Exercise: Why Pricing Under $\mathbb{P}$ Admits Arbitrage

Prerequisites: Risk-Neutral Measure, Binomial Tree Model

Problem

Reconsider the one-period binomial example from the lesson: $S_0 = 100$, up to $110$ or down to $90$, $r = 0$, call strike $K = 100$. The risk-neutral probability is $q = 0.5$.

Suppose a naïve trader uses $\mathbb{P}(\text{up}) = p = 0.7$ and prices the call at its "real-world expected payoff," getting $C^{\mathbb{P}} = 0.7 \cdot 10 + 0.3 \cdot 0 = 7$.

1. Construct an explicit arbitrage strategy: exhibit a self-financing portfolio that takes advantage of the mispricing. Your answer should specify the initial positions, the value at time $T$ in each state, and verify that the initial cost is strictly negative (or zero with positive payoff in at least one state).
2. Compute the profit per contract in each state ($\omega = \text{up}$ or $\text{down}$).
3. Explain in 2–3 sentences why the real-world probability $p = 0.7$ is irrelevant to the no-arbitrage price. What is the only information from the market that enters the price?
4. Now consider an incomplete-market variant: suppose the stock can move to $110$, $100$, or $90$ (three states) with $\mathbb{P}$-probabilities $(0.5, 0.2, 0.3)$. Argue that the no-arbitrage price of the call is not uniquely determined — exhibit two different risk-neutral probability triples $(q_u, q_m, q_d)$ that both make the discounted stock a martingale and produce different call prices.

Hint

For part 1, set up the replicating portfolio $(\phi, \psi)$ (stock holdings $\phi$, bond holdings $\psi$) that matches the call payoff in both states. Then buy the replicating portfolio and sell the mispriced call. For part 4, the martingale condition is one linear equation in three unknowns — a one-parameter family of solutions.