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The holder chooses at a specified time between a European call option and a European put option with the same later maturity and strike. Put-call parity makes its value equal to a call of the original maturity plus a put of the choice-time maturity with appropriately discounted strike.

The difference of the terminal European call option and European put option payoffs is $S_T^1-K$. Replicating that difference by one share and a borrowing position gives conditional put-call parityThus the call is more valuable exactly when the right side is positive. With a put selected at equality, the chooser option payoff isAt equality both choices have the same price, even though their eventual payoffs can differ. Put-call parity assumes no dividends or other intermediate stock cash flows, as in the stated asset-price model.

Put $K^{\prime }=K(1+r{)}^{t_0-T}$. At the choice time,Buy a European call option of strike $K$ and maturity $T$, together with a European put option of strike $K^{\prime }$ and maturity $t_0$. If the call is selected, the short-maturity put expires worthless. If the put is selected, sell the call and combine its value with the short-maturity put's cash payment; put-call parity makes this exactly enough to buy the maturity-$T$ put. This self-financing replication provesIt requires tradability of the priced options at the choice time, but not completeness of every other market claim. The identity also covers $t_0=0$ and $t_0=T$.

Write $q_{15},q_{12},q_9$ for the state probabilities under an equivalent martingale measure. Since cash is constant and the European put option pays only in the lowest state, the pricing equations areTheir unique solution is $q_9={\xi }_0/2$, $q_{15}={\xi }_0/2-2/3$, $q_{12}=5/3-{\xi }_0$. Every physical state has positive probability, so equivalence requires all three values to be strictly positive. ThusFor each price in this open interval the pricing kernel takes values $3q_{15},3q_{12},3q_9$ in the three equally likely states, proving absence of arbitrage.

The necessity, including the exclusion of endpoints, can also be checked directly. The three terminal payoff vectors of cash, the stock and the European put option form the invertible matrixEach unit state payoff therefore has a replicating strategy, whose initial cost is the corresponding $q$. A zero or negative $q$ supplies an arbitrage, so the endpoints are genuinely excluded.

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 $(4,1,0)$.

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.