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In Tomas Bjork's Arbitrage Theory in Continuous Time (or here), $\exists$ what seems to be 2 inconsistent definitions of arbitrage:

The first definition is for the single period Binomial model enter image description here

The second definition is for the multi period Binomial model enter image description here

The second suggests that there is a possibility of the portfolio value ending up zero while the first does not...

...Why?

Edit: Oh, I forgot to mention: My prof uses the latter definition to replace the first definition for the one-period. E said something about different conditions or something. (I'll ask about it during next consultation hours.)

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  • $\begingroup$ Thank you @DoubleTrouble! I can't believe I forgot to mention that. $\endgroup$ – BCLC Jul 4 '14 at 10:17
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    $\begingroup$ the first is an arbitrage portfolio, whereas the second is an arbitrage possibility, these two things do not define the same thing. $\endgroup$ – user1157 Jul 4 '14 at 14:29
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    $\begingroup$ @AskQuestion THANK YOU! I CAN'T BELIEVE I DIDN'T SEE THAT HAHAHAHAHA $\endgroup$ – BCLC Jul 6 '14 at 22:58
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My initial answer was incorrect, I was thinking to quickly (or slowly!?)

I agree with you that these two definitions are not consistent. The first definition is much more strict since it does not allow for any outcome $\omega \in \Omega = \{\omega_1, \omega_2\}$ such that $V_1^h(\omega)=0$. We only have 2 outcomes since we are considering the single period Binomial model.

As a side note, here are three equivalent definitions of an arbitrage portfolio $h$ (same notation as in Björk).

  1. $V_h^0 = 0$, $V_h^1 \geq 0$, and $\ E[V_h^1]>0.$
  2. $V_h^0 = 0$, $P(V_h^1 \geq 0)=1$, and $\ P(V_h^1 > 0)>0.$
  3. $V_h^0 = 0$, $V_h^1 \geq 0$, and $\ V_h^1 \neq 0$
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  • $\begingroup$ I don't get it. Why are there only 2 sample outcomes in Def 2.2? (Is it related to the fact that there are 2 outcomes in the one-period binomial model?) Are the definitions inconsistent are not? Oh, I forgot to mention: My prof uses the latter definition to replace the first definition for the one-period. E said something about different conditions or something. (I'll ask about it during next consultation hours.) $\endgroup$ – BCLC Jul 4 '14 at 10:22
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    $\begingroup$ Hello BCLC, I corrected my answer since the first one was incorrect. $\endgroup$ – DoubleTrouble Jul 4 '14 at 11:25
  • $\begingroup$ DoubleTrouble, isn't the second one supposed to be greater than 0? Thanks! $\endgroup$ – BCLC Jul 4 '14 at 11:32
  • $\begingroup$ Ah also, which is right then? Haha quant.stackexchange.com/questions/12917/… $\endgroup$ – BCLC Jul 4 '14 at 11:36
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    $\begingroup$ @BCLC: Yes, it should be >0. Thanks for pointing that out! Since they are definitions they cannot be right or wrong. However the second one makes more sense, and is the one most commonly used in the literature. $\endgroup$ – DoubleTrouble Jul 4 '14 at 11:46
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I agree with the question and not with the answer. Definition (2.2) means that $\omega$ for which $V^h_1(ω )=0$ is such that $P(\omega) = 0$, ie an event with measure (probability) 0.

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  • $\begingroup$ Sooooo...Def (2.15) is wrong...? Oh, I forgot to mention: My prof uses the latter definition to replace the first definition for the one-period. E said something about different conditions or something. (I'll ask about it during next consultation hours.) $\endgroup$ – BCLC Jul 4 '14 at 10:18
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    $\begingroup$ @KaapstadKwant: Thanks for pointing out my error! $\endgroup$ – DoubleTrouble Jul 4 '14 at 11:26
  • $\begingroup$ It's a pleasure! One can easily make that type of mistake when working with outcomes. $\endgroup$ – KaapstadKwant Jul 7 '14 at 6:58

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