# Two definitions of arbitrage in finite markets

I have read two definitions of the term an arbitrage opportunity in the literature*. Are they equivalent?

Consider a single period market model over the measurable space $\Omega = \{\omega_1, \dots, \omega_M\}$, comprising $n + 1$ assets $S^0, S^1, \dots, S^n$, of which $S^0$ is the risk-free asset with risk-free interest rate $R \geq 0$. A portfolio is an $n + 1$ tuple $(x_0, x_1, \dots, x_n) \in \mathbb{R}^{n+1}$.

Definition 1 A portfolio $(x_0, \dots, x_n)$ is an arbitrage opportunity iff

1. $x_0 S^0_0 + \cdots + x_n S^n_0 = 0$,
2. $x_0 S^0_1 + \cdots + x_n S^n_1 \geq 0$ for all $\omega \in \Omega$,
3. $x_0 S^0_1 + \cdots + x_n S^n_1 > 0$ for some $\omega \in \Omega$.

Definition 2 A portfolio $(x_0, \dots, x_n)$ is an arbitrage opportunity iff

$$x_1 (\frac{1}{1 + R} S^1_1 - S^1_0) + \cdots x_n (\frac{1}{1 + R} S^n_1 - S^n_0) \geq 0$$ for all $\omega \in \Omega$ with strict inequality holding for at least one $\omega \in \Omega$.

* Definition 1 is from Capiński & Kopp's "Discrete Models of Financial Markets" (Cambridge University Press 2012), whereas definition 2 is from Roman's "Introduction to the Mathematics of Finance: Arbitrage and Option Pricing", 2nd edition (Springer 2012).

They are equivalent. From Definition 1, note that $S_1^0 = S_0^0(1+R)$. Then \begin{align*} x_0 S_1^0 + \cdots x_n S_1^n &= x_0 S_0^0 (1+R)+ x_1 S_1^1 + \cdots x_n S_1^n\\ &= (-x_1 S_0^1 - \cdots -x_n S_0^n) (1+R)+ x_1 S_1^1 + \cdots x_n S_1^n\\ &= x_1 \left(S_1^1 - S_0^1(1+R) \right) + \cdots + x_n \left(S_1^n - S_0^n(1+R) \right)\\ &=(1+R)\bigg[x_1\Big(\frac{S_1^1}{1+R} - S_0^1 \Big) + \cdots + x_n\Big(\frac{S_1^n}{1+R} - S_0^n \Big) \bigg]. \end{align*} That is, definition 1 implies definition 2.
On the other hand, assume that definition 2 holds. Then, let $S_0^0 = 1$ and \begin{align*} x_0 = -x_1 S_0^1 - \cdots - x_n S_0^n. \end{align*} It is easy to check that definition 1 holds, by noting that $S_1^0 = 1+R$.
• why are you multiplying by $(1+R)$ all that terms on the second step? why is there a minus sign on $x_1S_0^1$? May 16 '16 at 16:19
• @simmy: It is based on the identity $x_0S_0^0 = -x_1 S_0^1 - \cdots -x_n S_0^n$ May 16 '16 at 17:13
• and the term $(1+R)$ on the last line that multiplies the rest?? May 16 '16 at 19:06
Short answer: They seem not equivalent. On the first definition, the risk-free rate is part of the portfolio, and enters with quantity $x_0$. In the second definition, they discount the $S_1^n$ quantities by the risk free rate, implying that you have an arbitrage only if the gain you get on your portfolio (at time 1), is greater than the gain you could have had by investing only in the risk free.