Pricing Claims in Complete Markets

From Attainability to Price

In a complete market, every contingent terminal payoff can be produced by trading in the available assets. The resulting price is not an arbitrary choice: it is the initial cost of a replicating strategy. In a finite-state market with a unique risk-neutral measure, that same cost is a discounted conditional expectation. The conditional version matters: it gives the claim's value at each intermediate date, not just at time zero.

Work on a finite filtered probability space $(\Omega,\mathcal F,(\mathcal F_n)_{n=0}^N,\mathbb P)$, where each node at date $n$ has positive probability and at least one successor. The measure $\mathbb Q$ is equivalent to $\mathbb P$, so every node that has positive $\mathbb P$-probability also has positive $\mathbb Q$-probability. There are $d$ traded risky assets with prices $S^1_n,\ldots,S^d_n$, and a strictly positive bank account with value $A_n$. Assume assets have no intermediate payouts. Write $\widetilde S^j_n=S^j_n/A_n$ for discounted prices. A risk-neutral measure here means that each discounted traded price, including the bank account's constant discounted price, is a $\mathbb Q$-martingale. Since the model has finitely many states and dates, all the conditional expectations below are finite sums. No Brownian motion or stochastic integral is needed in this discrete-time setting.

Definition: A claim is a finite terminal payoff $X$ measurable with respect to $\mathcal F_N$. A strategy is a sequence of holdings in the traded assets, chosen using information available at each date. It is self-financing if changes in holdings are paid for entirely by the portfolio, with no additional deposits or withdrawals. The claim is attainable if a self-financing strategy has terminal value $X$ in every terminal state.

We include the bank account as asset $0$, with $S^0_n=A_n$. At a node at date $n$, let its $m$ successor nodes be indexed by $k=1,\ldots,m$, and let $p_{n+1,k}\in\mathbb R^{d+1}$ be the vector of asset prices at successor $k$. A portfolio with holdings $\theta\in\mathbb R^{d+1}$ has value $\theta\cdot p_{n+1,k}$ at that successor. Its current cost is $\theta\cdot p_n$, where $p_n$ is the current asset-price vector.

Definition: A node is one-step complete if every vector of real successor values $y=(y_1,\ldots,y_m)$ can be written as $y_k=\theta\cdot p_{n+1,k}$ for some portfolio $\theta$. Equivalently, the $m$-by-$(d+1)$ matrix whose $k$th row is $p_{n+1,k}$ has rank $m$.

The equivalence in this definition is the elementary linear-algebra criterion for the map from holdings $\theta$ to successor values to be onto $\mathbb R^m$. If every node is one-step complete, backward construction gives a strategy for any finite terminal claim. The converse also holds in a tree with unrestricted trading: if a node's successor-payoff matrix is not onto, there is a terminal claim that cannot be replicated. A proof, including the necessary discounting from a successor value to its terminal payoff, is given below.

The Pricing and Replication Theorem

Theorem: Suppose the finite-state market has a strictly positive bank account, a risk-neutral measure $\mathbb Q$, and a one-step complete successor-payoff matrix at every node. For any terminal claim $X$, its value at date $n$ is $$ V_n=A_n\,\mathbb E^{\mathbb Q}\!\left[\frac{X}{A_N}\,\middle|\,\mathcal F_n\right]. $$ There is a self-financing strategy whose value is $V_n$ at every date and whose terminal value is $X$. The replication cost and the discounted conditional-expectation value agree at every node.

Proof. Define $V_N=X$, and for $n=N-1,\ldots,0$ define $V_n=A_n\mathbb E^{\mathbb Q}[V_{n+1}/A_{n+1}\mid\mathcal F_n]$. The tower property of conditional expectation gives $V_n=A_n\mathbb E^{\mathbb Q}[X/A_N\mid\mathcal F_n]$. At a node at date $n$, the values $V_{n+1}$ at its successors form a finite vector. One-step completeness supplies holdings whose successor portfolio values equal these values. Choose those holdings at that node. Doing this at every node gives a predictable trading strategy, and its value at each successor is the prescribed $V_{n+1}$; at date $N$ its value is $X$.

It remains to check that the portfolio's current cost is $V_n$, rather than an undiscounted expectation of its next-date value. Let $\theta_n$ be the holdings chosen at the node. Conditional on that node, the $\mathbb Q$-martingale property of each discounted asset price gives $$ \theta_n\cdot p_n =A_n\,\mathbb E^{\mathbb Q}\!\left[\frac{\theta_n\cdot p_{n+1}}{A_{n+1}}\,\middle|\,\mathcal F_n\right] =A_n\,\mathbb E^{\mathbb Q}\!\left[\frac{V_{n+1}}{A_{n+1}}\,\middle|\,\mathcal F_n\right] =V_n. $$ The first equality uses that $\theta_n$ is fixed by the information at date $n$, and the second uses the chosen successor values. Thus the strategy costs exactly $V_n$ at each node, is self-financing as it is rebalanced, and replicates $X$. All variables are finite on a finite state space, so the conditional expectations are well-defined. $\square$

The same calculation gives a useful converse criterion. If a node's successor matrix is not onto, choose $y\in\mathbb R^m$ outside its column space. For each terminal descendant of successor $k$, define the payoff $X=y_k A_N/A_{n+1,k}$, where $A_{n+1,k}$ is the bank-account value at that successor. At that successor, the discounted conditional expectation of $X$ is $y_k/A_{n+1,k}$, so its value is $y_k$. Any strategy replicating $X$ would therefore need to produce the vector $y$ at the successors, which is impossible because $y$ is outside the matrix's column space. Hence completeness requires one-step completeness at every node.

Worked Examples

Worked Example: A One-Step Call and Its Replicating Portfolio

At date $0$, let $A_0=1$, $A_1=11/10$, and $S_0=10$. At date $1$, the stock is $14$ in the up state and $8$ in the down state. Let $\mathbb Q$ assign probability $1/2$ to each state, and take the call payoff $X=(S_1-10)^+$.

The discounted stock is a $\mathbb Q$-martingale because its expected date-1 value is $\frac12(14)+\frac12(8)=11$, and $11/(11/10)=10=S_0/A_0$. The call pays $4$ up and $0$ down. Its price is $$ V_0=\frac{\mathbb E^{\mathbb Q}[X]}{A_1} =\frac{\frac12(4)+\frac12(0)}{11/10} =\frac{2}{11/10}=\frac{20}{11}. $$ To replicate, let $\theta$ be stock units and $\phi$ bank-account units. Matching the two successor payoffs requires $14\theta+(11/10)\phi=4$ and $8\theta+(11/10)\phi=0$. Subtracting gives $6\theta=4$, hence $\theta=2/3$. The down-state equation gives $(11/10)\phi=-16/3$, so $\phi=-160/33$.

Check: In the up state, $14(2/3)+(11/10)(-160/33)=28/3-16/3=4$. In the down state, $8(2/3)+(11/10)(-160/33)=16/3-16/3=0$. The initial cost is $10(2/3)-160/33=220/33-160/33=20/11$, equal to the discounted conditional-expectation price.

Worked Example: Backward Pricing and Holdings in a Two-Period Tree

Set $A_n=(11/10)^n$ for $n=0,1,2$, $S_0=100$, and let the stock move by a factor $3/2$ up or $2/3$ down at each date. At each node the risk-neutral up probability $q$ solves $q(3/2)+(1-q)(2/3)=11/10$. Thus $q=(11/10-2/3)/(3/2-2/3)=(13/30)/(5/6)=13/25$, and the down probability is $12/25$. Both are strictly positive. Consider the date-2 call payoff $X=(S_2-100)^+$.

The terminal stock values in the order $UU,UD,DU,DD$ are $225,100,100,400/9$. The corresponding payoffs are $125,0,0,0$. At the date-1 up node, the conditional expected payoff is $(13/25)125=65$, so its value is $65/(11/10)=650/11$. At the date-1 down node, both possible terminal payoffs are zero, so its value is zero. Discounting those date-1 values to date 0 gives $$ V_0=\frac{(13/25)(650/11)+(12/25)(0)}{11/10} =\frac{338/11}{11/10} =\frac{3380}{121}. $$

At the date-1 up node, matching successor values $125$ and $0$, at stock prices $225$ and $100$, gives stock holding $(125-0)/(225-100)=1$. The bank-account holding is $-100/A_2=-10000/121$, since $A_2=121/100$. Its current value is $150+(11/10)(-10000/121)=150-11000/121=7150/121=650/11$. At the date-1 down node, zero holdings replicate the two zero payoffs.

At date 0, the stock holdings $\theta$ and bank-account holdings $\phi$ must produce $650/11$ in the up node and $0$ in the down node. Therefore $150\theta+(11/10)\phi=650/11$ and $(200/3)\theta+(11/10)\phi=0$. Subtracting gives $(250/3)\theta=650/11$, so $\theta=39/55$. The second equation gives $(11/10)\phi=-520/11$, hence $\phi=-5200/121$. The initial cost is $100(39/55)-5200/121=8580/121-5200/121=3380/121$.

Check: In the up node, the date-1 portfolio value is $150(39/55)+(11/10)(-5200/121)=1170/11-520/11=650/11$. In the down node it is $(200/3)(39/55)-520/11=520/11-520/11=0$. The date-1 up-node hedge then gives $225-100=125$ in $UU$ and $100-100=0$ in $UD$; the date-1 down-node hedge gives zero in both remaining states. Thus every terminal payoff is matched.

Worked Example: A Three-State Claim in a Complete One-Step Market

There are three states, with $\mathbb P$-probabilities $1/3,1/3,1/3$ and equivalent $\mathbb Q$-probabilities $1/4,1/2,1/4$. Let $A_0=1$ and $A_1=5/4$. In the three states, two risky assets have date-1 payoff vectors $(2,0,0)$ and $(0,2,0)$. Their date-0 prices, required by the discounted martingale condition, are respectively $((1/4)2)/(5/4)=2/5$ and $((1/2)2)/(5/4)=4/5$. Let the claim payoff be $(4,2,0)$.

The discounted conditional-expectation price is $$ V_0=\frac{(1/4)4+(1/2)2+(1/4)0}{5/4} =\frac{2}{5/4}=\frac85. $$ Two units of the first risky asset and one unit of the second produce the payoff $(4,2,0)$. Their cost is $2(2/5)+1(4/5)=8/5$, matching the valuation. The three successor-payoff vectors of the bank account and the two risky assets span $\mathbb R^3$, so this node is one-step complete.

Check: The portfolio's state payoffs are $2(2,0,0)+(0,2,0)=(4,2,0)$. Discounting each payoff by $5/4$ and taking the $\mathbb Q$-expectation gives $(1/4)(16/5)+(1/2)(8/5)+(1/4)(0)=4/5+4/5=8/5$, the same price.

Why the Discounted Conditional Expectation Matters

For a self-financing strategy with value $W_n$, the discounted value $W_n/A_n$ is a $\mathbb Q$-martingale when the discounted traded asset prices are $\mathbb Q$-martingales. Indeed, holdings chosen at date $n$ are $\mathcal F_n$-measurable, and the conditional expectation of each next-date discounted asset price equals its current discounted price. A finite-state model makes these variables integrable, so the conditional-expectation identity is a martingale identity, not merely a local-martingale assertion.

If $W_N=X$, iterating the martingale identity gives $W_n/A_n=\mathbb E^{\mathbb Q}[X/A_N\mid\mathcal F_n]$. This explains why replication and risk-neutral valuation agree. In a complete market, the replicating strategy determines the price; the unique equivalent martingale measure gives the same price formula. The physical measure $\mathbb P$ describes actual probabilities in the model, whereas $\mathbb Q$ is used for valuation. Replacing $\mathbb Q$ by $\mathbb P$ in the pricing formula is not generally justified.

Common mistake: forgetting to discount a next-date value. In the first example, the $\mathbb Q$-expected date-1 call payoff is $2$, but the date-0 cost is not $2$. The correct calculation is $2/(11/10)=20/11$. More generally, if $Y$ is the value at date $n+1$, the current value is $A_n\mathbb E^{\mathbb Q}[Y/A_{n+1}\mid\mathcal F_n]$, not $\mathbb E^{\mathbb Q}[Y\mid\mathcal F_n]$. Discounted values have conditional expectation equal to the current discounted value; undiscounted values do not generally have that property.

Check Your Understanding

Use the theorem and the node-by-node replication equations to check each answer.

In the one-step call example, what is the price if the date-1 bank-account value is changed to $6/5$, while the discounted stock prices still define an equivalent martingale measure?
For a self-financing strategy with terminal value $X$, write the conditional-expectation formula for its date-$n$ value.
In the three-state example, calculate the price of payoff $(0,0,4)$ under the stated $\mathbb Q$ and bank-account values.
Why does a successor-payoff matrix with rank less than the number of successors prevent one-step replication of every successor-value vector?
Explain why a terminal payoff used to prescribe successor value $y_k$ can be set to $X=y_kA_N/A_{n+1,k}$ on that successor's terminal descendants.

Pricing Claims in Complete Markets

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