No-Arbitrage Condition in a Two-State One-Period Market
Summary
The document considers a one-period market with a single risky asset whose terminal price takes one of two positive multiples of its initial price, each with positive probability. With discounted and undiscounted prices treated as equal, the no-arbitrage condition is that the two possible gross price factors straddle one. If both outcomes are at or below the initial price, shorting the asset yields a nonnegative payoff and a positive payoff in at least one state. If both are at or above it, buying the asset gives the corresponding arbitrage argument.
The response establishes the direction that failure to straddle one permits arbitrage, including equality at a boundary, and calls this the converse of the requested implication. It does not fully spell out why straddling one rules out every admissible strategy, nor does it supply the requested example with a nontrivial initial information sigma-algebra. Thus it offers a useful construction but leaves parts of the original exercise unresolved.
Key ideas
- In the one-period two-state setup, no arbitrage requires one terminal price factor below one and the other above one.
- When both possible prices are no greater than the initial price, shorting the asset gives a candidate arbitrage.
- When both possible prices are no less than the initial price, buying the asset gives a candidate arbitrage.
- The response does not prove the sufficiency direction or provide an example involving nontrivial initial information.
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Full text
# Showing a basic market admits no arbitrage
# Showing a basic market admits no arbitrage
I'm learning the fundamentals of financial mathematics and came across the following problem I cannot solve
#### Setting
We work in $\left(\Omega, \mathcal{F},\left(\mathcal{F}_t\right)_{t=0}^1, \mathbb{P}\right)$. Let $d=1, T=1$ and assume the discounted price equals the non-discounted price.
Take $S_0^1 \in \mathbb{R}_{+}$, and $S_1^1 \in\left\{\alpha S_0^1, \beta S_0^1\right\}$ each with positive probability s.t. $0<\alpha<\beta$.
#### Task
I want to show that $\alpha<1<\beta $ iff there is no arbitrage. Additionally I'd like to find an example which shows that if $\mathcal{F}_0$ is not the trivial $\sigma$-algebra, then there exists an arbitrage.
#### Attempt
I know that for there to be an arbitrage I'd need to find $H_1$ s.t. $$ \mathbb{P}\left(H_1 \cdot \left(S_1^1-S_0^1\right) \geq 0\right)=1 \text { and } \mathbb{P}\left(H_1 \cdot \left(S_1^1-S_0^1\right)>0\right)>0 . $$ but I don't know what to base the proof on besides that. I would be grateful for any help!
## Answer by msantama (score 0, accepted)
https://quant.stackexchange.com/a/79192
I will take "the discounted price equals the non-discounted price" to mean the interest rate is zero.
Suppose $0 < \alpha < \beta \leq 1$ and consider the strategy of shorting one share for $S_0^1$ at $t = 0$. At $t = 1$, we can buy back the share for a price $S_1^1$. Observe our profit is non-negative with probability one and strictly positive with non-negative probability. A similar argument can be used to construct an arbitrage if $1 \leq \alpha < \beta$.
This proves if there is no arbitrage, then $\alpha < 1 < \beta$. It technically proves the converse.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.