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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 $H \cdot P_1<0$.

tergarg
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  • 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. –  Aug 16 '18 at 09:10
  • @tergarg: What book has the definition you quoted? –  Aug 16 '18 at 09:27
  • @BCLC, I'll post the question there, thanks. – tergarg Aug 16 '18 at 10:28
  • @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. – tergarg Aug 16 '18 at 10:31
  • @quasi, I found it in the notes for Advanced Financial Models by Mike Tehranchi, a master's course at Cambridge Uni – tergarg Aug 16 '18 at 10:31
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    @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. –  Aug 16 '18 at 10:48
  • @tergarg: However, as I see it, your objection is valid. To fix the author's condition $(1)$, write it instead as just $H\cdot P_0 \le H\cdot P_1$ almost surely. –  Aug 16 '18 at 10:54
  • @quasi The point about "consume" is good, but the impression I got with consume is that we do not necessarily spend it because we maximize $E[U(c_0,c_1)]$, where $c_0=x-H \cdot P_0$, $c_1=H \cdot P_1$ and $x$ is initial wealth. In other words we care about maximizing the utility of this wealth, i.e what we can do with it rather than spend it on assets again. In his second example he doesn't spend all the money, i.e he arrives to $H \cdot P_0 = -1$. http://www.statslab.cam.ac.uk/~mike/AFM/notes.pdf – tergarg Aug 16 '18 at 11:15
  • @tergarg: But there's no harm in buying a constant asset with any excess funds from the transactions at time $0$, and that resolves the issue, doesn't it? –  Aug 16 '18 at 11:24
  • @quasi True, that's the only explanation thus far. Thanks – tergarg Aug 16 '18 at 11:29
  • @tergarg: Equivalently, you can loan out any excess funds at time $0$ at an interest rate of zero That mirrors the scenario where you have to borrow at time $0$. –  Aug 16 '18 at 11:31

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