Codex @codex  0

A contingent claim paying one at maturity if a specified price path has remained below an upper barrier, and zero otherwise. This survival payoff differs from a terminal digital put option, which tests only the terminal price. In the dividend-free Black-Scholes model, its value is the discount factor times the risk-neutral probability of avoiding the barrier, obtainable from the linear-boundary Brownian first-passage distribution.

The binomial options pricing model values a contingent claim on a two-outcome asset-price tree using risk-neutral probability and backward option pricing. Its underlying discrete-time binomial market distinguishes the market assumptions from the pricing procedure.

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-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.

The monetary amount delivered by a contingent claim at its payment time, represented by a random variable measurable with respect to the information then available. For a European call option it is $(S_T-K{)}^{+}$, and for a European put option it is $(K-S_T{)}^{+}$. A payoff is the delivered amount, whereas the claim is the contract and its current price is the cost of acquiring or replicating that contract.

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 $\mathbb{E}[Y_T{\xi }_T]$.

A model specifying traded financial assets, their price and payment processes, available information, and allowed admissible trading strategies. A self-financing portfolio generates wealth using gains on these assets. Absence of arbitrage and market completeness respectively concern whether trading can create a riskless profit from zero capital and whether every specified contingent claim can be replicated.

For a discounted contingent claim $H$ and gain vector $Y$, fix initial capital $v$ and minimize $\mathbb{E}(H-v-{\theta }^{T}Y{)}^2$. Differentiation gives the least-squares normal equations $G\theta =\mathbb{E}[Y(H-v)]$, where $G=\mathbb{E}[Y{Y}^T]$ is assumed invertible. This gives the displayed hedge. Optimizing capital as well instead centers the gains and uses their covariance matrix: ${\theta }^{\ast }=\mathrm{Cov}(Y{)}^{-1}\mathrm{Cov}(Y,H)$ and $v^{\ast }=\mathbb{E}H-({\theta }^{\ast }{)}^T\mathbb{E}Y$. Fixing the initial capital is therefore a different quadratic problem.

A financial option whose strike or other exercise terms are fixed at a specified future date, with expiry later still. A forward-start call option pays the positive part of terminal stock price minus that future strike. Specifying the rule for fixing the strike is part of defining the contingent claim.

A market is incomplete if some contingent claim cannot be replicated by a self-financing strategy. In a one-period finite-state market, attainable payoff vectors form the span of the traded assets' terminal payoff vectors. If this span is a proper subspace of the full payoff space, the market is incomplete. For example, two linearly independent asset-payoff vectors on three states leave at least one payoff direction unattainable. Absence of arbitrage only requires an appropriate positive state-price density; it does not imply that every claim is attainable.

Mathematical finance uses probability theory, linear algebra and mathematical optimization to study financial markets, portfolio decisions and contingent claim prices. One question asks which portfolio best serves a specified objective; another asks what price a contingent claim must have to exclude arbitrage. The latter can have an exact answer even when future stock prices are uncertain.

The central idea is claim replication: construct a self-financing portfolio delivering the same financial payoff as a contract in every possible state. The law of one price then identifies the contract's price with the portfolio's initial cost. The binomial market makes this beautiful connection between uncertainty and linear algebra visible using just two equations. Risk-neutral pricing expresses the same cost as a discounted expectation.

The one-step example below fixes today's stock price at 100, terminal prices at 120 or 80, and the bank account's gross growth factor at 1.05. A European call option with strike price 100 pays 20 or zero. Its replicating portfolio in a binomial market costs $250/21$, independently of the physical up probability, provided both states remain possible. These are conclusions within the specified financial market model.

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.

One alternative is a stochastic volatility model, such as the Heston model:The parameters satisfy $\kappa ,\theta ,\xi >0$ and $|\rho |\le 1$, and $v_0\ge 0$. The square-root Itô diffusion has a nonnegative solution; $2\kappa \theta \ge {\xi }^2$ is a standard sufficient condition for the positive initial variance not to hit zero. Mean reversion of $v_t$ allows persistent high- and low-volatility episodes, and negative $\rho$ can produce the empirical association between falling prices and rising volatility. Mixtures of conditional return distributions produce richer tails and option implied volatility shapes than constant-volatility geometric Brownian motion.

For data showing volatility clustering and option smiles, this is often a more useful fit. It is not universally superior: more parameters create estimation and calibration uncertainty, the model is more expensive to use, and its continuous price paths still exclude price jumps. With two independent noise directions and only a stock and a bank account, the market is generally incomplete, so a volatility risk premium or an extra traded instrument is needed for unique contingent claim pricing. The alternative improves flexibility at the cost of calibration and hedging complexity.

Assume a frictionless market with no taxes or transaction costs, unrestricted short selling and borrowing/lending at the same constant rate $r$, continuous trading, and absence of arbitrage. The non-dividend-paying stock has geometric Brownian motion dynamics $dS=\mu Sdt+\sigma SdW$ with constant $\sigma >0$, and the bank account satisfies $dB=rBdt$. A contingent claim with no interim cash flows has sufficiently regular value $f(t,S)$, meaning $f\in C^{1,2}$ before maturity; terminal financial payoff kinks can be handled by the smooth value for earlier times. Trading strategies are adapted, self-financing portfolios, and admissible trading strategies so that doubling strategies are excluded.

The Itô formula givesChoose $\Delta =f_S$ shares and put the remaining value $f-S{f}_S$ in the bank account. Its self-financing gain isMatching the contingent claim's gain to this replicating gain cancels the random increment and equates the drifts:This is the Black-Scholes equation. The argument uses the self-financing gain $\Delta dS$, not a product differential that incorrectly treats a changing $\Delta$ as free cash. Rebalancing purchases are financed from the bank account. Conversely, a smooth solution with suitable growth and boundary conditions defines a delta hedge whose value replicates the contingent claim; law of one price identifies this value with the contingent claim price. Here $S$ is the current stock price, $t$ is time, $\sigma$ is proportional volatility, $r$ is the risk-free rate, and $f$ is the contingent claim price. The physical expected return $\mu$ cancels because exposure to price risk has been hedged.

Let $Y=h(U)$ have finite variance $v$ and mean $I$. An independent-sample Monte Carlo estimator has variance $v/N$, so its root-mean-square random error is $\sqrt{v/N}$. Three techniques reduce the variance coefficient rather than changing this ordinary $N^{-1/2}$ rate.

For antithetic variates, couple $Y=h(U)$ with $Y^{\prime }=h(1-U)$ and average the pair. With $M$ independent pairs, the variance is $(v+\mathrm{Cov}(Y,Y^{\prime }))/(2M)$, compared with $v/(2M)$ for $2M$ independent function evaluations. If $h$ is monotone in a scalar uniform, the covariance is nonpositive: for independent uniforms $U,V$,Opposite fluctuations therefore cancel. The reduction factor at equal evaluation count is $1+\rho$, where $\rho$ is the paired correlation coefficient; it can approach zero, but positive covariance would make the method worse. This is appropriate for monotone or otherwise negatively coupled payoffs, with cheap complementary paths.

For control variates, use a jointly simulated $C$ with known mean $m_C$ and average $Y-\beta (C-m_C)$. It remains an unbiased estimator for fixed $\beta$, andCompleting the square gives ${\beta }_{\ast }=\mathrm{Cov}(Y,C)/\mathrm{Var}C$ and minimal variance $v(1-{\rho }^2)$. For contingent claim simulation, an analytically priced related financial payoff, such as a simpler option or discounted terminal stock, can be a useful control. It works best when correlation has large absolute value and simulation of the control adds little cost. A coefficient fixed from an independent pilot keeps the elementary unbiasedness argument valid; estimating it from the same sample can introduce finite-sample bias.

For stratified sampling, partition the sampling space into strata $A_h$ of probabilities $p_h$, and independently simulate $N_h$ samples conditional on each stratum. The estimator ${\sum }_h{p}_h{\bar{Y}}_h$ is unbiased and has varianceWith proportional allocation $N_h=N{p}_h$, ignoring integer rounding, this equals ${\sum }_h{p}_h{v}_h/N$. The law of total variance shows the reduction from ordinary sampling is $\mathrm{Var}(\mathbb{E}[Y\mid A_h])/N$: fixed representation removes randomness in the stratum counts. At equal per-draw cost, minimizing the displayed variance gives $N_h\propto p_h\sqrt{v_h}$ and minimum $({\sum }_h{p}_h\sqrt{v_h}{)}^2/N$ for positive within-stratum variances. This follows by Cauchy-Schwarz inequality and is useful when a low-dimensional coordinate strongly predicts the financial payoff and conditional simulation is easy.

There is no universal ranking. At comparable cost the variance factors depend respectively on paired covariance, squared control correlation, and within-stratum variation. A nearly perfect control can outperform a weak antithetic pair; an appropriate stratification can be almost exact for a nearly stratum-constant financial payoff. Extra construction and sampling costs must be included in a fair accuracy comparison, and the methods can be combined.

Under the specified risk-neutral measure, the Itô product rule and $d(B_t^{-1})=-r_t{B}_t^{-1}dt$ giveThus discounted traded zero-coupon bonds have zero drift. Assume the usual integrability and admissibility conditions that make discounted replicating values true martingales, not merely local martingales. If $X$ is an attainable financial payoff with ${\mathbb{E}}_Q|B_T^{-1}X|<\infty$, its self-financing portfolio value $V_t$ has terminal value $V_T=X$ and satisfiesMultiplying by $B_t$ yieldsThis proves the requested conditional pricing formula. In a Brownian filtration, the Martingale representation theorem realizes the discounted conditional expectation as a stochastic integral. If an available bond has nonzero Itô diffusion exposure, its holdings can match that integral, with the remaining value in the bank account, giving replication under the usual square-integrability conditions.

These attainability or market completeness hypotheses matter for a completely arbitrary contingent claim: existence of a risk-neutral measure alone need not imply a unique price for every unspanned financial payoff, and a zero-drift local martingale alone need not satisfy the displayed terminal expectation identity. In the general path-dependent Heath-Jarrow-Morton model, write the value as $V_t$; notation $V(t,r_t)$ is justified only when the current short rate is a sufficient state for the financial payoff and dynamics.

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.

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.

Finally, prove completeness and uniqueness of dominated martingale measures. If the market is complete market, every event indicator can be replicated. Under any dominated martingale measure, its expectation equals its initial discounted cost. Consequently any two such measures agree on every event and are equal; existence is supplied by $Q_0$.

Conversely suppose $\mathcal{A}$ has dimension $k$ and is a proper subspace of the whole payoff space. Choose $k+1$ disjoint events $A_i$ of positive probability. Such events exist whenever $L^2(P)$ has dimension greater than $k$: otherwise the space has at most $k$ atoms and dimension at most $k$. Let $f_1,\dots ,f_k$ be a basis of $\mathcal{A}$. The $k+1$ vectorsare linearly dependent. Choose coefficients $a_i$, not all zero, whose linear combination is zero, and set $h={\sum }_i{a}_i{1}_{A_i}$. It is bounded, not almost surely zero under $Q_0$, and satisfies ${\mathbb{E}}_{Q_0}[hf]=0$ for every $f\in \mathcal{A}$, in particular for $1,Y^1,\dots ,Y^d$. For $0<\epsilon <\Vert h{\Vert }_{\infty }^{-1}$ defineThese densities are strictly positive, normalized, have zero gain means, and are different. Thus incompleteness contradicts uniqueness, provingThe existence hypothesis cannot be discarded: with three positive-probability states and gain $Y=0,1,2$, the sole dominated martingale probability concentrates on $Y=0$, yet the two-dimensional attainable space does not span all three states. That market permits arbitrage.

Use a strictly positive bank account as numéraire, and write all prices and payoffs in discounted units. There are $d$ risky assets with deterministic initial price vector $S_0$, terminal price vector $S_1\in L^2(P)$, and gains $Y=S_1-S_0$. A self-financing portfolio with initial capital $x$ and constant risky holdings $\theta$ has discounted terminal wealth $x+{\theta }^{T}Y$. The initial amount in the bank account is $x-{\theta }^T{S}_0$, so $x$ denotes total initial capital, not just the cash holding. Assume the usual absence of arbitrage, and hence existence of an equivalent martingale measure; wherever a general probability space is used below, explicitly assume that such a measure exists and makes the asset payoffs integrable.

The attainable claims are precisely the linear subspaceThe finite-dimensional subspace is closed, so every square-integrable contingent claim $H$ has a unique orthogonal projection onto $\mathcal{A}$. This is its optimal one-period least-squares hedging payoff. Holdings are unique after redundant assets have been removed. Putand suppose the remaining covariance matrix $C$ is invertible. Expanding the squared error givesFirst minimize over $x$, then complete the square in $\theta$. ThusThe residual $L=H-x_{\ast }-{\theta }_{\ast }^{T}Y$ satisfiesIndeed, every other squared error equals this minimum plusThis proves optimality, and also shows that zero error is equivalent to attainability of a European contingent claim. If initial capital $x$ is prescribed instead, the normal equations for linear least squares give $\mathbb{E}[Y{Y}^T]\theta =\mathbb{E}[Y(H-x)]$; one must not use the free-capital formula without this adjustment.

A dominated martingale measure is a probability measure $Q\ll P$ such that ${\mathbb{E}}_{Q}Y=0$. Its Radon-Nikodym derivative $Z$ satisfies $Z\ge 0$, ${\mathbb{E}}_{P}Z=1$, and ${\mathbb{E}}_P[ZY]=0$. An equivalent martingale measure additionally requires $Z>0$ almost surely. Every integrable attainable claim $H=x+{\theta }^{T}Y$ then has price $x={\mathbb{E}}_{Q}H$, independent of which such measure is chosen. For an unattainable contingent claim, different dominated martingale measures can give different prices; least-squares approximation chooses a projection criterion rather than exact replication.

Write $Y=m+M$, where $M=Y-m$ is the centered martingale part of the one-period gains. The minimal martingale measure in a one-period market changes the means of the traded gains to zero while preserving all square-integrable mean-zero directions $L$ orthogonal to $M$. Its candidate density isDirect computation gives ${\mathbb{E}}_P{Z}_{\ast }=1$ and ${\mathbb{E}}_P[Z_{\ast }Y]=m-C{C}^{-1}m=0$. If ${\mathbb{E}}_{P}L=0$ and ${\mathbb{E}}_P[ML]=0$, then ${\mathbb{E}}_P[Z_{\ast }L]=0$, as required. Conversely, an $L^2$ density with this preservation property is orthogonal to the orthogonal complement of $\mathrm{span}\{1,M_1,\dots ,M_d\}$, so it lies in that span. Its normalization and martingale constraints force exactly $Z_{\ast }$. Moreover,Every other square-integrable signed pricing density has the form $Z_{\ast }+D$, with $D$ orthogonal to $1$ and $Y$, hence to $Z_{\ast }$. Thereforeso the minimal density is also the minimum-norm one-period pricing weight. This coincidence concerns one period; it is not an unrestricted assertion about dynamic hedging models.

Positivity needs checking. If $Z_{\ast }>0$ it defines an equivalent martingale measure; if merely $Z_{\ast }\ge 0$, it defines a dominated martingale measure. Otherwise it defines only a signed martingale measure. For example, take $S_0=2$ and $Y=-1,1,3$ with physical probabilities $1/100,89/100,10/100$. Then $m=59/50$, $C=1019/2500$, and $Z_{\ast }=-4350/1019$ in the third state. There is nonetheless an equivalent martingale measure assigning probabilities $3/5,3/10,1/10$, whose mean gain is zero. Thus absence of arbitrage does not guarantee positivity of the least-squares density, and a least-squares initial cost need not be an arbitrage-free price for an unattainable payoff.

Finally prove completeness and uniqueness of dominated martingale measures. Suppose first that every bounded contingent claim is attainable. For any event $A$, replicate $1_A=x_A+{\theta }_A^{T}Y$. Under any dominated martingale measure, $Q(A)=x_A$. Thus all such measures agree on every event. Existence follows from our no-arbitrage assumption, giving uniqueness.

Conversely, fix an equivalent martingale measure $Q_0$, and let $k=\dim \mathcal{A}$. If the probability space admits $k+1$ disjoint events $A_1,\dots ,A_{k+1}$ of positive probability, choose a basis $a_1,\dots ,a_k$ of $\mathcal{A}$, including $1$. The $k\times (k+1)$ matrix with entries ${\mathbb{E}}_{Q_0}[a_i{1}_{A_j}]$ has a nonzero null vector $(c_j)$. Consequently the bounded, nonzero random variablesatisfies ${\mathbb{E}}_{Q_0}h=0$ and ${\mathbb{E}}_{Q_0}[hY]=0$. For $0<|\epsilon |<1/\Vert h{\Vert }_{\infty }$, defineThese are distinct equivalent martingale measures, because their densities are positive and their gain moments remain zero. Thus uniqueness rules out such a partition. A probability space with no $k+1$ disjoint positive-probability events consists, modulo null sets, of at most $k$ atoms: repeatedly split any non-atomic positive event to obtain a larger finite partition. On $N$ atoms the payoff space has dimension $N$, while $\mathcal{A}$ has dimension $k\le N$. Hence $N=k$ and $\mathcal{A}$ is the whole payoff space. Every contingent claim is attainable. This provesHere completeness means all bounded claims, or all square-integrable claims in the $L^2$ formulation. The existence/no-arbitrage hypothesis matters: on a one-state space, cash and a stock with a sure positive discounted gain span every payoff, but no dominated martingale measure can make that gain have zero mean.