T}X]+\mathbb{E}{\int }_0^T{\zeta }_t{c}_{t}dt\le w$. In a complete market, an integrable nonnegative terminal claim with full budget equality is replicable.

A terminal portfolio wealth floor is a prescribed lower bound, possibly a random variable, on terminal portfolio portfolio wealth. In a complete market, its state-price density cost cannot exceed initial portfolio wealth. Nonnegative admissibility replaces a possibly negative floor by its positive part.

If a positive terminal state-price density has essential infimum zero, let $A_n=\{{\zeta }_T<1/n\}$ and $X_n=w_0{1}_{A_n}/\mathbb{E}[{\zeta }_T{1}_{A_n}]$. Whenever $\mathbb{P}(A_n)>0$, these claims cost $w_0$ and satisfy $\mathbb{E}{X}_n>n{w}_0$. In a Brownian complete market with constant nonzero market price of risk, ${\zeta }_T$ has a lognormal distribution and all these events have positive probability. Thus bounded market coefficients and an increasing concave utility do not alone ensure a finite attained optimum: the linear utility $U(x)=x$ already fails. Usual inverse-marginal-utility formulas require further utility and integrability hypotheses.