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Here is a simple fact: In your notation, the model under consideration is complete if and only if the matrix \begin{bmatrix} 1+R & 1+R \\ d & u \end{bmatrix} is one-to-one, i.e. $d \neq u$. (Equivalently, its transpose is onto, which is what is shown in your quoted text.) No-arbitrage holds if and only if $d \leq 1+R \leq u$ and $d < u$. (By ...


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