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This question already has an answer here:

Definition. An arbitrage is a portfolio $H$ ∈ $R^n$ such that

• $H · P_0 ≤ 0 ≤ H · P_1$ almost surely, and

• $P(H · P_0 = 0 = H · P_1) < 1$.

where $P_0$ and $P_1$ $\in R^n$ represent the prices at time $t=0,1$ respectively.

Now, my question is why do we need the first condition. Suppose there is only one asset A which at time $t=0$ costs $3$. Then, at $t=1$ we have $P(A=2)=\frac{1}{2}$ and $P(A=1)=\frac{1}{2}$. The portfolio $H=-1$ should be an arbitrage because it yields certain profit with no risk attached but it isn't because $H \cdot P_1<0$.

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marked as duplicate by Bob Jansen Aug 16 '18 at 17:23

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  • $\begingroup$ I am not familiar with this definition of arbitrage. This is in a way similar to the way I see arbitrage: We have a portfolio at a time $t$: $V_t$ with $V_0=0$. With probability 1 (almost surely) we the portfolio $V_t \geq 0$. So we cannot lose money. Furthermore the probability of having a profit has to be non-zero: $P(V_t>0)>0$. Here it seems your second statement is essentially that the probability that the profit is $0$ is smaller than 1. The first statement ensures that our strategy is not losing money almost surely. $\endgroup$ – Jan Aug 16 '18 at 9:10
  • $\begingroup$ @tergarg: What book has the definition you quoted? $\endgroup$ – quasi Aug 16 '18 at 9:27
  • $\begingroup$ @BCLC, I'll post the question there, thanks. $\endgroup$ – tergarg Aug 16 '18 at 10:28
  • $\begingroup$ @Jan, I assume here $V_0$ is what I call $H \cdot P_0$. Both your definition and mine make intuitive sense, except that I do not see why the example that I gave is not an arbitrage when it is a sure way to make money. $\endgroup$ – tergarg Aug 16 '18 at 10:31
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    $\begingroup$ @tergarg: Note: The author requires you to "consume" the excess wealth at time $0$, so with the proceeds of the short sale in your example, you have to buy something. For simplicity, assume you can always buy a forever constant asset. Thus, in your example, if you throw in an asset with price forever constant at $1$, and if you use $H=(-1,3)$, conditions $(1)$ and $(2)$ are both satisfied. $\endgroup$ – quasi Aug 16 '18 at 10:48

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