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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.