One-period quadratic hedge (${\pi }^{\ast }=V^{-1}\mathrm{Cov}(S_1,\xi )$)

A one-period quadratic hedge minimizes $\mathbb{E}[(\xi -\varphi R-\pi \cdot S_1{)}^2]$ over deterministic holdings, where $R\ne 0$ is the riskless gross return, $\mu =\mathbb{E}{S}_1$, and the covariance matrix $V$ is invertible. For a square-integrable claim, the unique solution is ${\pi }^{\ast }=V^{-1}\mathrm{Cov}(S_1,\xi )$ and ${\varphi }^{\ast }=(\mathbb{E}\xi -{\pi }^{\ast }\cdot \mu )/R$. Centering separates the squared bias from the quadratic error. Completing the square gives the excess error as $(\pi -{\pi }^{\ast }{)}^{T}V(\pi -{\pi }^{\ast })$ after the bias is minimized. Unlike claim replication, the residual need not vanish.

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.

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.

A finite signed measure with total mass one under which discounted asset gains have zero integrals. Its density can be negative, so it need not be a probability measure or give arbitrage-free prices to every nonnegative contingent claim. A square-integrable signed density is useful in one-period least-squares hedging, even when it is not an equivalent martingale measure.

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.

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.