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On a finite state space, the Arrow–Debreu state price is the current price of a payoff equal to one in state $s$ and zero elsewhere. It equals the physical state probability times the state-price density in that state. With a riskless asset of deterministic prices $B_0,B_1$, the Arrow state prices sum to $B_0/B_1$. Normalizing them gives the risk-neutral probabilities, while division by the physical probabilities instead recovers the pricing density.

Two assets with linearly independent payoff vectors across two positive-probability states span all terminal payoffs. The state-price equations have a unique solution. A positive solution gives the state-price density after division by physical state probabilities. Arrow state prices sum to the riskless discount factor, while normalized Arrow state prices are risk-neutral probabilities; these three quantities must not be confused.

Let $z_u,z_d$ be the state-price-density values at the higher and lower risky payoff. Pricing the two assets givesSolving yieldsBoth values are positive, and the two independent equations make the solution unique. The payoff matrix has nonzero determinant, so this is a complete two-state market. The actual Arrow state prices, including physical probabilities, are $q_u=5/9$ and $q_d=5/18$. Their sum is $5/6$, the discount factor. Normalizing gives risk-neutral probabilities $2/3,1/3$. Thus $Z$ itself is not a probability density of mean one: its mean is $5/6$ because the riskless asset earns interest.

Let $b$ and $h$ be the numbers of units of the numéraire and the stock in a replicating strategy. The two terminal payoffs of the European call option are $2$ and $0$, soSolving these simultaneous equations gives $b=-6/35$ and $h=8/35$. The initial replication cost isAs an independent pricing check, let $q$ be the risk-neutral probability of the state $(15,20)$ under the numéraire measure. The discounted stock must have initial value one and terminal mean one:Risk-neutral valuation then gives $10(3/7)(2/15)=4/7$. The physical probabilities $1/2,1/2$ are not the numéraire-measure probabilities.