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A European call option gives its holder the right, without an obligation, to buy one unit of an underlying financial asset at the strike price $K$ on the expiry date $T$. European describes exercise timing, not geography: exercise is allowed only at $T$. Here the underlying is one stock share, $S_T$ is its price at expiry, and the financial payoff is expressed as a cash amount. Physical exercise delivers a share against payment of $K$; immediate resale gives the same net amount under the assumed frictionless market.

The holder exercises only when buying at the strike price is advantageous. Writing $x^{+}=\max (x,0)$ for the positive part, the terminal financial payoff isThe European call option is a reusable building block for claim replication: its financial payoff is zero below the strike price and rises by one unit for each unit of stock price above it. The corner at $K$ lets combinations of European call options construct other payoff shapes, as in static replication on a finite terminal support.

For strike price $K=100$, the following are deterministic calculations conditional on each displayed terminal stock price. They assign no probabilities to the three scenarios. The European put option has the opposite exercise right and the financial payoff $(K-S_T{)}^{+}$.

A financial payoff excludes the initial contract price. If the illustrative European call option costs $C_0=10$, its financial payoff at $S_T=120$ is $20$, and subtracting its initial cost gives $10$. If that cost is instead borrowed for one year at the constant continuously compounded interest rate $r=0.05$, the terminal net amount is $20-10e^{0.05}\approx 9.48728904$. At $S_T=80$ or $100$, the corresponding net amount is $-10e^{0.05}\approx -10.51271096$. Keeping all cash amounts at the same date avoids confusing a financial payoff with a financed gain.

Robert Shiller's Options Markets lecture, Yale ECON 252 (2011) introduces these exercise rights and payoff shapes.

A European put option gives its holder the right, without an obligation, to sell one unit of an underlying financial asset at the strike price $K$ only on the expiry date $T$. For one stock share, its terminal financial payoff is the positive partFor strike price $K=100$, terminal stock prices $80$, $100$, and $120$ give financial payoffs $20$, $0$, and $0$, respectively. At $S_T=80$, selling for $100$ rather than the market price $80$ supplies the difference $20$; at $S_T=120$, the holder can let the right expire. Equality at $S_T=100$ gives zero net exercise value. The European call option example plots both financial payoffs on the same axes.

Combining a European put option with the underlying stock share produces the terminal amount $S_T+(K-S_T{)}^{+}=\max (S_T,K)$. Its floor at the strike price is why a European put option can describe protection against a fall in terminal stock value. The protection has an initial cost and lasts to the specified expiry; the terminal floor alone says nothing about the financial return. Put-call parity replicates this same amount with a European call option and a zero-coupon bond.

Using the put-call parity example's price $P_0=10+100e^{-0.05}-100\approx 5.12294245$, the cost financed for one year at interest rate $r=0.05$ grows to $P_0{e}^{0.05}=100-90e^{0.05}\approx 5.38560133$. The terminal net amount is therefore approximately $14.61439867$ when $S_T=80$, and $-5.38560133$ when $S_T=100$ or $120$. These computations use the unrounded price and distinguish the financial payoff from the financed gain. Unlike an American put option, a European put option supplies no right to exercise before $T$.

A model specifying traded financial assets, their price and payment processes, available information, and allowed admissible trading strategies. A self-financing portfolio generates wealth using gains on these assets. Absence of arbitrage and market completeness respectively concern whether trading can create a riskless profit from zero capital and whether every specified contingent claim can be replicated.

Normalize the riskless financial asset to $B_0=1$, $B_1=b>0$, and write $S_0,S_1$ for the $s$ risky asset prices. Their discounted gain random vector is $Y=S_1/b-S_0$. A zero-cost portfolio holding $h\in {\mathbb{R}}^s$ in the risky assets and $-h\cdot S_0$ in the riskless asset has terminal value $bh\cdot Y$. Thus an arbitrage meansAn equivalent martingale measure is a probability measure $Q$ with the same null sets as $P$, under which $Y$ is integrable and ${\mathbb{E}}_{Q}Y=0$. The discounted risky prices are then martingales over the single trading period.

If such $Q$ exists, a nonnegative terminal gain has expectation zero under $Q$. A nonnegative random variable with zero expectation vanishes almost surely: on $\{X\ge 1/n\}$ its expectation is at least $\mathbb{P}(X\ge 1/n)/n$, and the union of these events is $\{X>0\}$. Equivalence transfers this conclusion to $P$, excluding arbitrage.

For the converse we prove the needed positive-density separation proof of the one-period asset-pricing theorem. First make the gain integrable without changing its null sets. Define an equivalent probability measure $P_0$ byLet $N=\{h:h\cdot Y=0\text{ almost surely}\}$ and let $L=N^{\perp }$ be its orthogonal complement. Taking a finite basis of $N$ shows $Y\in L$ almost surely. If $L=\{0\}$, the gain is zero and $P_0$ already gives the required measure. Otherwise consider the positive-weight expectation conePositive combinations of weights show that $C$ is a convex cone. It is open relative to $L$. Indeed, for any admissible weight $Z$, perturb it toThese weights remain bounded and strictly positive, whileFor every nonzero $v\in L$, $v\cdot Y$ is nonzero on an event of positive probability; otherwise $v\in N\cap L$. ConsequentlyThe matrix is finite because $\Vert Y{\Vert }^2/(1+\Vert Y\Vert )\le \Vert Y\Vert$. The kernel of $A_Z$ on $L$ is zero, so it defines an invertible matrix on the finite-dimensional vector space $L$. For any sufficiently small $\delta >0$, every $y$ with $\Vert y\Vert <\delta /\Vert A_Z^{-1}\Vert$ is the image of a vector in the radius-$\delta$ ball, since $\Vert A_Z^{-1}y\Vert <\delta$. Hence the perturbations give a neighborhood of zero, proving relative openness of $C$.

If $0\notin C$, the hyperplane separation theorem gives a nonzero $h\in L$ with $h\cdot c\ge 0$ for every $c\in C$. The separating constant can be taken to be zero because $C$ is a cone. Put $A=\{h\cdot Y<0\}$ and use the strictly positive weights $Z=1_A+\epsilon$. Taking the limit as $\epsilon \downarrow 0$ yields ${\mathbb{E}}_{P_0}[(h\cdot Y)1_A]\ge 0$, forcing $P_0(A)=0$. But $h\notin N$, so $P_0(h\cdot Y>0)>0$. This is an arbitrage, a contradiction. Hence $0\in C$: some bounded strictly positive $Z$ satisfies ${\mathbb{E}}_{P_0}[ZY]=0$. Normalize it by $dQ/d{P}_0=Z/{\mathbb{E}}_{P_0}Z$. This gives an equivalent martingale measure with integrable gains, establishing the fundamental theorem of asset pricing here without any further existence theorem.

Now retain only the $m$ positive-probability states of the finite probability space. Let $V\subseteq {\mathbb{R}}^m$ be the linear span of the discounted terminal asset payoffs; it contains the constant vector $1$. A complete market means $V={\mathbb{R}}^m$. An equivalent martingale measure is a vector $q$ of strictly positive state probabilities whose pairing with each asset payoff gives its initial price. By linearity it prices every attainable payoff. If $V={\mathbb{R}}^m$, the indicator functions of individual states are attainable, so these prices determine every $q_j$ uniquely.

If $V$ is proper, choose $0\ne v\in V^{\perp }$. Since $1\in V$, ${\sum }_j{v}_j=0$. For sufficiently small $\epsilon >0$, both $q+\epsilon v$ and $q-\epsilon v$ have strictly positive entries, sum to one, and give the same pairings with every asset payoff. They are distinct equivalent martingale measures. Therefore market completeness is equivalent to uniqueness of the equivalent martingale measure.

Use the riskless financial asset as numéraire, normalized to have initial value one and terminal value $R>0$. Write $S_0\in {\mathbb{R}}^s$ for the initial risky prices and $S_1$ for the finite, measurable terminal price random vector. Define the discounted gain vector $X=S_1/R-S_0$. A risky holding $h$ financed by the cash holding $-h\cdot S_0$ has zero initial cost and discounted terminal gain $h\cdot X$. An arbitrage is such a portfolio with $h\cdot X\ge 0$ almost surely and $\mathbb{P}(h\cdot X>0)>0$. Allowing nonpositive initial cost is equivalent: invest any initial surplus in the riskless financial asset to obtain a zero-cost arbitrage.

An equivalent martingale measure is a probability measure $Q$ with exactly the same null events as the physical probability measure $P$, under which discounted asset prices are integrable and satisfy ${\mathbb{E}}_Q[S_1/R]=S_0$. Equivalently, ${\mathbb{E}}_{Q}X=0$. Such a measure rules out arbitrage: every zero-cost portfolio has mean discounted gain zero, whereas a nonnegative gain positive with positive $P$-probability is also positive with positive $Q$-probability and has strictly positive $Q$-expectation.

For the converse, the probability space need not be finite, and the payoffs need not be integrable under $P$. Here is a direct Gaussian-damped martingale density construction. SetThe set $N$ is a linear subspace. Choosing a finite basis of $N$ shows that $X\in L$ almost surely: intersect the probability-one events on which each basis vector is orthogonal to $X$. If $h\in L\setminus \{0\}$, then $h\cdot X$ is not almost surely zero. Absence of arbitrage implies both $P(h\cdot X<0)>0$ and $P(h\cdot X>0)>0$, by applying the definition to $h$ and $-h$.

All the analytical steps in minimizing $F$ can be checked directly. On $|h|\le H$ the integrand is bounded by $e^{H^2/4}$, since $-|X{|}^2+H|X|\le H^2/4$. Its first two directional derivatives are bounded uniformly in $X$ by constants depending only on $H$ and the direction, because the functions $r^j{e}^{-r^2+(H+1)r}$ are bounded for $j=0,1,2$. The mean-value bound therefore proves continuity of $F$, and the Taylor theorem remainder bound, integrated over $P$, provesThus differentiation under the integral does not require an unproved physical moment assumption.

To prove coercivity, suppose $|h_k|\to \infty$ in $L$. A subsequence of $h_k/|h_k|$ converges to a unit vector $v\in L$. This elementary finite-dimensional compactness follows by successively taking convergent subsequences of the bounded coordinates. Since $P(v\cdot X<0)>0$, some event $A=\{v\cdot X\le -\epsilon ,|X|\le K\}$ has positive probability: the negative-gain event is the countable union of such events with $\epsilon =1/j$ and integer $K$. Eventually $(h_k/|h_k|)\cdot X\le -\epsilon /2$ on $A$, and henceIf coercivity failed, an unbounded sequence with bounded $F$ would have just such a contradictory subsequence. Consequently a minimizing sequence for $F$ is bounded and has a convergent subsequence; continuity makes its limit $h_{\ast }$ a minimizer. If $L=\{0\}$, simply take $h_{\ast }=0$. In every case the directional derivatives vanish at $h_{\ast }$. The vector ${\mathbb{E}}_P[X{e}^{-|X{|}^2-h_{\ast }\cdot X}]$ lies in $L$ and is orthogonal to $L$, so it is zero. Thereforeis a positive mean-one density defining an equivalent martingale measure. Integrability of $X$ under $Q$ follows from the same Gaussian bound with $j=1$. This establishes absence of arbitrage if and only if an equivalent martingale measure exists, without quoting a financial duality theorem or assuming a finite state space.

Now suppose the probability space is finite. Discard null states and enumerate its positive-probability atoms by $1,\dots ,m$. Every measurable terminal payoff is a vector in ${\mathbb{R}}^m$. Let $A$ be the $m\times (s+1)$ matrix whose columns are the discounted terminal payoffs of the traded financial assets, with first column $1$, and let $p=(1,S_0)$ be their initial price vector. A portfolio with holding vector $z$ replicates the discounted payoff $Az$, and an equivalent martingale measure is exactly a vector $q$ satisfyingThe first pricing equation includes ${\sum }_j{q}_j=1$. A complete market means $\mathrm{im}A={\mathbb{R}}^m$. If $q,q^{\prime }$ both price the financial assets, then $(q-q^{\prime })\cdot Az=0$ for every $z$. Completeness allows $Az$ to be each coordinate vector, so $q=q^{\prime }$.

Conversely, if the financial market is incomplete, there is a nonzero vector $v$ with $A^{T}v=0$. To see this without invoking a duality theorem, perform row elimination on the homogeneous system: its number of pivots is $\mathrm{rank}A<m$, leaving a free variable and hence a nonzero solution. The constant payoff column gives ${\sum }_j{v}_j=0$. Existence of an equivalent martingale measure was already proved, so choose one strictly positive vector $q$. Forboth $q+\epsilon v$ and $q-\epsilon v$ are distinct strictly positive probability vectors satisfying $A^T(q\pm \epsilon v)=p$. Thus uniqueness fails. On a finite arbitrage-free market, completeness is equivalent to uniqueness of the equivalent martingale measure.

Let $Y$ be the discounted gains of finitely many financial assets on any probability space. Tilting the physical probability measure by a normalized multiple of $(1+\Vert Y\Vert {)}^{-1}$ makes $Y$ integrable without changing null events. The positive-weight expectation cone alternative then says either a bounded strictly positive weight makes the weighted gain expectation zero, or a deterministic risky holding gives an arbitrage. Normalizing the weight yields an equivalent martingale measure. Conversely, under such a measure a nonnegative terminal gain with zero initial price has zero expectation, and must vanish almost surely. This proves the one-period fundamental theorem of asset pricing without assuming physical integrability of asset gains.