Why Risk-Neutral Probability Stays Constant in a Binomial Model
Summary
The note explains why the risk-neutral branch probability in a Cox–Ross–Rubinstein binomial tree is constant when the up and down multipliers are fixed. It distinguishes this probability from the option’s delta hedge, which is a position in the underlying used to replicate the derivative and can change as the option’s value and the stock price change.
The explanation derives the probability by requiring the stock price to be a martingale under the risk-neutral measure: its expected next-step value must equal its current value when rates are ignored. With a risk-free rate, the corresponding condition uses the discounted stock price. The same up and down multipliers at each step therefore imply the same probability at each step. The result depends on the model’s fixed parameters; it does not imply that a replicating hedge remains unchanged. The question’s displayed pricing expression appears to have notation issues, and the answer omits the rate-adjusted derivation.
Key ideas
- Risk-neutral probability is set by the stock’s expected growth condition under the pricing measure.
- With fixed up and down multipliers, the probability solving that condition is the same at every step.
- Risk-neutral probability is distinct from delta, which describes a replicating hedge.
- A risk-free rate changes the martingale condition to apply to the discounted stock price.
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# Answer by alexprice (score 2, accepted)
# 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.
## Answer by alexprice (score 2, accepted)
https://quant.stackexchange.com/a/54163
$q$ 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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.