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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!

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1 Answer 1

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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.

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