# Show discrete market is arbitrage free if and only if there exist no admissible arbitrage portfolios

Problem: Let S be a discrete market. Show S is arbitrage free if and only if there exist no admissible arbitrage portfolios.

1. Definition of Discrete Market: Let $$T$$ be a positive real number and $$N$$ be a positive integer. Let $$0=t_0. Let $$(\Omega, \mathscr{F}, P)$$ be a probability space, where $$\Omega$$ is finite, $$\mathscr{F}=2^{\Omega}$$, and $$P\{\omega\}>0$$ for every $$\omega \in \Omega$$. A discrete market is a $$(\boldsymbol{d}+1)$$-dimensional stochastic process $$S=\left\{\left(S_n^{(0)}, S_n^{(1)}, \ldots, S_n^{(d)}\right): n=0,1, \ldots, N\right\}$$ defined on the probability space $$(\Omega, \mathscr{F}, P)$$. The first entry $$S_n^{(0)}$$ is the price at time $$t_n$$ of a riskless asset and is deterministic, that is, $$S_0^{(0)}=1, \quad S_n^{(0)}=S_{n-1}^{(0)}\left(1+r_n\right), \quad n=1,2, \ldots, N,$$ where $$r_n>-1$$ is the risk-free interest rate in the $$n$$th period $$\left[t_{n-1}, t_n\right]$$.

For $$j \geq 1$$, the $$j$$ th entry $$S_n^{(j)}$$ is the price at time $$t_n$$ of the $$j$$ th risky asset and follows the following stochastic dynamics: $$S_0^{(j)}>0, \quad S_n^{(j)}=S_{n-1}^{(j)}\left(1+\mu_n^{(j)}\right), \quad n=1,2, \ldots, N$$ Here $$S_0^{(j)}$$ is a real number and $$\mu_n^{(j)}>-1$$ is a real random variable. The amount of information available in the market at time $$t_n$$ is represented by the $$\sigma$$-algebra $$\mathscr{F}_n$$ given by $$\mathscr{F}_0=\{\emptyset, \Omega\}$$ and $$\mathscr{F}_n=\sigma\left(\left(\mu_i^{(1)}, \mu_i^{(2)}, \ldots, \mu_i^{(d)}\right): i=1,2, \ldots, n\right), \quad n=1,2, \ldots, N .$$ We assume that $$\mathscr{F}_N=\mathscr{F}$$.

2. Definition of Arbitrage: Let $$\mathcal{A}$$ be the family of all self-financing, predictable portfolios of the discrete market. A portfolio $$\alpha \in \mathcal{A}$$ is called an arbitrage if $$V_0(\alpha)=$$ 0 and there exists $$n \geq 1$$ such that $$P\left\{V_n(\alpha) \geq 0\right\}=1$$ and $$P\left\{V_n(\alpha)>0\right\}>0$$. The discrete market is called arbitrage-free if $$\mathcal{A}$$ does not contain arbitrage portfolios.

3. Definition of Admissible Portfolio: A portfolio $$\alpha$$ is called admissible if for every $$n$$ in $$\{1,2, \ldots, N\}$$, we have $$P\left\{V_n(\alpha) \geq 0\right\}=1$$

4. Definition of Portfolio: A portfolio or strategy is a $$(d+1)$$-dimensional stochastic process $$\alpha=\left\{\left(\alpha_n^{(0)}, \alpha_n^{(1)}, \ldots, \alpha_n^{(d)}\right): n=0,1, \ldots, N\right\} .$$ The value of the portfolio $$\alpha$$ at time $$t_n$$ is $$V_n(\alpha)=\sum_{j=0}^d \alpha_n^{(j)} S_n^{(j)}$$

5. Self financing portfolio definition: A portfolio $$\alpha$$ is self-financing if for every n in $$\{1,2,...,N\}$$, we have $$V_{n-1}(\alpha)=\sum_{j=0}^d \alpha_n^{(j)} S_{n-1}^{(j)}$$

Question: For the if part of the problem, I was told to use method of contradiction, i.e. find a admissible arbitrage portfolio and lead to contradiction. Based on definitions, I haven't figured out how to construct such admissible arbitrage portfolio. I checked https://math.stackexchange.com/questions/2998811/show-that-the-market-is-arbitrage-free-if-a-s-011r-b and have not figured out how to apply it here. Thanks in advance.

Some attempt: I am trying to prove by contradiction that if S is arbitrage free, then there is no admissible portfolio. I am thinking about defining portfolio value $$\alpha$$ as following piecewise function:

$$V_n=0$$ when discrete market S is defined on $$\Omega \setminus F$$, where $$F \subset \Omega$$

$$V_n=0$$ when discrete market S is defined on $$F$$ when $$n for $$n=\{1,2,...,N\}$$

$$V_n=1$$ when discrete market S is defined on $$F$$ when $$n=N$$. This way it is an admissible arbitrage portfolio but it violates assumption that S is arbitrage free, i.e. there exists no $$n \geq 1$$ so that $$P(V_n(\alpha) \geq 0\}=1$$ and $$P\{V_n(\alpha)>0\}>0$$.Not sure if this approach is general or not.

• What is the definition of "$S$ is arbitrage free"? I would think that the definition of "arbitrage free" would be "there exist no admissable arbitrage portfolios." Sep 20, 2023 at 14:45
• @user6247850 I think that definition is near the end of point 2 Sep 20, 2023 at 14:59
• @neveryield Unless I've misunderstood, the two things you'd have to prove are (1) if there is no admissible arbitrage portfolio (AAP), then there is no arbitrage portfolio (AP), and (2) if there is no AP, then there is no AAP. It sounds like you are having trouble with (1), and your approach is to say "suppose there was an AAP but no AP" -- in which case you could show that this AAP also fits the definition of an AP to find your contradiction. Have I completely misunderstood? Sep 20, 2023 at 15:33
• @Rylan no, you are right that's what i am trying to show if S is arbitrage free, then there exists no admissible arbitrage portfolios by method of contradiction. Sep 20, 2023 at 23:49
• @neveryield OK, so for that, using the definition of arbitrage free we want to show that if there are no arbitrage portfolios then there are no admissible arbitrage portfolios. If you want to show it by contradiction, you could assume there’s no arbitrage portfolios but there is an admissible arbitrage portfolios. I hope restating the setup in this way helps you see how we can prove that this is indeed a contradiction Sep 21, 2023 at 6:56