State-Price Bounds and Arbitrage in a Two-Asset Market

Summary

This exercise asks whether a market with two assets and three states admits arbitrage, and how to bound the price of a risk-free asset. State prices assign a price to receiving a unit payoff in each state; multiplying the payoff matrix by the state-price vector gives the asset prices. The proposed solution uses the corrected price ordering from the instructor to solve for a family of state-price vectors, parameterized by one component.

For the stated interval of that parameter, all state prices are positive, which supports the conclusion that the corrected market has no arbitrage under the usual state-price criterion. Summing the state prices gives the range for the risk-free asset’s price. The source notes that the original problem’s asset prices were transposed, so its conclusion applies to the corrected version. It gives the result but little derivation, and the bounds depend on the specified payoff matrix and corrected prices.

Key ideas

• State prices reproduce asset prices when multiplied by the assets’ state-contingent payoffs.
• A positive state-price vector is consistent with no arbitrage in this finite-state setup.
• The risk-free asset price is the sum of state prices when it pays one unit in every state.
• The stated no-arbitrage conclusion and price range rely on correcting the original asset-price ordering.

Tags

# Valuation functional


# Valuation functional












> Consider an economy with $J = 2$ assets and $S = 3$ states. The $J\times S$ payoff matrix for the two assets is $$X = \begin{pmatrix} 0 & 3 & 3\\ 1 & 1 & 0\\ \end{pmatrix}$$ and the asset prices $P' = (5/9,2)$. Determine whether there are arbitrage opportunities in this market. In addition, find the minimum and maximum prices for the risk free asset.

Attempted solution: We have $$P = X q = \begin{pmatrix} 0 & 3 & 3\\ 1 & 1 & 0\\ \end{pmatrix} \begin{pmatrix} q_1\\ q_2\\ q_3 \end{pmatrix} = \begin{pmatrix} 3q_2 + 3 q_3\\ q_1 + q_2 \end{pmatrix}$$ I believe we need to fix $q_2$ but the solution states that $q = (5/9 - q_2,q_2,6/9 - q_2)$. For $q_2\in (0,5/9)$ all $q$'s are positive so there is no arbitrage. $q_f \in (6/9,11/9)$.

I am not sure how we arrive at this solution any suggestions are greatly appreciated.

## Answer by Wolfy (score 2, accepted)

https://quant.stackexchange.com/a/33236

There was an error in my professors question he changed $$P = \begin{pmatrix} 2\\ 5/9 \end{pmatrix}$$ Thus when we fix $q_2$ we get $$q = (5/9 - q_2,q_2,6/9 - q_2)$$ Thus for $q_2\in (0,5/9)$ all $q$'s are positive so there is no arbitrage. Then clearly $q_f = (6/9,11/9)$

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.