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Assume the bank account stays positive and write ${\tilde{S}}_n=S_n/B_n$ and ${\tilde{V}}_n=V_n/B_n$. Dividing the time-$n$ rebalancing equality by $B_n$ givesThe discounted bank account has constant price one. This is exactly the self-financing equality in the discounted market. Equivalently, subtract the corresponding valuation at $n-1$ to obtain the discounted wealth equation in discrete timeThis algebra uses only positivity of $B_n$, and does not require it to be deterministic.

The usual replication argument needs two hypotheses implicit in the pricing conclusion: the strategy must replicate the payoff, $V_N=X$, and its discounted gains must be integrable. Bounded predictable risky holdings suffice for the latter. Under these conditions, the equivalent martingale measure $\mathbb{Q}$ and part (c) giveThus discounted wealth is a martingale, and repeated conditioning givesFor a normalized initial bank account, $B_0=1$, deterministic terminal bank value, and trivial initial information, this reduces to the intended risk-neutral pricing formulaIf $B_N$ is random it must remain inside the expectation. If a claim is not replicable, the existence of an equivalent martingale measure alone does not make its arbitrage-free price unique.

The discounted wealth martingale integrability condition cannot simply be dropped. For a concrete counterexample, set $B_n=1$, $S_0=3/2$, $S_1=1+U$, and $S_2=1+U+\epsilon /2$, where $U$ is uniform on $(0,1)$ and $\epsilon$ an independent fair sign. These positive bounded prices form a martingale in the natural filtration. Start with zero holdings; at time one buy $1/U$ shares financed by borrowing $S_1/U$. This is previsible and self-financing, but its terminal wealth is $\epsilon /(2U)$, whose absolute expectation is infinite. It is therefore not a martingale. The stated pricing proof applies to the standard integrable replication class, rather than all unconstrained previsible strategies.

The common short-rate market price of risk follows from absence of arbitrage, with the securities driven by the same Brownian motion. Take two traded derivatives $V_1,V_2$ with drifts $m_1,m_2$ and diffusion coefficients $s_1,s_2$. At a given state hold $s_2$ units of the first and $-s_1$ of the second, and place $-s_2{V}_1+s_1{V}_2$ in the bank account. The portfolio has zero current value and zero diffusion exposure. Its instantaneous excess drift isIf this were nonzero, selecting its positive orientation and maintaining the local risk-cancelling self-financing hedge would generate a gain without risk from zero wealth. Under the usual trading and regularity assumptions this violates absence of arbitrage. Thus the expression vanishes, and at nonzero exposuresAt zero exposure the correct relation is $m-rV=\theta s=0$, without dividing by zero. This is a condition on traded securities: the physical drift of the nontraded short rate alone does not determine $\theta$.

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 $R^n{w}_0$.

A self-financing portfolio changes its holdings using only its existing wealth. If it holds $g_t$ shares and $h_t$ units of a traded bond or bank account $B_t$, its value $V_t=g_t{S}_t+h_t{B}_t$ obeys the gain identityThe changes in holdings do not add or withdraw capital. Set $T=t_0$, $B_t=e^{-\rho (T-t)}$, so $d{B}_t=\rho B_{t}dt$, and let $d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t$, with $\sigma >0$. Write $V_t=p(S_t,t)$ and suppress the arguments of the smooth holdings $g,h$.

Differentiate $p=xg+B_{t}h$:The Itô formula minus the proposed portfolio gains givesUniqueness of the local-martingale and finite-variation decomposition shows that self-financing requires $p_x=g$. Since $S_t$ has full support on $(0,\infty )$ for each $t>0$, smoothness gives the corresponding functional identity for every positive $x$, and continuity extends it to the interval endpoints. ThusDifferentiating this identity once more in $x$ gives $g_x+x{g}_{xx}+B_t{h}_{xx}=0$, so $p_{xx}=g_x$. Substitute both identities into the drift condition to obtainConversely these equations give $p_x=g$, $p_{xx}=g_x$ and zero drift and diffusion in the displayed gain difference. They therefore prove the claimed self-financing conditions for smooth stock and bond holdings in both directions.

Using $B_{t}h=p-xg$ in the gain identity and setting $g=p_x$, self-financing impliesThis is the Black-Scholes equation, independent of the physical drift $\mu$. Conversely, if a smooth $p$ satisfies it, define its delta hedge byThe preceding Itô formula verifies that these holdings are self-financing.

For prescribed arbitrary holdings, the value equation alone is insufficient. The corrected equivalence is the Black-Scholes value equation needs delta-compatible holdings: a specified portfolio is self-financing if and only if its value solves the equation and $g=p_x$. Indeed $p=0$, $g=1$, $h=-x/B_t$ satisfies the value equation but its proposed gains are $d{S}_t-\rho S_{t}dt$, which have nonzero Brownian part while $dp=0$. Thus the printed final equivalence is valid as an existence assertion for the delta hedge, not for every decomposition of the same value. No change to the valid two holdings equations is needed.

With continuous underlying stock dividends, the stock's gain is $d{S}_t+\theta S_{t}dt$, so self-financing becomes $d{V}_t=g_t(d{S}_t+\theta S_{t}dt)+h_{t}d{B}_t$. Under the risk-neutral measure the ex-dividend stock drift is $(\rho -\theta )S_t$. Matching its diffusion still gives $g=p_x$, and matching its gain drift gives the Black-Scholes equation with continuous stock dividendsOnly the coefficient of $x{p}_x$ changes. The $-\rho p$ discount term remains, because the claim is financed at the bank rate; there is no separate source term for a dividend paid by the claim itself. The dividends here are paid by the underlying stock.

A one-factor short-rate model uses a single scalar state, normally the short rate itself. Under the physical probability measure writeThe second process is the continuous-time bank account. Specifying the physical diffusion alone does not determine bond prices: it must be accompanied by the market price of risk $\lambda (t,r)$, or equivalently by pricing-measure dynamics. If the Girsanov theorem applies to the likelihood $\mathcal{E}(-\int \lambda dW)$, then under the risk-neutral measureChanging measure alters drift, not quadratic variation. The coefficients need conditions for a nonexplosive solution and for the relevant discounted prices and measure-change density to be true martingales.

A unit zero-coupon bond maturing at $T$ has priceThe Markov property makes current time and rate sufficient state variables. Apply Itô formula to $P(t,r_t;T)/B_t$. Its drift must vanish, giving the short-rate bond pricing equationConversely, under the appropriate boundary and integrability conditions the Feynman-Kac formula identifies its solution with the displayed discounted expectation. For a terminal rate-dependent payoff $H(r_T)$, the same equation has terminal condition $H$ instead of one. A coupon bond is a sum of such unit-bond prices weighted by the coupon payments.

The common market price of risk can also be derived from trading, instead of being introduced arbitrarily for each maturity. Under the physical measure a bond's diffusion exposure is $\eta P_r$ and its excess drift above bank financing is $P_t+\mu P_r+{\eta }^2{P}_{rr}/2-rP$. For two nondegenerate traded bonds, combine their holdings to cancel their Brownian exposure. Absence of arbitrage requires that combination to earn the bank rate; hence their excess-drift-to-exposure ratios must coincide. Calling that common ratio $\lambda$ yields the pricing equation with ${\mu }_Q=\mu -\eta \lambda$ for every maturity. A derivative $V(t,r)$ is then hedged locally by holding $V_r/P_r$ units of a nondegenerate bond and financing the remaining value through the bank account. Their Brownian exposures agree, and substituting both pricing equations proves the drift gain identity. This is short-rate diffusion hedging; it requires $\eta P_r\ne 0$, rather than a bare assertion that one factor automatically gives completeness everywhere.

The entire yield curve is generated consistently from these bond prices. The continuously compounded yield and the instantaneous forward rate areThese encode different summaries: the yield averages the forward rate over the remaining lifetime. Bond prices must satisfy $P(t,t)=1$ and be positive; they need not be below one in models permitting negative rates.

The Vasicek model has constant pricing dynamicsSolving the linear equation gives $r_{t+u}=b+(r_t-b)e^{-au}+\eta {\int }_t^{t+u}{e}^{-a(t+u-v)}d{W}_v^Q$. Consequently, with $\tau =T-t$ and $B(\tau )=(1-e^{-a\tau })/a$,This conditional normal random variable has variance ${\eta }^2{\int }_0^{\tau }B(u{)}^{2}du$. Its normal exponential moment yieldsEvaluating the elementary integral givesThis explicit form makes estimation and valuation convenient. The rate is Gaussian, its conditional mean reverts to $b$, and its long-run variance is ${\eta }^2/(2a)$. Its disadvantage, when nonnegative rates are desired, is a positive probability of negative rates.

The CIR model replaces constant volatility by square-root volatility:The standard solution stays nonnegative. The Feller positivity condition for the CIR model is $2ab\ge {\eta }^2$ in this parameterization; with a positive initial rate it makes zero inaccessible. Below this threshold zero can be reached, although the process does not cross into negative rates. For an affine diffusion bond pricing ansatz $P=A(\tau )e^{-B(\tau )r}$, substitution into the bond equation and matching its constant and rate coefficients givesWriting $d=\sqrt{a^2+2{\eta }^2}$, factor the quadratic in the first equation and integrate from $B=0$; integrating the second equation then gives the CIR bond pricing formulasThe square-root form produces an affine term structure while respecting nonnegative rates. These parameters are pricing-measure parameters; historical mean-reversion estimates alone do not determine them without a risk-premium specification.

Both preceding constant-parameter models constrain the family of initial curves. The Hull-White model allows a deterministic time-dependent drift:For a smooth desired initial forward curve $f_0(t)$, its exact forward-curve fit in the Hull-White model is obtained by takingTo derive this rather than merely assert calibration, write $r_t=x_t+\varphi (t)$, where $d{x}_t=-a{x}_{t}dt+\eta d{W}_t^Q$, $x_0=0$. The conditional Gaussian-integral calculation at time zero givesso its forward curve is $\varphi (T)-{\eta }^{2}B(T{)}^2/2$. Choose $\varphi (T)=f_0(T)+{\eta }^{2}B(T{)}^2/2$. Differentiating this shift and adding $a\varphi (t)$ gives exactly the displayed $\beta (t)$. Thus $P(0,T)=\exp (-{\int }_0^T{f}_0(u)du)$ for every maturity, while $a$ and $\eta$ govern the stochastic fluctuations. This Gaussian model still allows negative rates.

Finally, one-factor models are economical and often give explicit bond prices, but their single Brownian driver is restrictive. Bond diffusion exposures $v_i=\eta P_{i,r}$ give the instantaneous covariance matrix $v{v}^T$, of rank at most one. Nondegenerate maturity changes are instantaneously perfectly correlated, with correlation $+1$ or $-1$, though their finite-horizon correlations need not equal one. A single factor therefore cannot independently describe level, slope and curvature shocks to the whole yield curve. Multiple factors or a full forward-rate model offer greater flexibility. A valid one-factor model must specify the pricing drift, fit the initial curve to the required accuracy, and handle boundary and martingale conditions; naming a rate diffusion alone is not a complete pricing model.

Since $d(B_t^{-1})=-r_t{B}_t^{-1}dt$, multiplying the given bond dynamics by $B_t^{-1}$ cancels the drift and gives $d{Z}_t=Z_t\Sigma (t,T)d{W}_t^Q$. For an integrable discounted terminal payoff, the risk-neutral value process is the discounted-payoff martingaleFor an attainable claim this also follows directly from its self-financing discounted wealth being a true martingale, with terminal value $B_T^{-1}X$. In a Brownian market, the Brownian martingale representation theorem provides $d{M}_t={\psi }_{t}d{W}_t^Q$; a traded bond with nonzero diffusion exposure can hedge it by holding ${\psi }_t/(Z_t{\Sigma }_t)$ bond units, with the remainder in the bank account. Thus, under the usual attainability and integrability assumptions, this construction gives the replicated price as well as the risk-neutral pricing formula.

Because $B_t$ is ${\mathcal{F}}_t$-measurable,In a general Heath-Jarrow-Morton model, writing this value as $V(t,r_t)$ requires an additional state-reduction assumption. The conditional expectation is valid without asserting that the short rate alone is the full state.

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.

Take a non-dividend-paying stock with physical dynamics $d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t$ and continuous-time bank account $B_t=e^{\rho t}$. Write $T=t_0$ and assume the candidate claim price $p(x,t)$ is $C^{2,1}$ before maturity, with enough growth control for the hedge to be admissible. The terminal condition is $p(x,T)=f(x)$.

The Itô formula gives the claim's price changeA self-financing portfolio worth $p$ with $\Delta$ shares has bank-account value $p-\Delta S_t$. Its gain isTo replicate the claim, the Brownian coefficients must agree, forcing the delta hedge $\Delta =p_x$. Equating the remaining drift coefficients cancels the physical drift $\mu$ and yieldsThe cancellation explains why the Black-Scholes equation contains the interest rate, not the physical stock drift. Conversely, if an admissible solution of this equation is given, holdings $p_x$ in the stock and $(p-x{p}_x)/B_t$ in the bank account have value $p$ and gains exactly $dp$. They are therefore a self-financing portfolio replicating the terminal payoff. Absence of arbitrage forces the claim to have this price. Merely decomposing the current value into arbitrary holdings would not justify self-financing; the gain identity and delta hedge are essential.

Now suppose the claim holder receives the cash rate $k(S_t,t)$ before $T$. The price $p$ is the ex-dividend value, so its total gain is $dp+kdt$. The replicating portfolio pays the same amounts out of its wealth, with no external injections. Thus its gain identity isAgain the Brownian coefficients force $\Delta =p_x$. Comparing drifts now gives the Black-Scholes equation with claim dividends:In particular the cash-flow source has a positive sign in the left-hand side. It is not an underlying-stock dividend yield and does not replace the drift $\rho x$ by a dividend-adjusted drift.

For another justification, under the risk-neutral measure $Q$, discount the claim's cum-dividend gain. The Itô formula shows thatexactly when the displayed equation holds. Subject to the usual integrability making this a true martingale, conditional expectation at maturity givesPositive interim payments therefore add positive value, which also checks the source sign. For instance, a constant payment rate $k_0$ and zero terminal payoff have price $k_0(1-e^{-\rho (T-t)})/\rho$, interpreted as $k_0(T-t)$ at $\rho =0$; substitution satisfies $p_t-\rho p+k_0=0$.

Use ${\theta }_t$ for the dollar amount invested in the stock; the number of shares is ${\theta }_t/S_t$. The bank account contains $w_t-{\theta }_t$ dollars. With no intermediate consumption, the self-financing portfolio equation isIf instead $\theta$ denotes share holdings, replace each dollar ${\theta }_t$ here by $S_t{\theta }_t$. The equation says that trading only reallocates existing portfolio wealth; gains come from cash interest and stock returns.

For the standard unconstrained complete-diffusion formulation, assume the completed natural Brownian filtration, initial wealth $w_0>0$, and nonnegative admissible wealth. Define the market price of risk ${\kappa }_t=({\mu }_t-r_t)/{\sigma }_t$. The boundedness assumptions make $\kappa$ bounded. The state-price density normalized at zero isIts money-market-adjusted process $Z_t=e^{{\int }_0^t{r}_{s}ds}{\zeta }_t$ is a true martingale by the Novikov condition. Under $d\mathbb{Q}=Z_{T}d\mathbb{P}$, the Girsanov theorem gives $W_t^{\mathbb{Q}}=W_t+{\int }_0^t{\kappa }_{s}ds$ as a Brownian motion, and the discounted stock has zero drift.

The Itô product rule applied to the wealth equation givesFor nonnegative admissible wealth this is a nonnegative local martingale, hence a supermartingale. Therefore every affordable terminal payoff $X=w_T$ satisfies the state-price budget constraint $\mathbb{E}[{\zeta }_{T}X]\le w_0$. Nonzero volatility and the Brownian martingale representation theorem allow replication of nonnegative claims with finite state-price cost.