# Why are the risk neutral probabilities constant in the Cox Rubinstein model when delta needs to be changed at each time step

Consider the Cox Rubinstein binomial pricing model with N steps, with stock price change given by parameters u and d so that at step $$i$$ we have $$S_{i+1} = uS_{i}$$ or $$S_{i+1} = dS_{i}$$ with $$0\leq i \leq N$$. Let $$r$$ be the risk free rate. As usual suppose we have a cash instrument growing at the risk free rate, and assuming the no arbitrage condition we have that a call option price is give as $$C$$ = $$\sum_{i=0}^{N}$$ $$N\choose i$$ $$\max(S_0 q^{i}(1-q)^{N-i}u^{i}d^{N-i} -K,0)\frac{1}{r^N}$$, where $$K$$ is the option strike, $$q$$ is the risk neutral probability and $$S_0$$ is the initial stock price.

To me, the above implies that on each branch of the tree has the same probability (risk neutral) of $$q$$ or $$q-1$$. When working out the value of $$q$$ though, if we use a replicating argument we see that $$q$$ corresponds to the delta hedge? My understanding was that this hedge has to be adjusted at each time step, but this is inconsistent with the above. Clearly I think I am missing something here - is it because I am assuming a constant risk free rate throughout? Thanks for your help.

• Is the pricing formula for $C$ correct? I thought the discounting factor should be $(1+r)^N$ instead of $r^N$? Here I assume that $q$ is the risk-neutral probability to get upward movement of stock price. Jun 7, 2020 at 3:37

$$q$$ is not delta hedge. $$q$$ is determined from the fact that $$S_i$$ is a martingale i.e. for $$S_0$$
$$S_0=E(S_1)=quS_0+(1-q)dS_0$$ (if no rates)
This equation gives the same $$q$$ , dependent only on $$u$$ and $$d$$ , if calculated for $$S_0$$ , $$S_1$$ etc , thus $$q$$ is the same for all steps.