Finding a One-Period Binomial Arbitrage by Comparing Asset Payoffs
Summary
The document shows how to construct an arbitrage portfolio in a one-period binomial market when an asset’s price lies outside the no-arbitrage bounds implied by its current price and possible future payoffs. In the example, with a zero-rate bank account, an asset priced at 10 pays either 5 or 9. Selling one unit and investing the proceeds in the bank account produces a zero-cost position with a positive payoff in either state.
The example clarifies that finding an arbitrage means selecting portfolio holdings whose initial cost is zero and whose terminal value is never negative, with a strictly positive value in at least one state. It gives the general condition that an asset’s down-state payoff must not exceed its current value grown at the risk-free rate, which in turn must not exceed its up-state payoff. The construction applies to the stated model; transaction costs, short-sale constraints, and more complex market assumptions are not discussed.
Key ideas
- An arbitrage portfolio has zero initial cost and a nonnegative payoff in every state, with a positive payoff in at least one state.
- When an asset’s possible payoffs violate the binomial no-arbitrage bounds, a trade can lock in a gain.
- In the example, shorting the mispriced asset and investing the proceeds in the bank account yields positive payoffs in both states.
- The payoff bounds must bracket the asset’s current price compounded at the risk-free rate.
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# How to find arbitrage in a one period binomial model? # How to find arbitrage in a one period binomial model? Here is the setup: one period binomial model The market has three assets: bank account $B$, asset $S$, and asset $K$ $r = 0$, so payoff for asset invested at bank at any $t$ is $1$ $S_0 = 100, S_u = 120$ and $S_d = 80.$ $K_0 = 10, K_u = 15$ and $K_d = 5$ Lets say $K_u$ is different and I want to find a type 1 (type A) arbitrage. How can I do this? I know that there are certain conditions for type 1, $V_0=0, p(V_t \ge 0) = 1$ and $p(V_T > 0)$ So $V_T = a + b K_T + c S_T$ $V_0 = a + b K_0 + c S_0 = a + 10b + 100c$ So what does it mean to "find" a type 1 arbitrage? Does it mean to construct a portfolio? So am I finding the values of $a,b,c$? I can let $V_0 = 0$, so this is one equation and I guess I need $V_T$ (up and down case) to be $ V_T \ge 0$? But in this case my up case for K_T is some deterministic $x \ne 15$. How do I formulate this problem? Does this approach make sense? Am I understanding the problem? ## Answer by Andrea (score 0) https://quant.stackexchange.com/a/81110 If $K_u$ is lower that 10 (let's say 9), you have an arbitrage (not sure about the type convention). And from one arbitrage, you can create infinite others. At 0, sell 1 unit of K and buy 10 units of B (assuming $B_0=1$, but its value is irrelevant), total cost of this is 0. At time 1, $10B$ are worth 10 and $-1K$ is worth either $-5$ or $-9$. So the PV of your portfolio is either $1$ or $5$. So, you paid nothing and got $1$ or $5$ after one period: arbitrage! The absence of arbitrage in a binomial model (any dimension) is $A_d \le A_0 (1+r) \le A_u$ for every asset $A$.
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