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A complete market can replicate every contingent claim in its specified payoff class: for every terminal ${\mathcal{F}}_T$-measurable payoff $X$, there is an admissible self-financing portfolio whose terminal wealth equals $X$ almost surely. Equivalently, in an arbitrage-free finite-state discrete-time model the equivalent martingale measure is unique. In notation,Market completeness is a replication property. The following price comparisons also use the usual absence of arbitrage, which the source leaves implicit: without it the initial cost of claim replication need not be unique. For example, in a deterministic one-period model with cash worth one at both dates and a stock worth one initially but zero finally, all terminal claims are replicable using cash. Adding any stock holding changes the initial cost without changing the terminal payoff. This is a complete model with arbitrage, so it has no well-defined unique replication price.

Under the usual no-arbitrage interpretation from part (a), let $Q^N$ be an equivalent martingale measure for the strictly positive numéraire $N$. Then $M_t=S_t/N_t$ is a martingale. Define the discounted European call option payoff $Y_t=(S_t-K{)}^{+}/N_t=(M_t-K/N_t{)}^{+}$. Because $N_{t+1}\ge N_t$ and $K>0$, we have $K/N_{t+1}\le K/N_t$. The conditional Jensen inequality therefore givesFor the middle step, the strike $K/N_t$ is fixed conditionally on ${\mathcal{F}}_t$; alternatively use $\mathbb{E}[X^{+}\mid {\mathcal{F}}_t]\ge \max \{\mathbb{E}[X\mid {\mathcal{F}}_t],0\}$. Thus $Y$ is a submartingale. Risk-neutral valuation in units of $N$ gives $C(T,K)=N_0{\mathbb{E}}^{Q^N}{Y}_T$, with the usual fixed initial prices. ConsequentlyThe source's word “increasing” means nondecreasing: a European call option can have zero payoff at successive maturities, so strict increase is not guaranteed. Integrability is understood in the claim class for which the stated replication prices exist.

The positive-part function is convex, so pathwise Jensen inequality givesTo compare the Asian option price with European call option prices at different dates, carry each earlier payoff forward using the numéraire. Purchase $1/T$ of each replicating strategy for maturity $t$, and, when its payoff is received, reinvest it in $N$ until $T$. This is a self-financing portfolio, with terminal wealthThe first inequality uses $N_T\ge N_t$ and nonnegative payoffs. Since the Asian option is replicable in the complete market, absence of arbitrage makes its replication cost no larger than this superhedging cost. ThereforeEquivalently, divide the first pathwise bound by $N_T$, use $(S_t-K{)}^{+}/N_T\le (S_t-K{)}^{+}/N_t$, and take expectations under the numéraire equivalent martingale measure. Merely averaging earlier payoffs without reinvestment would miss the difference in payment dates.

For a finite-state, finite-horizon discrete-time frictionless market with a strictly positive numéraire, the fundamental theorem of asset pricing states no arbitrage if and only if there exists an equivalent martingale measure under which all discounted asset prices are martingales. Equivalence means the same null events as the physical measure. In this setting market completeness is equivalent to uniqueness of the equivalent martingale measure, and a replicable payoff $H$ has price $S_t^0{\mathbb{E}}_Q[H/S_T^0\mid {\mathcal{F}}_t]$. Without finite-state hypotheses, stronger integrability and closure conditions are needed; this is the standard multi-period course version.

Use discrete dates $j=0,\dots ,N$, a bank account $B_j=R^j$ with $R=1+r>0$, and a positive stock with each one-period gross return either $u$ or $d$, where $0<d<u$. On the natural tree assume both moves have positive physical probability. The Cox--Ross--Rubinstein model is arbitrage-free precisely whenOtherwise a stock/bank position gives nonnegative profit in both states and positive profit in at least one. Under the condition, the unique equivalent martingale measure has independent upward moves with conditional probabilityIndeed $qu+(1-q)d=R$, making the discounted stock a martingale. Both branches have positive probability under this measure, ensuring equivalence. The uniqueness/completeness claim is for the stock-generated binomial tree; an enlarged filtration carrying additional untraded risk would change that claim.

A pricing kernel is a strictly positive random variable $R$ such that $R{P}_1^i$ is integrable andIt is a one-period state-price density, converting a terminal payoff into its initial price. Strict positivity means $R>0$ almost surely, not merely $R\ge 0$. When cash has price $1$ at both dates, this identity forces $\mathbb{E}R=1$, so $d\mathbb{Q}=Rd\mathbb{P}$ defines an equivalent martingale measure.

Write $q_{15},q_{12},q_9$ for the state probabilities under an equivalent martingale measure. Since cash is constant and the European put option pays only in the lowest state, the pricing equations areTheir unique solution is $q_9={\xi }_0/2$, $q_{15}={\xi }_0/2-2/3$, $q_{12}=5/3-{\xi }_0$. Every physical state has positive probability, so equivalence requires all three values to be strictly positive. ThusFor each price in this open interval the pricing kernel takes values $3q_{15},3q_{12},3q_9$ in the three equally likely states, proving absence of arbitrage.

The necessity, including the exclusion of endpoints, can also be checked directly. The three terminal payoff vectors of cash, the stock and the European put option form the invertible matrixEach unit state payoff therefore has a replicating strategy, whose initial cost is the corresponding $q$. A zero or negative $q$ supplies an arbitrage, so the endpoints are genuinely excluded.

Assume the usual positive initial stock price and strike, so the logarithm is defined, and interpret $U$ as a classical solution of the displayed partial differential equation. For a fixed complex $z$, put $F_t(z)=S_t^{z}U(t,{\sigma }_t,z)$, defining $S_t^z=\exp (z\log S_t)$. The Itô formula for the two Brownian motions with correlation coefficient $\rho$ givesThe partial differential equation cancels the entire drift. By the permission to treat the resulting local martingales as true martingales, and the terminal condition, $F_t(z)=\mathbb{E}[S_T^z\mid {\mathcal{F}}_t]$.

The bounded spot volatility ensures $\mathbb{E}{S}_T^{x_0}<\infty$: the Itô formula for a real power and localization bound its moment by $S_0^{x_0}\exp (x_0(x_0-1)\Vert \sigma {\Vert }_{\infty }^{2}T/2)$. Rewrite the proposed integrand asNow $|F_t(x_0+iy)|\le \mathbb{E}[S_T^{x_0}\mid {\mathcal{F}}_t]$, an integrable bound independent of $y$. Conditional Fubini's theorem and part (b) therefore implyThis route justifies the contour exchange without assuming bounds on $U$ uniform in all complex $z$. Cash is constant, $S$ is a true martingale under the stated allowance, and the displayed $C$ is a true martingale. The original measure is thus an equivalent martingale measure for all three assets. The fundamental theorem of asset pricing gives the market has no arbitrage under the usual admissible trading convention.

With constant coefficients, $\lambda =(\mu -r)/\sigma$ is constant, so the stochastic exponential $Z_t=e^{rt}{Y}_t$ is a true martingale. Under its equivalent martingale measure $\mathbb{Q}$, the stock follows the Black-Scholes model $d{S}_t=S_t(rdt+\sigma d{W}_t^{\mathbb{Q}})$. The Gaussian exponential moment gives the value of the power optionThe option delta is ${\partial }_{s}V=V/(2s)$, so the replicating strategy holdsHalf the wealth value is in the stock and half in the bank account, with continuous rebalancing. To check the self-financing portfolio property under the original measure, the Itô formula gives $d{V}_t=\frac{1}{2}(r+\mu )V_{t}dt+\frac{1}{2}\sigma V_{t}d{W}_t={\theta }_{t}d{S}_t+{\eta }_{t}d{B}_t$. Also $V_T=\sqrt{S_T}$ and $V_t>0$.

The payoff is unbounded, so the bounded-payoff restriction of part (c) is not invoked automatically; the Black-Scholes model has the necessary finite Gaussian moments, and the explicit strategy attains $V_0=\mathbb{E}[Y_T\sqrt{S_T}]$. The supermartingale bound from part (b) therefore proves its minimality. The cost is independent of the physical drift $\mu$.

The price process is a geometric Brownian motion withPutand define the equivalent measure $Q$ on ${\mathcal{F}}_T$ by the exponential Brownian martingaleThe Cameron-Martin-Girsanov theorem says that $W_t^Q=B_t+\vartheta t$ is Brownian motion under $Q$. Thereforeso $e^{-rt}{S}_t$ is a $Q$-martingale. This is the unique equivalent martingale measure, since cancellation of the single Brownian risk forces the displayed drift shift.

By risk-neutral valuation,Writing $W_T^Q=\sqrt{T}Y$ with $Y\sim N(0,1)$ gives

The finite-horizon discrete-time fundamental theorem of asset pricing states that the market has no arbitrage if and only if there exists a probability measure $\mathbb{Q}$ equivalent to $\mathbb{P}$ under which every integrable discounted asset price $X^i$ is a martingale. Such a $\mathbb{Q}$ is an equivalent martingale measure.

An equivalent martingale measure is a probability measure $\mathbb{Q}$ on the terminal sigma-algebra such that $\mathbb{Q}$ and the physical measure $\mathbb{P}$ are mutually absolutely continuous and every asset price discounted by the chosen numéraire is a $\mathbb{Q}$-martingale with respect to the market filtration.

The fundamental theorem of asset pricing for a finite discrete-time market states that the market is free of arbitrage if and only if it admits an equivalent martingale measure. Subject to the usual finite-market hypotheses, the arbitrage-free market is complete if and only if that measure is unique.

For $0\le t<T$, set $\tau =T-t$ andUnder the equivalent martingale measure, conditional log-normality givesThe risk-neutral pricing value of the digital call option is consequentlyAt $t=T$ this converges to the stated payoff away from $S_T=K$, with the payoff convention specifying the boundary value.

The delta hedge holds the derivative of the claim value with respect to the current stock price. Sincethe number of risky-asset units for $t<T$ isThis is the Black-Scholes digital option formula. The hedge becomes singular close to maturity near the strike, reflecting the discontinuity of the payoff.

Set $Y_n=\log Z_n$ andDefine a probability measure ${\mathbb{P}}^{\ast }$ by the positive densityThe density has expectation one and is strictly positive, so ${\mathbb{P}}^{\ast }$ is equivalent to $\mathbb{P}$. By exponential tilting, the variables $Y_k$ remain independent under ${\mathbb{P}}^{\ast }$ and have distribution $N({\mu }_{\ast },{\sigma }^2)$. Thus $Z_k$ has a log-normal distribution andFor the discounted stock ${\tilde{S}}_n=S_n/S_n^0$,Hence ${\mathbb{P}}^{\ast }$ is an equivalent martingale measure.

There is no arbitrage. Indeed, part (iii) supplies an equivalent martingale measure, so the fundamental theorem of asset pricing implies absence of arbitrage. Directly, the discounted terminal value of a zero-cost admissible self-financing strategy is a ${\mathbb{P}}^{\ast }$-martingale with mean zero; it cannot be nonnegative almost surely and positive with positive probability.

The model has a unique equivalent martingale measure $\mathbb{Q}$ under whichand the discounted stock is a martingale. The time-zero value of an integrable time-$T$ contingent claim $C$ is therefore

At the optimal terminal wealth, normalized marginal utility defines an equivalent martingale measure $Q$ and hence a marginal utility priceThis is an arbitrage-free price, so $p_Q(Y)\in I$. Concavity of utility implies that the finite-quantity indifference bid does not exceed its marginal price:Consequently

For $S_t=S_0{e}^{\mu t+\sigma W_t}$, shifting Brownian drift bymakes $S_t=S_0{e}^{(r-{\sigma }^2/2)t+\sigma W_t^Q}$ under an equivalent martingale measure $Q$. The discounted stock is then a martingale.