No-Arbitrage Pricing and Replication in the Binomial Model

Pricing a One-Period Claim by Replication

In a one-period binomial model, a stock can move to one of two prices. A claim whose payoff depends on that move can be priced by constructing a portfolio that pays exactly the same amount in either outcome. This approach makes the price a consequence of the available trading opportunities: if two portfolios have the same future payoffs, they must have the same current cost in a market without arbitrage.

Use a probability space $(\Omega,\mathcal{F},\mathbb{P})$ with $\Omega=\{u,d\}$, $\mathcal{F}$ the set of all subsets of $\Omega$, and $\mathbb{P}(\{u\})=p$, $\mathbb{P}(\{d\})=1-p$, where $0<p<1$. The filtration is $\mathcal{F}_0=\{\varnothing,\Omega\}$, $\mathcal{F}_1=\mathcal{F}$. The stock-price process $(S_n)_{n=0,1}$ takes values in $S=\{S_0,S_0u,S_0d\}$: $S_0>0$ is its known initial price, and $S_1=S_0u$ in outcome $u$ and $S_1=S_0d$ in outcome $d$, with $0<d<u$. A risk-free cash investment has value $1$ at time $0$ and $R>0$ at time $1$. A portfolio holds $\Delta$ shares and $b$ units of cash, so it costs $\Delta S_0+b$ initially and is worth $\Delta S_1+bR$ at time $1$. A claim pays $C_u$ or $C_d$ in the two outcomes.

Definition: An arbitrage is a portfolio with initial cost at most zero, terminal value nonnegative in both outcomes, and terminal value strictly positive in at least one outcome. The market is arbitrage-free if no such portfolio exists.

The physical probability $p$ describes the outcomes under $\mathbb{P}$, but the replication calculation below does not require its value. What matters for no-arbitrage pricing is whether both outcomes can be assigned strictly positive pricing weights consistent with the stock and cash returns.

The Replicating Portfolio and Pricing Formula

To replicate the claim, choose $\Delta$ and $b$ so that the portfolio has the claim’s payoff in each outcome. Subtracting the down-state equation from the up-state equation determines the stock holding; substituting it back determines the cash holding.

Theorem (One-period replication and pricing): Suppose $S_0>0$, $R>0$, and $0<d<u$. For any real payoffs $C_u,C_d$, there is a unique portfolio replicating them. Its holdings and initial cost are $$ \Delta=\frac{C_u-C_d}{S_0(u-d)},\qquad b=\frac{uC_d-dC_u}{R(u-d)},\qquad V_0=\Delta S_0+b. $$ If also $d<R<u$, define $q=(R-d)/(u-d)$. Then $0<q<1$, the market is arbitrage-free, and $$ V_0=\frac{qC_u+(1-q)C_d}{R}. $$

Proof. Replication requires $\Delta S_0u+bR=C_u$ and $\Delta S_0d+bR=C_d$. Subtracting gives $\Delta S_0(u-d)=C_u-C_d$, which uniquely determines $\Delta$ because $S_0(u-d)>0$. Substitution into the down-state equation gives $b=(uC_d-dC_u)/(R(u-d))$. These holdings satisfy both equations, and any replicating portfolio must solve them, so the portfolio is unique.

Now suppose $d<R<u$. Then $q=(R-d)/(u-d)>0$, and $1-q=(u-R)/(u-d)>0$, with $q+(1-q)=1$. For any portfolio with terminal values $X_u=\Delta S_0u+bR$ and $X_d=\Delta S_0d+bR$, direct substitution gives $$ qX_u+(1-q)X_d =\Delta S_0\bigl(qu+(1-q)d\bigr)+bR =R(\Delta S_0+b), $$ because $qu+(1-q)d=R$. If its initial cost is at most zero and both terminal values are nonnegative, the left side is nonnegative, whereas the right side is nonpositive. Equality must hold, and since both weights are strictly positive, $X_u=X_d=0$. The two terminal equations then imply $\Delta=0$ and $b=0$, so the portfolio is not an arbitrage. Finally, for the replicating portfolio, the same identity gives $RV_0=qC_u+(1-q)C_d$, proving the pricing formula. $\square$

The weight $q$ is called the one-period risk-neutral probability. It is the probability that makes the stock’s weighted expected terminal price equal to its risk-free growth: $qS_0u+(1-q)S_0d=RS_0$. The formula above is a consequence of replication and does not require that $q$ equal the physical probability $p$.

Worked Example: Replicating a Claim with Exact Cash Accounting

Let $S_0=80$, $u=\tfrac32$, $d=\tfrac34$, and $R=1$. A claim pays $C_u=18$ and $C_d=6$. First check the no-arbitrage condition: $0<\tfrac34<1<\tfrac32$. The terminal stock prices are $80\cdot\tfrac32=120$ and $80\cdot\tfrac34=60$.

The stock holding is $$ \Delta=\frac{18-6}{120-60}=\frac{12}{60}=\frac15. $$ The cash holding is $$ b=\frac{\tfrac32\cdot6-\tfrac34\cdot18}{1\cdot(\tfrac32-\tfrac34)} =\frac{9-\tfrac{27}{2}}{\tfrac34} =\frac{-\tfrac92}{\tfrac34}=-6. $$ Thus the initial cost is $\tfrac15\cdot80-6=16-6=10$.

Check: In the up outcome, the terminal portfolio value is $\tfrac15\cdot120-6=24-6=18$. In the down outcome, it is $\tfrac15\cdot60-6=12-6=6$. Both payoffs match. The risk-neutral probability is $q=(1-\tfrac34)/(\tfrac32-\tfrac34)=(\tfrac14)/(\tfrac34)=\tfrac13$, so the pricing formula gives $\bigl(\tfrac13\cdot18+\tfrac23\cdot6\bigr)/1=6+4=10$, the same initial cost.

Why the Return Condition Is Necessary

The strict inequalities $d<R<u$ are not merely convenient for the pricing formula. If they fail, a stock-and-cash portfolio produces an arbitrage. When $R$ is at least as large as $u$, investing in cash can dominate the stock in both outcomes; when $R$ is at most $d$, the stock can dominate cash. The strictness matters: at equality one outcome may have zero excess payoff, but the other still has a positive one.

Proposition: Under $S_0>0$, $R>0$, and $0<d<u$, the one-period market is arbitrage-free if and only if $d<R<u$.

Proof. The case $d<R<u$ was proved in the theorem. If $R\ge u$, take $\Delta=-1$ and $b=S_0$: short one share and invest $S_0$ in cash. Its initial cost is $-S_0+S_0=0$, while its terminal values are $S_0(R-u)\ge0$ and $S_0(R-d)>0$. This is an arbitrage. If $R\le d$, take $\Delta=1$ and $b=-S_0$: buy one share financed by borrowing $S_0$. Its initial cost is $S_0-S_0=0$, while its terminal values are $S_0(u-R)>0$ and $S_0(d-R)\ge0$. This is also an arbitrage. These cases cover every $R$ outside the strict interval, including both endpoints, proving the equivalence. $\square$

Worked Example: An Arbitrage When Cash Outgrows Both Stock Outcomes

Let $S_0=40$, $d=\tfrac12$, $u=\tfrac32$, and $R=2$. The stock ends at $40\cdot\tfrac32=60$ in the up outcome and $40\cdot\tfrac12=20$ in the down outcome. Since $R=2\ge\tfrac32=u$, short one share and invest $40$ in cash.

The initial cost is $-40+40=0$. The terminal value is $40\cdot2-60=20$ in the up outcome and $40\cdot2-20=60$ in the down outcome. Both values are nonnegative and both are strictly positive, so this is an arbitrage. The attempted risk-neutral probability would be $q=(2-\tfrac12)/(\tfrac32-\tfrac12)=\tfrac32$, which is not a probability; this failure reflects the arbitrage rather than a valid pricing measure.

Call-Option Bounds from Payoff Comparisons

Replication prices can also be checked by comparing payoffs state by state. If two payoffs are ordered in both outcomes, their replicating prices have the same order when $d<R<u$: the pricing formula uses positive weights $q$ and $1-q$. For a European call with strike $K\ge0$, the terminal payoff is $C_1=(S_1-K)^+$, where $x^+=\max(x,0)$. In each outcome, $C_1\ge0$, $C_1\ge S_1-K$, and $C_1\le S_1$. Applying the positive-weight pricing formula to these inequalities gives the standard bounds.

Corollary (One-period call bounds): If $d<R<u$, the no-arbitrage price $C_0$ of a call with strike $K\ge0$ satisfies $$ \max\!\left(S_0-\frac{K}{R},0\right)\le C_0\le S_0. $$

Proof. Since $C_1\ge0$, its price $C_0=[qC_u+(1-q)C_d]/R$ is nonnegative. Since $C_1\ge S_1-K$, the same formula gives $C_0\ge[qS_0u+(1-q)S_0d-K]/R=(RS_0-K)/R=S_0-K/R$. Together these yield the lower bound. Since $C_1\le S_1$, the formula gives $C_0\le[qS_0u+(1-q)S_0d]/R=S_0$. $\square$

Worked Example: Pricing a Call and Checking Its Bounds

Take $S_0=100$, $u=\tfrac32$, $d=\tfrac34$, $R=1$, and strike $K=90$. The condition $\tfrac34<1<\tfrac32$ holds. The terminal stock prices are $150$ and $75$, so the call pays $60$ in the up outcome and $0$ in the down outcome. The risk-neutral probability is $q=(1-\tfrac34)/(\tfrac32-\tfrac34)=\tfrac13$.

Replication gives $$ \Delta=\frac{60-0}{150-75}=\frac{60}{75}=\frac45,\qquad b=\frac{\tfrac32\cdot0-\tfrac34\cdot60}{1\cdot(\tfrac32-\tfrac34)} =\frac{-45}{\tfrac34}=-60. $$ The initial cost is $\tfrac45\cdot100-60=80-60=20$. The risk-neutral formula independently gives $(\tfrac13\cdot60+\tfrac23\cdot0)/1=20$.

Check: The replicating portfolio pays $\tfrac45\cdot150-60=120-60=60$ up and $\tfrac45\cdot75-60=60-60=0$ down. The call bounds are $\max(100-90/1,0)=10\le20\le100$. Thus the computed price lies within both bounds.

A Common Calculation Pitfall

Common mistake: In the first example, it is easy to multiply $\tfrac34\cdot18$ incorrectly as $\tfrac{27}{4}$. That gives the false cash numerator $9-\tfrac{27}{4}=\tfrac94$, and then the false holding $b=3$. Substitution exposes the error: with $\Delta=\tfrac15$ and $b=3$, the up-state portfolio value is $\tfrac15\cdot120+3=27$, not $18$, and the down-state value is $\tfrac15\cdot60+3=15$, not $6$. The correct multiplication is $\tfrac34\cdot18=\tfrac{54}{4}=\tfrac{27}{2}$, giving $b=(9-\tfrac{27}{2})/(\tfrac34)=-6$. Always substitute both holdings into both terminal payoff equations; this checks the arithmetic and the replication.

Replication is especially useful in this two-outcome, one-period setting because it determines a unique price without estimating the physical probability of an up move. The no-arbitrage condition ensures that the resulting pricing weights are positive. In later settings with multiple periods, the same idea is applied at successive dates rather than by solving just one pair of payoff equations.

Check Your Understanding

For each question, use the state-by-state payoff equations and check any proposed price by substitution.

For $S_0=50$, $u=2$, $d=\tfrac12$, and $R=1$, find $q$. Does the model satisfy the no-arbitrage condition?
A claim pays $12$ up and $4$ down in the market of Question 1. Find $\Delta$, $b$, and its initial price.
For a call with $S_0=80$, $K=70$, and $R=1$, what are the general no-arbitrage lower and upper bounds?
If $R=d$, explain why the stock-financed-by-borrowing portfolio in the proposition still gives an arbitrage.
Why does the pricing formula not require the physical up probability $p$ to equal $q$?