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Use discrete dates $j=0,\dots ,N$, a bank account $B_j=R^j$ with $R=1+r>0$, and a positive stock with each one-period gross return either $u$ or $d$, where $0<d<u$. On the natural tree assume both moves have positive physical probability. The Cox--Ross--Rubinstein model is arbitrage-free precisely whenOtherwise a stock/bank position gives nonnegative profit in both states and positive profit in at least one. Under the condition, the unique equivalent martingale measure has independent upward moves with conditional probabilityIndeed $qu+(1-q)d=R$, making the discounted stock a martingale. Both branches have positive probability under this measure, ensuring equivalence. The uniqueness/completeness claim is for the stock-generated binomial tree; an enlarged filtration carrying additional untraded risk would change that claim.

In the one-period building block of the Cox--Ross--Rubinstein model, the money-market account grows by a deterministic factor $R>0$, while the stock grows independently at each time byThus $S_t^0=R^t{S}_0^0$ and $S_t^1=S_0^1{\prod }_{j=1}^t{Y}_j$. Equivalently, the discounted stock has multipliers $u=U/R$ and $d=D/R$.

The model has no arbitrage precisely when the risk-free return lies strictly between the stock returns:Indeed, failure of either strict inequality gives a one-period arbitrage, while under the strict inequalities the probabilitylies in $(0,1)$. Taking the moves independent under $Q$ with up probability $q$ makes the discounted stock a martingale, since $qu+(1-q)d=1$. This is the unique Risk-neutral probability in the Cox--Ross--Rubinstein model, hence the unique equivalent martingale measure.