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Write $(a,b,c)$ for holdings of cash, the stock and the European put option, and let $\gamma =-V_0$ be initial consumption. At the given price, $a=-10b-c-\gamma$, and the three terminal payoffs, in descending order of the stock price, areNonnegative terminal payoffs require $c+\gamma \le 2b$ and $c-\gamma \ge b$. Together with $\gamma \ge 0$, these imply $b\ge 2\gamma \ge 0$. If $b=0$ they force $\gamma =c=a=0$, which is not an arbitrage. If $b>0$, the highest-state payoff is at least $3b>0$, so every such choice is an arbitrage. Consequently the complete set isThe pure-investment arbitrages are exactly the choices $\gamma =0$:For example, $(a,b,c)=(-11,1,1)$ costs zero and pays $(4,1,0)$.

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.

Let $V_t$ denote wealth measured in units of the strictly positive numéraire. A predictable self-financing portfolio is an arbitrage ifEquivalently, one may permit $V_0\le 0$ provided one of the two gains is strict.

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.

Suppose an arbitrage existed. Its discounted terminal wealth would satisfy $V_T\ge 0$ $\mathbb{P}$-almost surely and $\mathbb{P}(V_T>0)>0$. Equivalence of $\mathbb{Q}$ and $\mathbb{P}$ gives the same two statements under $\mathbb{Q}$, soBut part (c), after the usual admissibility or localization needed for integrability, givesa contradiction. Hence the existence of the stated martingale measure rules out arbitrage.

At maturity, $P_T^T=1$. If $P_t^T\le 0$ on an event $A\in {\mathcal{F}}_t$ of positive probability, buying one bond on $A$ has nonpositive cost and certain payoff $1$ on $A$ at $T$; any negative purchase cost can also be consumed or retained. This is an arbitrage. Therefore absence of arbitrage implies

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.

In the one-period building block of the Cox--Ross--Rubinstein model, the money-market account grows by a deterministic factor $R>0$, while the stock grows independently at each time byThus $S_t^0=R^t{S}_0^0$ and $S_t^1=S_0^1{\prod }_{j=1}^t{Y}_j$. Equivalently, the discounted stock has multipliers $u=U/R$ and $d=D/R$.

The model has no arbitrage precisely when the risk-free return lies strictly between the stock returns:Indeed, failure of either strict inequality gives a one-period arbitrage, while under the strict inequalities the probabilitylies in $(0,1)$. Taking the moves independent under $Q$ with up probability $q$ makes the discounted stock a martingale, since $qu+(1-q)d=1$. This is the unique Risk-neutral probability in the Cox--Ross--Rubinstein model, hence the unique equivalent martingale measure.

Let $\varphi =({\varphi }_n^0,{\varphi }_n^1,\dots ,{\varphi }_n^d)$ be a predictable self-financing portfolio, and let $V_n^{\varphi }={\varphi }_n\cdot S_n$ be its value. An arbitrage is such a strategy with

An arbitrage is a zero-initial-wealth admissible self-financing strategy with $V_T\ge 0$ almost surely and $\mathbb{P}(V_T>0)>0$ for some finite horizon $T$.

A risky-asset portfolio $\phi$ financed through the risk-free asset has zero-cost discounted gainIt is an arbitrage when $G_{\phi }\ge 0$ almost surely and $\mathbb{P}(G_{\phi }>0)>0$.

If $G\ge 0$ were an arbitrage gain, then $X_1^{\ast }+tG$ would be attainable from the same initial wealth for every $t>0$. Since $U$ is increasing,almost surely, with strict inequality on a set of positive probability under the usual strictly increasing meaning of utility. This contradicts optimality of $X_1^{\ast }$. Hence existence of an optimizer implies no-arbitrage.

For a symmetric matrix,
$\mathrm{range}V=(\ker V{)}^{\perp }$. Since $q\notin \mathrm{range}V$, there is
$\eta \in \ker V$ with ${\eta }^{T}q\ne 0$. For any target excess mean $h$, setThen ${\theta }^{T}q=h$ andThus the required minimum is zero for every $m,x$.

Choose the sign of $\eta$ so that ${\eta }^{T}q>0$, buy the risky portfolio $\eta$, and finance it by borrowing its time-zero cost in the bank. Its initial wealth is zero, while its terminal excess payoff has variance zero and positive mean ${\eta }^{T}q$, so it is a strictly positive constant almost surely. This is an arbitrage.

A portfolio $H\in {\mathbb{R}}^n$ is an arbitrage when $H\cdot P_0\le 0$, $H\cdot P_1\ge 0$ almost surely, and at least one inequality supplies a strict gain: either $H\cdot P_0<0$ or $\mathbb{P}(H\cdot P_1>0)>0$. It is a terminal-consumption arbitrage when $H\cdot P_0=0$, $H\cdot P_1\ge 0$ almost surely, and the terminal inequality is strict with positive probability.

The stock tree has $S_0=5$, first-period values $6,4$, and terminal successors $7,5$ from $6$ and $5,3$ from $4$. When $r=1/6$, the risk-free growth factor is $R=7/6$. At the upper time-one node, investing $6$ in the risk-free asset produces $7$ at time two, while one share bought for $6$ produces only $7$ or $5$.

Use the following predictable self-financing portfolio. Hold nothing initially. If $S_1=6$, short one share and invest the proceeds $6$ in the risk-free asset; if $S_1=4$, continue to hold nothing. Its terminal payoff ison the upper branch, and zero on the lower branch. It costs zero, is never negative, and is positive with positive probability, so it is an arbitrage.

An arbitrage is a finite-horizon previsible strategy with no positive initial cost, nonnegative cash flows at every date, and a strictly positive cash flow with positive probability at some date, after liquidation. Equivalently, one may require zero initial value and a nonnegative terminal gain that is positive with positive probability, after retaining intermediate cash flows in a cash account.

A pure-investment arbitrage has zero initial cost, no intervening consumption, and a nonnegative terminal payoff that is strictly positive with positive probability.