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If $S_1$ has a multivariate normal distribution with mean $\mu$ and invertible covariance $V$, and $g$ has bounded gradient, Gaussian integration by parts gives $\mathrm{Cov}(S_1,g(S_1))=V\mathbb{E}[\nabla g(S_1)]$. Consequently the risky holdings of its one-period quadratic hedge equal the expected payoff gradient. The initial cost is $R^{-1}\mathbb{E}[g(S_1)]+(S_0-R^{-1}\mu )\cdot \mathbb{E}[\nabla g(S_1)]$. Bounded gradient ensures linear growth and square integrability, even if the payoff is not bounded.

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 $\mathbb{E}[Z_{\ast }H]$.

The initial cost of the one-period quadratic hedge is $\mathbb{E}[{\rho }^{\ast }\xi ]$. This pricing weight satisfies $\mathbb{E}[{\rho }^{\ast }R]=1$ and $\mathbb{E}[{\rho }^{\ast }{S}_1]=S_0$. Among square-integrable weights satisfying these equations it has the smallest squared norm:For another admissible weight $\rho$, its difference from ${\rho }^{\ast }$ is orthogonal to $1$ and all coordinates of $S_1$, hence to ${\rho }^{\ast }$. The Pythagorean theorem in an inner-product space gives $\mathbb{E}[{\rho }^2]=\mathbb{E}[({\rho }^{\ast }{)}^2]+\mathbb{E}[(\rho -{\rho }^{\ast }{)}^2]$. This weight can be signed; positivity, and hence interpretation as a state-price density, requires additional assumptions.

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.

Write $R=1+r$, $m=\mathbb{E}{\xi }_1$ and $c=\mathrm{Cov}(S_1,{\xi }_1)=\mathbb{E}[(S_1-\mu )({\xi }_1-m)]$. The riskless gross return must be nonzero for the requested formulas; in the usual positive-bank-account model $R>0$. If $R=0$, the bank holding cannot affect the terminal payoff, and the later moment equation for pricing $B_0=1$ is impossible. We use the intended nondegenerate case $R\ne 0$.

The claim and asset prices are square integrable. Centering separates the bias from the stochastic error:For fixed $\pi$, the unique minimizing bank holding removes the bias. The invertible covariance matrix $V$ is positive definite, and completing the square givesConsequently the unique one-period quadratic hedge isIts minimum expected squared error is $\mathrm{Var}({\xi }_1)-c^T{V}^{-1}c$. This is a least-squares hedge, not necessarily exact claim replication.

Put $h=S_0-\mu /R$. The initial capital of the one-period quadratic hedge isDefine the minimum-norm one-period pricing weightIt is square integrable, and $\mathbb{E}[(S_1-\mu ){\xi }_1]=c$, soThe weight depends only on the asset prices and their first two moments, not on the claim. It can be negative, so this identity alone does not supply an equivalent martingale measure or a positive state-price density.