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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.

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  • $\begingroup$ 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. $\endgroup$
    – Idonknow
    Jun 7, 2020 at 3:37

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$q$ 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.

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