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I will use this theorem 3.2 from the book "Quantitative modeling of Derivative Securities" by Marco Avellandea:

Theorem 3.2 - Assume that there is no arbitrage, i.e. there exists a risk neutral probability $\pi$. Then, the market is complete if and only if there is a unique risk-neutral probability, i.e. the system of linear equation $$p = D\pi$$ has a unique solution where $D$ is are the different future scenarios in the market.

Show that in any model if there are two distinct risk-neutral probabilities, then there are infinite number of them.

Attempted solution: Consider a one-period binomial model:

State 1: $S_0 u$ with risk-neutral probability $\pi_u$

State 2: $S_0 l$ with risk-neutral probability $\pi_l$

Assume $\pi_u,\pi_l$ are two distinct risk-neutral probabilities, i.e., $\pi_u + \pi_l \neq 1$. Then, this violates theorem 3.2, the market is not complete since there cannot exist a unique risk-neutral probability, i.e. the system of linear equations $$p = D\pi$$ does not have a unique solution. Thus if there is not a unique solution then there are an infinite number under our example.

I am not sure if this suffices, any suggestions is greatly appreciated.

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  • $\begingroup$ Two risk neutral probabilities means $p=D\pi_1=D\pi_2$ i.e $\pi$ is not unique. $\endgroup$ – Alex C Feb 2 '16 at 2:55
  • $\begingroup$ I see, so then if it is not unique then there is infinitely many of them correct? $\endgroup$ – Wolfy Feb 2 '16 at 2:58
  • $\begingroup$ If $p=D\pi_1=D\pi_2$, then $p=D\big[\lambda \pi_1 + (1-\lambda)\pi_2\big]$, for any $0 < \lambda <1$; that is, there are infinitely many risk-neutral probabilities. $\endgroup$ – Gordon Feb 4 '16 at 19:24
  • $\begingroup$ @Gordon Hey could you comment on my question that I just posted you seem to be the only one that understands this stuff $\endgroup$ – Wolfy Feb 5 '16 at 20:05
  • $\begingroup$ @MorganWeiss: Many people here are able to comment on, depending on their timing and chances. $\endgroup$ – Gordon Feb 5 '16 at 20:09

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