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1

In general, if one can create a portfolio with the same payoff as the derivative, their prices must be equal. This is also called "Law of One Price". Here an excerpt from my script: Here EMM = Equivalent Martingale Measure (Q), NA = No-Arbitrage.


0

A market is arbitrage-free, if riskneutral measure Q exists (under which the discounted stockprice becomes martingale). A market is complete, when the riskneutral measure Q is unique. Therefore, any market with a riskneutral measure Q is arbitrage-free, and if Q is unique it is also complete. The riskneutral probabilities $q$ are unique for the binomial ...


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The only reason we are able to solve the system is because of $d<u$ which follows from $d<1+R<u$, which follows from absence of arbitrage by Prop 2.3


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You can view the arbitrage-free statement as being about infinitely liquid trading. If one has to trade $x$ by paying a half-spread $\nu$, and you have a trivial payoff $\Phi(\cdot) \equiv 1+R$ then there is no solution. You would be trying to solve $$ (1+R)x - \nu + suy = (1+R)x = (1+R)x - \nu + sdy $$


1

Is the one in red supposed to be the proof of the Pricing Principle 1? Or merely an intuitive explanation? It is not a proof. The explanation/reasoning in this paragraph lets the author state the pricing principle. It has hints on how to prove Prop 2.9 (for instance, see the line ...no difference between holding the claim and the portfolio...). If ...


1

I believe this is the way that Björk proposes, however I believe "my" way below is more elegant. The trick in Björk's case is to realize that in each "iteration" we get an expeted value: $$V_0(0) = \frac{1}{1+R}\left(q_u V_{1}(1) + q_d V_1(0) \right) \\ = \frac{1}{(1+R)^{2}} \left( q_u^2 V_2(2) + 2q_uq_d V_2(1) + q_d V_2(0) \right) \\ = ...


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"the first is an arbitrage portfolio, whereas the second is an arbitrage possibility, these two things do not define the same thing. – Ask Question 2 days ago"


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I think you are seeking a pragmatic definition of arbitrage instead of a theoretical definition. For practical definitions, there are two kinds of arbitrage: statistical arbitrage and deterministic arbitrage. Suppose you have a lottery with 10 identical tickets only. Each ticket sells for \$10. The single, grand prize of the lottery is \$1000. Hence, the ...


0

They are definitions, so it makes no sense in asking which one is correct. However, the second one is the one that makes most sense, and it is the one you will see in most literature.


2

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.


1

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 ...


1

Excluding trade costs (which is a big assumption), you would need to consider margin.. http://www.finra.org/Investors/smartInvesting/AdvancedInvesting/MarginInformation/p005922 Assuming 1MM Short with 25% margin, you would need 250K in a margin account. This would be marked to market and you might get a margin call if your shorts climb.


-5

An arbitrage strategy is by definition to start with 0 net investment (and riskfree profit at the end). In theory, arbitrage strategies cannot exist.



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