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In a Brownian complete market with constant coefficients and nonsingular $\sigma$, maximize expected CRRA utility of terminal portfolio wealth divided by a geometric stock index. Writing $p=1/R$ and $q=n^{-1}1$, the pointwise marginal condition gives $w_T^{\ast }={\lambda }^{-p}{\zeta }_T^{-p}{J}_T^{1-p}$. Its state-price budget constraint determines ${\lambda }^{-p}=w_0/\mathbb{E}[({\zeta }_T{J}_T{)}^{1-p}]$. Conditional lognormal moments and replication then give the displayed constant dollar fractions. The first term is ordinary optimal risk exposure; the second hedges the random benchmark. The bank holds the remaining fraction, which can be negative when borrowing is allowed.

In a one-period complete market with a strictly positive numéraire, at most one positive state-price density prices every traded asset. Suppose $\rho ,{\rho }^{\prime }$ do so. For each event $A$, replicate the contingent claim $S_1^{(0)}{1}_A$. Linearity of the pricing equations givesBoth weighted densities are integrable because they price the numéraire. An integrable random variable whose integral over every event is zero vanishes almost surely: take the events where it is positive and negative. Therefore $(\rho -{\rho }^{\prime })S_1^{(0)}=0$, and strict positivity of the numéraire gives $\rho ={\rho }^{\prime }$. This argument avoids assuming that the unweighted densities themselves are integrable.

Concavification replaces a nonconcave incentive utility function by its least concave majorant. In a complete market with an atomless state-price density, the optimizing payoff may avoid all intervals where the majorant strictly exceeds the original utility, making the relaxed optimum exactly attainable. At a pricing atom corresponding to a linear segment, a lottery between contact points can attain the same value at unchanged cost.

In a complete market with a unit cash asset and normalized strictly positive state-price density $Z$, the optimal positive terminal portfolio wealth is $x/Z$ whenever it is replicable in the admissible class. The state-price budget constraint is an equality: $\mathbb{E}[Z(x/Z)]=x$. The utility conjugate inequality gives $\mathbb{E}\log H\le \log x-\mathbb{E}\log Z$ for every affordable positive claim. This value can be infinite, because $\mathbb{E}(\log Z{)}^{+}<\infty$ while $\mathbb{E}(\log Z{)}^{-}$ need not be finite. Even completeness for bounded claims suffices to approximate the value: set $b_n=\mathbb{E}[Z/(Z\vee n^{-1})]$ and $H_n=x/[b_n(Z\vee n^{-1})]$. Each claim is bounded and costs $x$, while its expected logarithmic utility tends to the displayed value by monotone convergence after an integrable lower bound is subtracted.

In a one-period complete market, an invertible payoff Gram matrix identifies the unique holdings replicating each claim. The pricing weight is $Z=P_0^T{Q}^{-1}{P}_1$ and the random portfolio kernel is $W=Q^{-1}{P}_1$. Positive definiteness gives uniqueness of holdings; positivity of $Z$ requires an additional no-arbitrage condition.

For a positive state-price density $Z$ pricing every attainable payoff, a feasible positive terminal payoff $W^{\ast }$ is optimal at its budget if its marginal utility is $yZ$ for some scalar $y>0$. The pointwise conjugate inequality gives $\mathbb{E}U(W)\le \mathbb{E}\hat{U}(yZ)+yx$ for every competitor; the marginal-utility identity makes equality hold for $W^{\ast }$. Finite states make integrability straightforward. In a complete market an interior optimum also necessarily has this proportionality, by variations preserving the state-price budget.

Fix a finite horizon $N$, a filtration $({\mathcal{F}}_n)$, a positive bank account $B_n$ with predictable one-step growth, ex-dividend risky prices $S_n$, and asset dividends $D_n$ paid at date $n$. Put $s_n=S_n/B_n$ and $d_n=D_n/B_n$. An equivalent martingale measure $Q$ makes the discounted cum-dividend gainsmartingales; equivalently ${\mathbb{E}}_Q[s_n-s_{n-1}+d_n\mid {\mathcal{F}}_{n-1}]=0$. It is these gains, not necessarily the discounted ex-dividend prices alone, which enter the definition.

Let ${\theta }_n$ be risky holdings chosen at date $n-1$ and carried to date $n$. Let $C_n$ be the portfolio's net cash distribution to its owner at date $n$, and $V_n$ its remaining value after that distribution and rebalancing. A self-financing portfolio with these distributions has the budget identityTo see this, value the previous stock and bank holdings at the new prices, add the asset dividends, subtract the owner distribution, and observe that the discounted bank holding is unchanged. Rebalancing then alters holdings but not value. An asset dividend reinvested in the portfolio is already in the gain term; it is not also an owner distribution.

For a predictable strategy whose gains are integrable under $Q$ (bounded holdings in a finite-state market suffice), conditioning the gain increment gives zero. Therefore discounted remaining wealth plus cumulative discounted owner payments is a martingale. Conditioning its terminal value proves the multi-period dividend pricing identityThe relation holds under every equivalent martingale measure for which these integrability conditions hold. Each payment is discounted from its own payment date, not from one common terminal date. The expression inside the expectation is in bank account units; multiplication by $B_n$ returns current currency units.

When the portfolio is liquidated and all its final value is paid out, $V_N=0$ after liquidation and that value is included in $C_N$. The displayed identity then involves only future discounted distributions. Without liquidation, the terminal value must remain. A bank account holding with no coupons before redemption has positive value despite having no intervening dividends; omitting its redemption or terminal holding value would plainly give the wrong result. In an infinite horizon, deleting the terminal term instead requires a dividend-price transversality condition, namely convergence of its conditional discounted expectation to zero.

For a traded asset the same proof gives its discounted expected dividends plus terminal sale value. For a replicated claim paid only at $N$, it gives $V_n=B_n{\mathbb{E}}_Q[H/B_N\mid {\mathcal{F}}_n]$. In a complete market this prices every admissible claim uniquely. In an incomplete market only the portfolio's attainable payout stream has this common value under all measures; an arbitrary unattainable payoff can have different expectations under different measures. For a merely dominated measure the conditional identity holds $Q$-almost surely; it makes no assertion on physical states assigned zero mass by $Q$. External capital injections must also be counted, as negative owner distributions. Thus the valuation principle follows from the gain martingale and the trading budget, together with the terminal convention and integrability, rather than from unconditional averaging of arbitrary future cash flows.

Normalize the riskless financial asset to $B_0=1$, $B_1=b>0$, and write $S_0,S_1$ for the $s$ risky asset prices. Their discounted gain random vector is $Y=S_1/b-S_0$. A zero-cost portfolio holding $h\in {\mathbb{R}}^s$ in the risky assets and $-h\cdot S_0$ in the riskless asset has terminal value $bh\cdot Y$. Thus an arbitrage meansAn equivalent martingale measure is a probability measure $Q$ with the same null sets as $P$, under which $Y$ is integrable and ${\mathbb{E}}_{Q}Y=0$. The discounted risky prices are then martingales over the single trading period.

If such $Q$ exists, a nonnegative terminal gain has expectation zero under $Q$. A nonnegative random variable with zero expectation vanishes almost surely: on $\{X\ge 1/n\}$ its expectation is at least $\mathbb{P}(X\ge 1/n)/n$, and the union of these events is $\{X>0\}$. Equivalence transfers this conclusion to $P$, excluding arbitrage.

For the converse we prove the needed positive-density separation proof of the one-period asset-pricing theorem. First make the gain integrable without changing its null sets. Define an equivalent probability measure $P_0$ byLet $N=\{h:h\cdot Y=0\text{ almost surely}\}$ and let $L=N^{\perp }$ be its orthogonal complement. Taking a finite basis of $N$ shows $Y\in L$ almost surely. If $L=\{0\}$, the gain is zero and $P_0$ already gives the required measure. Otherwise consider the positive-weight expectation conePositive combinations of weights show that $C$ is a convex cone. It is open relative to $L$. Indeed, for any admissible weight $Z$, perturb it toThese weights remain bounded and strictly positive, whileFor every nonzero $v\in L$, $v\cdot Y$ is nonzero on an event of positive probability; otherwise $v\in N\cap L$. ConsequentlyThe matrix is finite because $\Vert Y{\Vert }^2/(1+\Vert Y\Vert )\le \Vert Y\Vert$. The kernel of $A_Z$ on $L$ is zero, so it defines an invertible matrix on the finite-dimensional vector space $L$. For any sufficiently small $\delta >0$, every $y$ with $\Vert y\Vert <\delta /\Vert A_Z^{-1}\Vert$ is the image of a vector in the radius-$\delta$ ball, since $\Vert A_Z^{-1}y\Vert <\delta$. Hence the perturbations give a neighborhood of zero, proving relative openness of $C$.

If $0\notin C$, the hyperplane separation theorem gives a nonzero $h\in L$ with $h\cdot c\ge 0$ for every $c\in C$. The separating constant can be taken to be zero because $C$ is a cone. Put $A=\{h\cdot Y<0\}$ and use the strictly positive weights $Z=1_A+\epsilon$. Taking the limit as $\epsilon \downarrow 0$ yields ${\mathbb{E}}_{P_0}[(h\cdot Y)1_A]\ge 0$, forcing $P_0(A)=0$. But $h\notin N$, so $P_0(h\cdot Y>0)>0$. This is an arbitrage, a contradiction. Hence $0\in C$: some bounded strictly positive $Z$ satisfies ${\mathbb{E}}_{P_0}[ZY]=0$. Normalize it by $dQ/d{P}_0=Z/{\mathbb{E}}_{P_0}Z$. This gives an equivalent martingale measure with integrable gains, establishing the fundamental theorem of asset pricing here without any further existence theorem.

Now retain only the $m$ positive-probability states of the finite probability space. Let $V\subseteq {\mathbb{R}}^m$ be the linear span of the discounted terminal asset payoffs; it contains the constant vector $1$. A complete market means $V={\mathbb{R}}^m$. An equivalent martingale measure is a vector $q$ of strictly positive state probabilities whose pairing with each asset payoff gives its initial price. By linearity it prices every attainable payoff. If $V={\mathbb{R}}^m$, the indicator functions of individual states are attainable, so these prices determine every $q_j$ uniquely.

If $V$ is proper, choose $0\ne v\in V^{\perp }$. Since $1\in V$, ${\sum }_j{v}_j=0$. For sufficiently small $\epsilon >0$, both $q+\epsilon v$ and $q-\epsilon v$ have strictly positive entries, sum to one, and give the same pairings with every asset payoff. They are distinct equivalent martingale measures. Therefore market completeness is equivalent to uniqueness of the equivalent martingale measure.

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.

The complete market makes every state-contingent consumption vector attainable. Its budget is ${\sum }_s{p}_s{c}_s=100$, and expected logarithmic utility under the equally likely states is $\frac{1}{4}{\sum }_s\log c_s$, with $c_s>0$. Increasing utility makes the budget bind. For a Lagrange multiplier $\eta$, the first-order conditions areSubstitution in the budget gives $1/\eta =100$, so $\eta =0.01$ andEvery state receives an expenditure $p_s{c}_s=25$, rather than equal consumption. Strict concavity of logarithmic utility makes this feasible stationary point the unique global maximizer. The physical probabilities enter the utility objective; the supporting prices enter the budget, and they must not be interchanged.

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 $k=\delta$.

Use the riskless financial asset as numéraire, normalized to have initial value one and terminal value $R>0$. Write $S_0\in {\mathbb{R}}^s$ for the initial risky prices and $S_1$ for the finite, measurable terminal price random vector. Define the discounted gain vector $X=S_1/R-S_0$. A risky holding $h$ financed by the cash holding $-h\cdot S_0$ has zero initial cost and discounted terminal gain $h\cdot X$. An arbitrage is such a portfolio with $h\cdot X\ge 0$ almost surely and $\mathbb{P}(h\cdot X>0)>0$. Allowing nonpositive initial cost is equivalent: invest any initial surplus in the riskless financial asset to obtain a zero-cost arbitrage.

An equivalent martingale measure is a probability measure $Q$ with exactly the same null events as the physical probability measure $P$, under which discounted asset prices are integrable and satisfy ${\mathbb{E}}_Q[S_1/R]=S_0$. Equivalently, ${\mathbb{E}}_{Q}X=0$. Such a measure rules out arbitrage: every zero-cost portfolio has mean discounted gain zero, whereas a nonnegative gain positive with positive $P$-probability is also positive with positive $Q$-probability and has strictly positive $Q$-expectation.

For the converse, the probability space need not be finite, and the payoffs need not be integrable under $P$. Here is a direct Gaussian-damped martingale density construction. SetThe set $N$ is a linear subspace. Choosing a finite basis of $N$ shows that $X\in L$ almost surely: intersect the probability-one events on which each basis vector is orthogonal to $X$. If $h\in L\setminus \{0\}$, then $h\cdot X$ is not almost surely zero. Absence of arbitrage implies both $P(h\cdot X<0)>0$ and $P(h\cdot X>0)>0$, by applying the definition to $h$ and $-h$.

All the analytical steps in minimizing $F$ can be checked directly. On $|h|\le H$ the integrand is bounded by $e^{H^2/4}$, since $-|X{|}^2+H|X|\le H^2/4$. Its first two directional derivatives are bounded uniformly in $X$ by constants depending only on $H$ and the direction, because the functions $r^j{e}^{-r^2+(H+1)r}$ are bounded for $j=0,1,2$. The mean-value bound therefore proves continuity of $F$, and the Taylor theorem remainder bound, integrated over $P$, provesThus differentiation under the integral does not require an unproved physical moment assumption.

To prove coercivity, suppose $|h_k|\to \infty$ in $L$. A subsequence of $h_k/|h_k|$ converges to a unit vector $v\in L$. This elementary finite-dimensional compactness follows by successively taking convergent subsequences of the bounded coordinates. Since $P(v\cdot X<0)>0$, some event $A=\{v\cdot X\le -\epsilon ,|X|\le K\}$ has positive probability: the negative-gain event is the countable union of such events with $\epsilon =1/j$ and integer $K$. Eventually $(h_k/|h_k|)\cdot X\le -\epsilon /2$ on $A$, and henceIf coercivity failed, an unbounded sequence with bounded $F$ would have just such a contradictory subsequence. Consequently a minimizing sequence for $F$ is bounded and has a convergent subsequence; continuity makes its limit $h_{\ast }$ a minimizer. If $L=\{0\}$, simply take $h_{\ast }=0$. In every case the directional derivatives vanish at $h_{\ast }$. The vector ${\mathbb{E}}_P[X{e}^{-|X{|}^2-h_{\ast }\cdot X}]$ lies in $L$ and is orthogonal to $L$, so it is zero. Thereforeis a positive mean-one density defining an equivalent martingale measure. Integrability of $X$ under $Q$ follows from the same Gaussian bound with $j=1$. This establishes absence of arbitrage if and only if an equivalent martingale measure exists, without quoting a financial duality theorem or assuming a finite state space.

Now suppose the probability space is finite. Discard null states and enumerate its positive-probability atoms by $1,\dots ,m$. Every measurable terminal payoff is a vector in ${\mathbb{R}}^m$. Let $A$ be the $m\times (s+1)$ matrix whose columns are the discounted terminal payoffs of the traded financial assets, with first column $1$, and let $p=(1,S_0)$ be their initial price vector. A portfolio with holding vector $z$ replicates the discounted payoff $Az$, and an equivalent martingale measure is exactly a vector $q$ satisfyingThe first pricing equation includes ${\sum }_j{q}_j=1$. A complete market means $\mathrm{im}A={\mathbb{R}}^m$. If $q,q^{\prime }$ both price the financial assets, then $(q-q^{\prime })\cdot Az=0$ for every $z$. Completeness allows $Az$ to be each coordinate vector, so $q=q^{\prime }$.

Conversely, if the financial market is incomplete, there is a nonzero vector $v$ with $A^{T}v=0$. To see this without invoking a duality theorem, perform row elimination on the homogeneous system: its number of pivots is $\mathrm{rank}A<m$, leaving a free variable and hence a nonzero solution. The constant payoff column gives ${\sum }_j{v}_j=0$. Existence of an equivalent martingale measure was already proved, so choose one strictly positive vector $q$. Forboth $q+\epsilon v$ and $q-\epsilon v$ are distinct strictly positive probability vectors satisfying $A^T(q\pm \epsilon v)=p$. Thus uniqueness fails. On a finite arbitrage-free market, completeness is equivalent to uniqueness of the equivalent martingale measure.

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.

Normalize by a riskless bank account with $B_0=1$, $B_1=R>0$. For $d$ risky assets let $Y=S_1/R-S_0$ be the vector of discounted gains, and let $H$ denote a discounted square-integrable contingent claim. An initial total capital $x$ and stock holding $\vartheta$ have discounted terminal portfolio wealth $x+{\vartheta }^{T}Y$: the bank holding finances the initial purchase of the stocks. Thus the space of attainable claims isIt is finite-dimensional and hence closed. The market is complete market when every square-integrable claim is in this span; with finitely many assets this is also equivalent to replication of every bounded claim.

A dominated martingale measure $Q$ is a probability measure with $Q\ll P$ and ${\mathbb{E}}_{Q}Y=0$, with the gains integrable under $Q$. Its Radon-Nikodym derivative satisfies $Z\ge 0$, $\mathbb{E}Z=1$ and $\mathbb{E}[ZY]=0$. An equivalent martingale measure has $Z>0$ almost surely. A signed martingale measure obeys the same normalization and pricing equations without requiring nonnegativity. The distinction matters for quadratic hedging. For the completeness assertion use the usual no-arbitrage hypothesis, expressed here by existence of an equivalent martingale measure $Q_0$.

The unrestricted one-period quadratic hedge minimizesPut $m=\mathbb{E}Y$, $C=\mathrm{Cov}(Y)$ and $k=\mathrm{Cov}(Y,H)$. Remove redundant gains so that $C$ is invertible. Differentiating in $x$ gives $x=\mathbb{E}H-{\vartheta }^{T}m$. The remaining squared error isThereforeThe residual is orthogonal in $L^2(P)$ to $1$ and every gain, making the fitted claim the orthogonal projection onto a finite-dimensional subspace onto $\mathcal{A}$. Its minimum mean-square error is $\mathrm{Var}(H)-k^T{C}^{-1}k$, and it vanishes exactly when $H$ is attainable. If initial capital $x$ is fixed instead, the normal equation isThis separates hedging with prescribed capital from simultaneous optimization of capital and risky holdings.

The minimal martingale measure in a one-period market is characterized by leaving unchanged the expectation of every centered square-integrable $L$ orthogonal to the martingale part $Y-m$. For square-integrable pricing densities this forces $Z-1$ into the span of $Y-m$: the orthogonal complement consists exactly of those $L$. The pricing equation then determinesIndeed $\mathbb{E}{Z}_{\ast }=1$ and $\mathbb{E}[Z_{\ast }Y]=m-C{C}^{-1}m=0$. For every such orthogonal $L$, $\mathbb{E}[Z_{\ast }L]=0$. Conversely a normalized density preserving all these $L$ must have the displayed form, since its coefficients must satisfy $C\beta =-m$. The least-squares initial capital is consequently $x_{\ast }=\mathbb{E}[Z_{\ast }H]$.

For any other square-integrable signed pricing density $Z$, put $L=Z-Z_{\ast }$. Both densities price $1$ and all gains, so $L$ is orthogonal to their span, including $Z_{\ast }$. HenceEquality holds only when $Z=Z_{\ast }$. Densities with infinite second moment cannot improve this finite minimum. This proves the requested minimum over dominated martingale measures whenever $Z_{\ast }\ge 0$, and over all square-integrable signed martingale measures without that extra condition. If $Z_{\ast }>0$, it is also an equivalent martingale measure.

Absence of arbitrage alone does not guarantee that the minimal density is a probability density. For a concrete example, let a single discounted gain take values $-1,1,10$ with probabilities $1/10,4/5,1/10$. Then $m=17/10$, $C=801/100$ and $Z_{\ast }(10)=-610/801$. Nevertheless pricing probabilities $(109/200,89/200,1/100)$ are strictly positive and give mean gain zero, so an equivalent martingale measure exists. This is negative minimal density in an arbitrage-free one-period market. If the term martingale measure is reserved for probabilities, the printed minimum assertion needs the extra nonnegativity hypothesis; otherwise it is the signed minimum-norm assertion just proved. Minimizing over the nonnegative pricing densities instead is a different constrained projection problem.