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In an arbitrage-free one-period market with finitely many assets and an equivalent martingale measure, market completeness is equivalent to uniqueness of the dominated martingale measure. Completeness prices every indicator function, forcing two measures to agree on every event. Conversely, if the attainable payoff span has dimension $k$ but is not the whole payoff space, choose $k+1$ disjoint positive-probability events. Their moment vectors against a basis of attainable payoffs are linearly dependent. A nonzero bounded linear combination $h$ of their indicators consequently has zero pricing moments. Multiplying an existing equivalent density by $1+\epsilon h$, for sufficiently small positive and negative $\epsilon$, constructs distinct equivalent martingale measures. This proves the converse without assuming that the original space was finite. On a general probability space, completeness here means replication of all bounded claims, or equivalently all square-integrable claims when asset payoffs are square-integrable.

Let $m=\mathbb{E}Y$ and $C=\mathrm{Cov}(Y)$ be an invertible covariance matrix of discounted gains. The displayed signed martingale measure leaves unchanged the mean-zero random variables orthogonal to the martingale part $Y-m$: if $\mathbb{E}L=0$ and $\mathbb{E}[(Y-m)L]=0$, then $\mathbb{E}[Z_{\ast }L]=0$. Conversely, preservation of all these orthogonal directions forces the density to lie in the span of $1,Y-m$, and its pricing constraints give the displayed formula. If $Z_{\ast }>0$ it is an equivalent martingale measure; if $Z_{\ast }\ge 0$ it is a dominated martingale measure. Positivity is an extra condition, not a consequence of absence of arbitrage alone. In one period this density also minimizes $\mathbb{E}{Z}^2$ among square-integrable signed pricing densities, because every other such density differs by a vector orthogonal to $1,Y-m$. The initial capital in the unrestricted one-period quadratic hedge of a discounted payoff $H$ is $\mathbb{E}[Z_{\ast }H]$.

Let a discounted gain $Y$ take values $-1,1,10$ with physical probabilities $1/10,4/5,1/10$. Then $m=17/10$ and $C=801/100$, so the minimal martingale measure in a one-period market has signed density $Z_{\ast }(10)=-610/801$. Nevertheless $(109/200,89/200,1/100)$ is a strictly positive pricing probability with zero mean gain. Thus absence of arbitrage does not ensure positivity of the minimal density. The minimum-norm identity holds for signed pricing densities; interpretation as a dominated martingale measure requires nonnegativity.

Work in discounted units with a riskless asset of value one, trivial initial information and finitely many risky assets. Let $Y=S_1-S_0\in L^2(P;{\mathbb{R}}^d)$ be their discounted gains. A terminal attainable claim has the form $v+{\theta }^{T}Y$, whose initial cost is $v$. Their spanis a finite-dimensional closed subspace of a Hilbert space. For $H\in L^2(P)$, one-period least-squares hedging is its orthogonal projection onto $\mathcal{H}$.

Let $m=\mathbb{E}Y$, $\Sigma =\mathrm{Cov}(Y)$ and $b=\mathrm{Cov}(Y,H)$. Remove redundant risky directions so that $\Sigma$ is invertible. Under absence of arbitrage, a gain direction with zero variance is a constant and must be zero, so this reduction loses no genuine trading opportunity. For freely chosen capital and holdings, minimizingfirst gives $v=\mathbb{E}H-{\theta }^{T}m$. The remaining centered error isCompleting the square yieldsThe residual $L=H-v^{\ast }-({\theta }^{\ast }{)}^{T}Y$ satisfies $\mathbb{E}L=0$ and $\mathbb{E}[LY]=0$, the least-squares normal equations. Exact claim replication is possible precisely when this residual is zero. If capital is fixed at $v$, its normal equation instead gives the fixed-capital quadratic hedge in a one-period marketThus fixed-capital and freely optimized hedging are distinct problems.

A dominated martingale measure is a probability measure $Q\ll P$ with integrable gains and ${\mathbb{E}}_{Q}Y=0$. Its density $Z$ obeys $Z\ge 0$, $\mathbb{E}Z=1$, $\mathbb{E}[ZY]=0$. An equivalent martingale measure additionally has $Z>0$ almost surely. Every such measure prices an attainable discounted claim $v+{\theta }^{T}Y$ at $v$, but different measures may give different expectations to an unattainable claim.

The minimal martingale measure in a one-period market is defined by preserving the mean-zero martingale directions orthogonal to the gain innovation $Y-m$. Among square-integrable densities it isIndeed $\mathbb{E}{Z}_{\ast }=1$, $\mathbb{E}[Z_{\ast }Y]=m-\Sigma {\Sigma }^{-1}m=0$, and for $\mathbb{E}L=0$, $\mathbb{E}[L(Y-m)]=0$ we have $\mathbb{E}[Z_{\ast }L]=0$. Conversely preservation of all these orthogonal directions puts $Z$ in $\mathrm{span}\{1,Y-m\}$; normalization and zero gain expectations then determine $Z_{\ast }$. Any other square-integrable signed pricing density differs from $Z_{\ast }$ by a vector in ${\mathcal{H}}^{\perp }$, so $Z_{\ast }$ also has the smallest $L^2$ norm. Its evaluation of a claim isThis is the capital of the optimal quadratic hedge, not automatically an arbitrage-free price. The formula may define only a signed martingale measure: $Z_{\ast }$ is a genuine dominated measure only if it is nonnegative, and equivalent only if strictly positive. For example, gains $(-1,1,2)$ with probabilities $(1/10,4/5,1/10)$ have $m=9/10$, $\Sigma =49/100$ and $Z_{\ast }(2)=-50/49$. Yet probabilities $(11/20,7/20,1/10)$ are all positive and have zero gain mean. Thus absence of arbitrage does not force a positive minimal density.

For the market completeness assertion, assume absence of arbitrage and hence existence of an equivalent martingale measure $Q_0$. In finite states this is the usual fundamental theorem of asset pricing; the one-period version also holds with finitely many assets on a general probability space. Market completeness means that every bounded claim is attainable, equivalently $\mathcal{H}=L^2(P)$ in this square-integrable setting. We now prove completeness and uniqueness of dominated martingale measures.

If the model is complete, replicate every indicator $1_A$. Two dominated martingale measures both evaluate it at its unique claim replication cost, and therefore assign every event $A$ the same probability. They are identical.

Conversely let $k=\dim \mathcal{H}$ and suppose the model is incomplete. There exist $k+1$ disjoint events $A_j$ of positive probability. Otherwise the probability space would have at most $k$ atoms: any non-atomic positive event could be split to increase the number. Its whole payoff space would then have dimension at most $k$, forcing equality with $\mathcal{H}$ and market completeness. Choose a basis $h_1,\dots ,h_k$ of $\mathcal{H}$. Its members are integrable under $Q_0$, because they are combinations of the constant and the gains. The $k+1$ vectorsare linearly dependent. Hence a nonzero bounded $g={\sum }_j{a}_j{1}_{A_j}$ satisfies ${\mathbb{E}}_{Q_0}[gh]=0$ for every $h\in \mathcal{H}$, in particular $h=1,Y_i$. For $0<\epsilon <1/\Vert g{\Vert }_{\infty }$, defineThese strictly positive densities integrate to one, preserve every gain expectation, and give distinct equivalent martingale measures. They are also dominated by $P$. Thus uniqueness is impossible in an incomplete market, proving the equivalence under the stated no-arbitrage hypothesis.

That hypothesis is essential. On three positive-probability states, the gain $(0,1,2)$ has the unique dominated martingale measure, concentrated on the first state, but its attainable span has dimension only two. The market is incomplete and admits arbitrage. This explains why uniqueness among merely dominated probabilities must not be used without an equivalent pricing measure or a no-arbitrage assumption.

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.