New answers tagged binomial-tree
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 ...
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
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 $$ (1+R)x - \nu + suy = (1+R)x = (1+R)x - \nu + sdy $$
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