Normalizing the Risk-Neutral Probability in a CRR Model

Summary

The exchange explains why a probability measure defined over paths in a Cox–Ross–Rubinstein binomial model sums to one. Each path receives a weight determined by the number of upward moves, using the risk-neutral probability p* and its complement. Paths with the same number of upward moves have equal weight, and there are as many such paths as the corresponding binomial coefficient.

Grouping the path weights by their number of upward moves turns the sum over all paths into a binomial expansion. The binomial theorem then shows that the total mass is one. The setup assumes the model’s risk-neutral probability is valid, as ensured in the question by the condition that the risk-free return lies between the down and up returns. The answer establishes normalization; it does not cover other properties needed to prove a full arbitrage-free pricing model.

Key ideas

• In the CRR model, each path’s probability weight depends on its count of upward moves.
• Paths with the same number of upward moves have identical weights.
• The number of paths with a given count is the corresponding binomial coefficient.
• Grouping paths by upward-move count reduces total probability to a binomial expansion.
• The stated return condition keeps the risk-neutral probability between zero and one.

Tags

# CRR model arbitrage free


# CRR model arbitrage free












I'm currently studying this proof

In this proof the author defines a probability measure

$$P^*[\{\omega\}]=(p^*)^{k(\omega)}(1-p^*)^{T-k(\omega)}$$ on $$\Omega=\{\omega=(y_1,\ldots,y_T)|y_i=\pm1\}$$

where $p^*=(r-a)/(b-a)$ and $k(\omega)$ is the number of ones in $\omega$.

$a<r<b$.

Unfortunately I can't prove that $P^*$ is indeed a probability measure.

$P^*[\{\omega\}]\ge0$ is clear.

I don't see $P^*(\Omega)=1$

$P^*(\Omega)=\sum_{\omega_i \in \Omega}(p^*)^{k(\omega_i)}(1-p^*)^{T-k(\omega_i)}$, but I don't know how to continue

## Answer by Andrew (score 1)

https://quant.stackexchange.com/a/45661

If you consider $\omega$ and $\tilde{\omega}$ with $k(\omega)=k(\tilde{\omega})$ it holds that $P^*(\omega)=P^*(\tilde{\omega})$. Now instead of summing up over every $\omega_i \in \Omega$ you can sum up from $n =0 ... T$ and count the elements with $k(\omega_i)= n$. There are $\dbinom{T}{n}$ elements in $\Omega$ which fullfill $k(\omega_i)= n$.

Therefore $P^*(\Omega)=\sum_{\omega_i \in \Omega}(p^*)^{k(\omega_i)}(1-p^*)^{T-k(\omega_i)} = \sum_{n=0}^T \dbinom{T}{n} (p^*)^n(1-p^*)^{T-n}=1$ as a result from the binomial theorem.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.