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In a discrete-time binomial market with $B_t=(1+r{)}^t$ and one-step stock factors $1\pm \epsilon$, use the risk-neutral probability $p=(1+r/\epsilon )/2$, where $0\le r<\epsilon <1$. For the exercise payoff $S_t^2$, the discounted one-step reward multiplier has meanThe Snell envelope of a multiplicative process gives the value $V_t=S_t^2{\lambda }^{T-t}$. Continuation strictly exceeds exercise before maturity, so the optimal exercise time is $T$. The replicating portfolio in a binomial market holds $2S_t{\lambda }^{T-t-1}$ shares on the next step, found by subtracting the two next-state values and dividing by the stock-price difference. At time zero the price is $S_0^2{\lambda }^T$.

A convex terminal payoff has a convex pricing extension in the binomial market, because risk-neutral pricing is a positive weighted sum of rescaled copies of the payoff. Let the stock factors be $u,d$, the riskless gross factor be $R$, and $q=(R-d)/(u-d)$. The replicating portfolio in a binomial market satisfiesThe weights sum to one. Convexity orders the secant slopes on the adjacent intervals $[d^{2}s,uds]$ and $[uds,u^{2}s]$, proving the displayed inequalities. The inequalities need not be strict: an affine payoff has constant stock holding.

Mathematical finance uses probability theory, linear algebra and mathematical optimization to study financial markets, portfolio decisions and contingent claim prices. One question asks which portfolio best serves a specified objective; another asks what price a contingent claim must have to exclude arbitrage. The latter can have an exact answer even when future stock prices are uncertain.

The central idea is claim replication: construct a self-financing portfolio delivering the same financial payoff as a contract in every possible state. The law of one price then identifies the contract's price with the portfolio's initial cost. The binomial market makes this beautiful connection between uncertainty and linear algebra visible using just two equations. Risk-neutral pricing expresses the same cost as a discounted expectation.

The one-step example below fixes today's stock price at 100, terminal prices at 120 or 80, and the bank account's gross growth factor at 1.05. A European call option with strike price 100 pays 20 or zero. Its replicating portfolio in a binomial market costs $250/21$, independently of the physical up probability, provided both states remain possible. These are conclusions within the specified financial market model.

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.

Let the bank account be $B_r=R^r$, where $R>0$ is the one-period riskless gross return. In the binomial market, $S_{r+1}$ is either $u{S}_r$ or $d{S}_r$, with $0<d<1<u$ and both successors having positive physical probability. The information at time $r$ consists of the first $r$ moves. A trading strategy chooses its holdings using only this information. A self-financing portfolio funds rebalancing from its current value; there is no external contribution.

Absence of arbitrage requires $d<R<u$. If $R\le d$, borrowing to buy the stock yields a nonnegative gain, strictly positive on the up branch; if $R\ge u$, shorting the stock and investing its price yields the reverse arbitrage. When $d<R<u$, define the risk-neutral probability in a binomial marketGiving each successive up move conditional probability $q$ defines an equivalent martingale measure $Q$, since every finite path has positive probability and ${\mathbb{E}}_Q[S_{r+1}\mid {\mathcal{F}}_r]=R{S}_r$.

For a contingent claim with successor values $V_u,V_d$ at a node of current stock value $s$, solve the two replication equations. The number of shares and current cash value areThen $\Delta us+RC=V_u$ and $\Delta ds+RC=V_d$, so the replicating strategy costsHold $C/B_r$ bank-account units over that period. Starting with the terminal payoff and applying these equations backwards supplies an adapted self-financing strategy replicating every path-dependent contingent claim. This proves market completeness. A different claim price would permit an arbitrage by buying the cheaper claim or replica and selling the dearer one. Iterating the backward pricing relation gives the unique risk-neutral pricing formulaThe physical up probability affects real-world frequencies but does not enter this replication price.

For a terminal payoff $f(S_n)$, write $k=n-r$. Its current price is the restriction to the available nodes of the functionIf $f$ is convex, each summand is convex in $s$ and each weight is positive; summing their defining convexity inequalities proves that $g_r$ is convex. This provides an explicit convex extension between the actual stock nodes, so “convex on the nodes” is unambiguous. If $f$ is strictly convex, each positive scaling preserves strict convexity, so $g_r$ is strictly convex as well.

For the replicating portfolio in a binomial market, the stock holding isFor $r+1<n$, let $F=g_{r+2}$. The next down and up holdings are the secant slopes of $F$ on the adjacent intervals with endpoints $d^{2}s,uds,u^{2}s$:For any convex function and $a<b<c$, its defining inequality at $b=(1-t)a+tc$ rearranges to $(F(b)-F(a))/(b-a)\le (F(c)-F(b))/(c-b)$. The inequality is strict for a strictly convex function. Applying this to the three stock values orders the two next holdings. Meanwhile the backward pricing relation gives, by subtracting its values at $us$ and $ds$,The coefficients are strictly positive and sum to one. Therefore the current share holding lies between the next two holdings:Both inequalities are strict for a strictly convex payoff. For an affine payoff the share holding is constant, explaining the required weak interpretation. These are numbers of shares, not the fraction of portfolio wealth invested in the stock.

Let the one-period riskless gross return be $R>0$, with $0<d<R<u$, and let the physical up probability be $\pi \in (0,1)$. The binomial market has risk-neutral probabilityUnder its equivalent martingale measure $Q$, successive multipliers $J_k$ are independent, taking $u,d$ with probabilities $q,1-q$. The risk-neutral pricing function has the extensionFor every fixed positive multiplier $A$, the function $s\mapsto f(As)$ is convex when $f$ is convex. Taking a positive weighted sum preserves convexity. More explicitly, for $0\le \lambda \le 1$,Thus the pricing extension is convex, and so are its values on the attainable stock-price grid.

The replicating portfolio in a binomial market holdsshares over the next interval. If $r+1<n$, put $A={\Delta }_{r+1}(us)$ and $D={\Delta }_{r+1}(ds)$. These are the secant slopes of $g_{r+2}$ on $[uds,u^{2}s]$ and $[d^{2}s,uds]$, respectively. For a convex function, the slope of a secant on a later adjacent interval is no smaller than that on the earlier interval: apply the defining convexity inequality to the middle point $uds$, then rearrange. Hence $A\ge D$. Using backward option pricing twice givesThe two coefficients are positive and sum to one. Therefore convex-payoff delta monotonicity in a binomial market yieldsIn particular an up move cannot reduce the stock holding. “Increases” here means nondecreases: an affine payoff gives equality, so strict increase cannot be claimed for every convex payoff.

For the expected utility maximization problem, let $P$ be the physical measure and $Z=dQ/dP$ on terminal paths. If a path has $j$ up moves, its binomial-market probability density isEvery terminal payoff is replicable in this complete market. Buying a positive payoff $H$ with initial wealth $w_0$ therefore imposes the state-price budget constraint ${\mathbb{E}}_{Q}H=R^n{w}_0$. Since the utility function is increasing, the full budget is used. Its derivative and second derivative arePut $a=\gamma /(\gamma -1)$. The first-order condition $v^{\prime }(H_{\ast })=\lambda Z$ for a positive multiplier $\lambda$ gives $H_{\ast }=C{Z}^{-a}$. The budget determines $C$ uniquely. Independence of the successive moves givesThus the optimal claim, as a function of the number of up moves, isBecause $S_n=S_0{u}^j{d}^{n-j}$ uniquely determines $j$, this is also a terminal-stock-price payoff, attainable by backward option pricing and claim replication. It rewards states relatively more probable under the investor's physical measure than under the pricing measure. If $\pi =q$, the optimum is the riskless payoff $R^n{w}_0$.

To prove global optimality rather than merely assert a stationary point, use the tangent inequality for the strictly concave function $v$:Taking physical expectations makes the last term zero for any fully financed competing claim, and nonpositive for any claim costing less. Therefore ${\mathbb{E}}_{P}v(H)\le {\mathbb{E}}_{P}v(H_{\ast })$, with equality only when $H=H_{\ast }$ almost surely. Finiteness causes no difficulty on the finite binomial tree, and $H_{\ast }>0$ in every terminal state.

Consider a binomial market with dates $0,\dots ,n$, bank account $B_k=b^k$ for $b>0$, and stock price $S_{k+1}=S_{k}u$ or $S_{k}d$, where $0<d<u$. Both moves have positive physical probability, and the information at date $k$ is the history of the first $k$ moves. Absence of arbitrage requires and is implied by $d<b<u$. If $b\le d$, borrowing cash to buy stock has nonnegative payoff with a strict gain in one state; if $b\ge u$, the reverse trade does. In the interior range, the risk-neutral probability of an up move isIt makes ${\mathbb{E}}_Q[S_{k+1}\mid {\mathcal{F}}_k]=b{S}_k$, so $S_k/B_k$ is a martingale. This proves absence of arbitrage by pricing discounted gains under the equivalent measure.

At a node with stock price $s$, let the desired next-period portfolio values be $V_u,V_d$. Solve the two replication equations $\Delta su+\beta B_{k+1}=V_u$ and $\Delta sd+\beta B_{k+1}=V_d$. The replicating portfolio in a binomial market hasIts current value isRebalance at the next node using exactly the value supplied by the previous holdings, so the strategy is self-financing. Backward induction from any terminal contingent claim $H$ gives its hedge and the unique priceThus the finite binomial market is a complete market. The construction works for path-dependent claims on the full history tree, even when their values do not recombine merely as functions of the current stock price. For $H=f(S_n)$, independence under $Q$ gives the explicit time-zero priceThe stock hedge is a difference quotient of successor values, the discrete analogue of the delta hedge.

The backward option pricing recursion iswith terminal value $V(N,s)=g(s)$. At a node with $S_{n-1}=s$, define the replicating stock holding byThis is a predictable process because it depends only on information available at time $n-1$.

Indeed, if $X_{n-1}=V(n-1,s)$, substituting the risk-neutral formula for $q$ into the self-financing update shows separately in the up and down states thatAn induction over $n$ therefore gives $X_N=V(N,S_N)=g(S_N)$. The required initial capital is

Now $R=1+r=9/8$. The local risk-neutral up probabilities solve $RS=q{S}_u+(1-q)S_d$. They areThe terminal payoff of the European put option with strike $5$ is $0,0,2$ at stock prices $7,5,3$. Backward option pricing gives time-one valuesand henceThe first-period stock holding in the replicating portfolio in a binomial market isThus the hedge initially shorts $2/9$ of a share.