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Suppose an unknown has prior distribution $\Pi$ and its conditional observation law $P_u$ has jointly measurable Radon-Nikodym derivative $L(u,m)$ relative to a fixed observation law $P_0$. If $L>0$ for $\Pi (du)P_0(dm)$-almost every pair, thendefines the posterior distribution for almost every observation. Indeed, $\int L(u,m)P_0(dm)=1$ for almost every $u$, so Tonelli theorem gives $\int Z(m)P_0(dm)=1$. Hence $Z$ is finite almost everywhere, and positivity of $L$ and Fubini's theorem give $Z>0$ almost everywhere. The data law is $Z{P}_0$, and integration of the proposed posterior against this data law recovers the joint law. This proves the conditional distribution property and also shows that the data law is an equivalent probability measure to $P_0$.

With independent up moves of physical probability $p$ and risk-neutral probability $q$, the Radon-Nikodym derivative on the first $n$ periods is the displayed likelihood ratio, where $N_n$ counts up moves. Its conditional one-step multiplier has physical expectation one, so $L_k$ is a positive mean-one martingale. The terminal state-price density is $L_n/B_n$ for deterministic bank-account value $B_n$.

For nonnegative integrable $f$, the finite measure $A\mapsto {\int }_{A}fd\mu$ restricted to a sub-sigma-algebra ${\Sigma }_0$ is absolutely continuous with respect to the restricted base measure. Its Radon-Nikodym derivative is ${\Sigma }_0$-measurable and has the required integral identities. Positive and negative parts extend the construction to real integrable functions; real and imaginary parts extend it to complex functions. The result is unique almost everywhere.

For a Radon-Nikodym derivative $L_T\ge 0$ with $\mathbb{E}{L}_T=1$, its conditional-expectation process is a nonnegative martingale. If $L_T>0$ almost surely, it defines an equivalent probability measure. A stochastic exponential satisfying the Novikov condition is a common positive density process.

A density ratio compares two probability density functions relative to the same underlying measure. More generally it is a Radon-Nikodym derivative when one probability measure has absolute continuity of measures with respect to the other.

For discounted one-period asset gains $Y$, a dominated martingale measure is a probability measure $Q$ absolutely continuous with respect to the physical probability measure $P$, under which the gains have zero expectation. 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 has $Z>0$ almost surely. Absolute continuity permits some positive-probability physical states to receive zero pricing probability; equivalence does not.

For reference Brownian motion $R=x+W$, stopped at $T=t\wedge {\tau }_{(a,b)}$ with $0<a<x<b$, this expression is the Radon-Nikodym derivative of the stopped Bessel process law of dimension $\delta$. The bounded stopped drift verifies the Novikov condition. The comparison laws must have the same initial value and the same stopping rule.

The intensity measure of a point process $\Pi$ is its expected counting measure:It may take the value $+\infty$.^[ref] Countable additivity follows by applying the monotone convergence theorem to the counts of finite unions of pairwise disjoint observation windows. If $\mu$ is locally finite and absolutely continuous with respect to Lebesgue measure, its Radon-Nikodym derivative is the intensity function of a point process $\lambda$:This is an expected number per unit volume, not a probability density function for a randomly selected point. For example, a single deterministic point has $\mu ={\delta }_0$, a Dirac measure, and no Lebesgue measure density. The density representation requires its own assumption.^[ref]

For a bounded Borel set $W$, $\mathbb{E}[N_W]=\mu (W)$, but local finiteness alone does not ensure that this expected value is finite. Conversely, $\mu (W)<\infty$ implies $N_W<\infty$ almost surely, even when $W$ is unbounded. A constant finite intensity function of a point process on all of ${\mathbb{R}}^d$ has infinite total intensity measure, although every bounded window has finite expected value.

On each finite horizon the displayed Radon-Nikodym derivative changes Wiener measure into the law of $dX=dB-\mu Xdt$. Stop at coordinate magnitude n so the Novikov condition applies. Under each stopped tilted measure the coordinate solves the drifted equation, and the Gronwall inequality with the Doob L2 maximal inequality bounds its maximal second moment uniformly in n. The probability of reaching n therefore tends to zero; this proves that the unstopped density has expectation one. No unrestricted large-time Novikov estimate is needed.

Separate $L$ into a diffusion generator and the constant killing term $-1/2$. The appropriate stochastic differential equation and its explicit geometric Brownian motion solution areThe second-order term in the Itô formula cancels the drift of $\log |X|$ when $x\ne 0$; if $x=0$, the solution stays zero.

Fix a terminal time $t$. Apply the Itô formula to $e^{-s/2}u(t-s,X_s)$ for $0\le s\le t$. Its drift isAfter localization this is a local martingale, and it is bounded because $u$ is bounded. It is therefore a true martingale. Taking expectations at its endpoints gives the Feynman-Kac formula

For $t>0$ and $x\ne 0$, the change of variable $y=x{e}^z$ in the normal density produces the killed geometric Brownian heat kernel with respect to Lebesgue measure:For $xy>0$, an equivalent form isIn particular,The apparent missing $1/y$ in this last expression is absorbed into the completed square, together with the killing factor; it is not a transcription error. The total mass is $e^{-t/2}$, as expected for killing at rate $1/2$. At $x=0$ the fundamental kernel is instead the measure $e^{-t/2}{\delta }_0(dy)$; it has no Radon-Nikodym derivative with respect to Lebesgue measure. Thus $u(t,0)=e^{-t/2}g(0)$. At $t=0$ the kernel is ${\delta }_x$, and continuity of bounded $g$ gives the initial condition. This also specifies the fundamental solution at the degenerate point omitted by a density-only formula.

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.

Saddlepoint approximation methods retain the nonquadratic shape of a distribution through its cumulant-generating function. Suppose $X_1,\dots ,X_n$ are independent and identically distributed, $K(t)=\log E(e^{t{X}_1})$ is finite on an open interval containing zero, and $K^{\prime \prime }>0$. For an interior target mean $x$, solve the saddlepoint equation $K^{\prime }(t)=x$. Introduce exponential tilting byUnder the tilted distribution the mean and variance are $K^{\prime }(t)=x$ and $K^{\prime \prime }(t)$. For the product sample the Radon-Nikodym derivative is $e^{t{\sum }_i{X}_i-nK(t)}$. Consequently the exact density relation for the sample mean isA local central limit approximation to the tilted density at its own mean gives $f_{\overline{X},t}(x)\simeq \sqrt{n/(2\pi K^{\prime \prime }(t))}$. HenceFor the sum $S_n$, the equivalent expression is $f_{S_n}(s)\simeq e^{nK(t)-ts}/\sqrt{2\pi n{K}^{\prime \prime }(t)}$, with $K^{\prime }(t)=s/n$. Under the usual smooth nonlattice density conditions this approximation has relative error of order $n^{-1}$ for a fixed interior target. In the tilted Edgeworth expansion the order $n^{-1/2}$ odd term vanishes at the mean. The exponent retains the Legendre-transform rate $I(x)=tx-K(t)$, rather than a quadratic Taylor approximation to that rate, explaining the useful tail and skewness behavior. Normalization is not automatic, and accuracy statements need their regularity and range restrictions.

For a normal distribution, $K(t)=\mu t+{\sigma }^2{t}^2/2$ and $t=(x-\mu )/{\sigma }^2$. Substitution recovers exactly the normal density of the sample mean. For an exponential distribution of rate $\lambda$, $K(t)=-\log (1-t/\lambda )$, $t=\lambda -1/x$ and $K^{\prime \prime }(t)=x^2$. The approximation becomesIt has exactly the gamma density shape. Replacing its leading constant by the exact normalizing constant makes it exact, illustrating why exactness of shape and exactness of the unnormalized approximation are separate questions.

For a vector sample mean, tilt by $e^{t^{\mathsf{T}}y-K(t)}$, solve $\nabla K(t)=x$, and use the local multivariate normal density at the tilted mean. This givesJoint and marginal saddlepoint densities can also be divided to approximate conditional densities, for example after conditioning on a nuisance sufficient statistic. The p* approximation uses likelihood and fitted-information coordinates to obtain a closely related approximation to a maximum-likelihood estimator's density, usually conditional on an ancillary statistic.

Tail probabilities need a corresponding integration approximation. The Lugannani-Rice saddlepoint tail approximation usesand givesHere $\Phi ,\varphi$ are the standard normal distribution functions. The upper-tail correction has the opposite sign: $1-\Phi (w)+\varphi (w)(1/u-1/w)$. Inverting a cumulative distribution function introduces a pole in addition to the density's saddlepoint; treating both supplies the correction term, which is why simply integrating a normal approximation at the untilted mean is less accurate. At $x=K^{\prime }(0)$ the apparent singularities cancel. With $v=K^{\prime \prime }(0)$ and $k=K^{\prime \prime \prime }(0)$, expansions give $w=t\sqrt{nv}(1+kt/(3v)+O(t^2))$ and $u=t\sqrt{nv}(1+kt/(2v)+O(t^2))$, so $1/w-1/u\to k/(6\sqrt{n}{v}^{3/2})$. For the normal distribution the correction is zero everywhere.

These methods require care for lattice distributions, where summation and lattice corrections replace the density inversion, and near boundaries or singular covariance matrices. A distribution with no moment-generating function in a neighborhood of zero does not satisfy the starting assumptions. A successful saddlepoint approximation is therefore a structured refinement of regular asymptotic inference, not a formula guaranteed uniformly in every tail or for every model.

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.

Take a riskless gross return $R>0$ per period and a stock whose independent multiplicative moves are $u$ and $d$, with physical up probability $p\in (0,1)$. Assume $0<d<R<u$. With $K_k$ up moves by date $k$,A self-financing portfolio chooses each stock holding from information available before the next move, and invests the remainder in the bank account. At any node the two possible next payoffs can be matched by two holdings because $u\ne d$. Backward induction therefore makes the binomial market complete.

The unique risk-neutral probability of an up move is determined by the discounted-stock martingale equation:The risk-neutral measure assigns product probabilities to the successive moves. A path with $K=K_n$ up moves has physical probability $p^K(1-p{)}^{n-K}$ and pricing probability $q^K(1-q{)}^{n-K}$. Its Radon-Nikodym derivative is thereforeSince $S_n=S_0{d}^n(u/d{)}^K$, eliminate $K$ to obtain the requested terminal-price expression:The pricing density depends only on the terminal stock price even though different paths to that price have different intermediate histories.

For $w_0>0$, a nonnegative terminal wealth $X$ with initial cost $w_0$ satisfiesThe utility function has derivative $v^{\prime }(x)=x^{-(\gamma -1)/\gamma }$. Put $\eta =\gamma /(\gamma -1)>1$. Its first-order condition makes $v^{\prime }(X)$ proportional to $Z$, so $X=C{Z}^{-\eta }$. Choosing $C$ to satisfy the pricing budget constraint gives the binomial power-utility terminal wealthAll states have positive probability and the state space is finite, so this payoff is positive, integrable and attainable. The normalizing constant is explicit: independence givesCombining these two formulas with the terminal-price expression for $Z$ gives $X_{\ast }$ entirely as a function of $S_n$.

To prove global optimality, strict concavity gives $v(X)-v(X_{\ast })\le v^{\prime }(X_{\ast })(X-X_{\ast })$. Because $v^{\prime }(X_{\ast })=C^{-1/\eta }Z$, the expectation of the right side is zero for any other payoff satisfying the same pricing budget constraint. Equality requires $X=X_{\ast }$ almost surely. Backward replication supplies its self-financing strategy, so this static optimum is attainable by trading. If $w_0=0$ and terminal wealth must be nonnegative, the only feasible terminal wealth is zero. For $p=q$, the density is one and the optimizer is the riskless wealth $R^n{w}_0$.

For Borel sets $A_1,\dots ,A_n$, multiply the finite-time Radon-Nikodym derivative into the product law of the increments:The factorization, extended from rectangles to the product sigma-algebra, proves independence and identical distributions. Their common law is the exponential tiltingFor $\lambda >0$, $|x|e^{\lambda x}$ is bounded on $x\le 0$, while $x{e}^{\lambda x}\le C_{\epsilon }{e}^{(\lambda +\epsilon )x}$ on $x\ge 0$. Choosing a small two-sided neighborhood of $\lambda$ still contained in $(0,\infty )$ gives an integrable dominating function for the derivative. Differentiation under the integral sign then yieldsThe mean under one-sided exponential tilting needs an endpoint qualification. At $\lambda =0$, finiteness of positive exponential moments makes $X_1^{+}$ integrable, but gives no integrability of $X_1^{-}$. The formula holds with the right derivative and the extended mean ${\varphi }_{+}^{\prime }(0)=\mathbb{E}{X}_1\in [-\infty ,\infty )$. To see this, apply monotone convergence to $(1-e^{-h{X}_1^{-}})/h$ and dominated convergence to $(e^{h{X}_1^{+}}-1)/h$ as $h\downarrow 0$. A finite mean at zero is not guaranteed by the printed hypotheses: $\mu (\{-j\})=6/({\pi }^2{j}^2)$, $j\ge 1$, has finite $\varphi (\lambda )$ for every $\lambda \ge 0$ but $\mathbb{E}{X}_1=-\infty$.

First take $k$ to be a nonnegative integer. The exponential tilting constructed from the independent increments restricts to ${\mathbb{P}}_n^{\lambda }$ on each ${\mathcal{F}}_n$. Since its Radon-Nikodym derivative is strictly positive, finite-time change of measure givesUnder the tilted law, $L_n=(M_n^{\lambda }{)}^{-1}$ is a martingale, becauseUse the stopped likelihood ratio under exponential tilting at this finite horizon. In detail, $\{{\tau }_k=j\}\in {\mathcal{F}}_j$, soHere the last expression is defined only on the displayed event, so no value of $M_{\infty }^{\lambda }$ is being assumed. The upward skip-free random walk has integer increments bounded above by one, which imply $S_{{\tau }_k}=k$ on a finite hit: the preceding value is at most $k-1$, and there is no upward overshoot. ConsequentlyAt $\lambda ={\lambda }_0$, the mean under one-sided exponential tilting is finite and positive. The strong law of large numbers gives $S_n/n\to {\varphi }^{\prime }({\lambda }_0)>0$ under ${\mathbb{P}}^{{\lambda }_0}$, so every integer level is hit almost surely. Since $\varphi ({\lambda }_0)=1$, passing to $n\to \infty$ givesLet $H={\sup }_{n\ge 0}{S}_n$. The event $\{H\ge k\}$ is exactly the finite-hit event: an integer sequence with all its values below $k$ has supremum at most $k-1$. Also $\mathbb{P}(H=\infty )={\lim }_k{e}^{-{\lambda }_{0}k}=0$. Subtraction of consecutive tail probabilities gives the geometric maximum of an upward skip-free random walk:This is the geometric distribution counting failures before the first success, with success parameter $1-e^{-{\lambda }_0}$.

The PDF writes only $k\ge 0$, without explicitly declaring integer levels. If it is read as allowing real $k$, then ${\tau }_k={\tau }_{\lceil k\rceil }$ and the displayed factor must be $e^{-\lambda \lceil k\rceil }$; the hitting probability is likewise $e^{-{\lambda }_0\lceil k\rceil }$. For example, increments $+1$ and $-1$ with probabilities $1/3$ and $2/3$ have ${\lambda }_0=\log 2$: hitting level $1/2$ has probability $1/2$, not $2^{-1/2}$. The integer interpretation supplies the intended formula.

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.

Write $\eta (A)={\int }_A{X}_{\infty }d\lambda$. This is a finite positive measure. For $A\in {\mathcal{F}}_n$, the conditional expectation inequality givesThe union $\mathcal{A}={\bigcup }_n{\mathcal{F}}_n$ is an algebra of sets generating $\mathcal{B}([0,1))$, since half-open dyadic intervals generate all open sets and hence all Borel sets. The class $\{A:\eta (A)\le \mu (A)\}$ is closed under increasing and decreasing limits, by continuity of these finite measures. The Monotone class theorem extends the inequality from $\mathcal{A}$ to every Borel set. Approximation by nonnegative simple functions and monotone convergence then giveThe difference $\nu =\mu -\eta$ is therefore a finite nonnegative measure, not merely a signed measure. For $A\in {\mathcal{F}}_n$,Consequently its restricted Radon-Nikodym derivative isChoose the cell-average versions of all the conditional expectations. Then $Y_n$ is constant on each level-$n$ cell and equals $2^n\nu (D_k^n)$ there, so these versions are nonnegative at every point.

Suppose under the physical measure $P$ the up indicators are independent with probability $p\in (0,1)$. If $N_k$ is the number of up moves through date $k$, a particular history has probability $p^{N_k}(1-p{)}^{k-N_k}$ under $P$ and $q^{N_k}(1-q{)}^{k-N_k}$ under $Q$. Thus the binomial-market probability density, the Radon-Nikodym derivative on ${\mathcal{F}}_k$, isIt is strictly positive. Its next multiplier has conditional expectation $p(q/p)+(1-p)((1-q)/(1-p))=1$, so $L_k$ is a mean-one martingale and $L_k={\mathbb{E}}_P[L_n\mid {\mathcal{F}}_k]$. These facts verify normalization and equivalence of the two probability measures.

The state-price density for terminal cash is ${\zeta }_n=L_n/B_n=b^{-n}{L}_n$. In particular $V_0={\mathbb{E}}_P[{\zeta }_{n}H]$. At an intermediate date the conditional change-of-measure identity givesIf physical up probabilities depend on the past, replace the constant-probability expression by the product over periods of $q/p_k$ on an up move and $(1-q)/(1-p_k)$ on a down move. Each multiplier still has conditional mean one, provided $0<p_k<1$, and the pricing measure retains conditional up probability $q$.