Reconciling Risk-Neutral Probabilities in the Binomial Option Model

Summary

The note explains why two formulas for risk-neutral probabilities in a binomial option-pricing model may appear unequal. One expression gives the up-move probability directly, while the cited algorithm stores discounted up and down weights. Comparing them requires accounting for the discount factor and the reciprocal-move assumption that the down factor equals the inverse of the up factor.

The answer derives the algorithm’s down weight and then its up weight, showing how multiplying those weights by the one-period growth factor recovers probabilities. It also notes that the formulas include a dividend-yield term, which disappears when that yield is set to zero. The explanation addresses the specific parameterization in the question; the equivalence relies on the stated relationship between up and down factors and consistent treatment of discounting and yield.

Key ideas

• The algorithm’s array entries are discounted probability weights rather than raw probabilities.
• With reciprocal up and down factors, the down weight can be rewritten to match the binomial model’s probability expression after discounting is accounted for.
• Multiplying each discounted weight by the period growth factor gives the corresponding risk-neutral probability.
• Dividend yield affects the relationship and must be set consistently when comparing formulas.

Tags

# binomial option pricing model - problem with risk-neutral probability


# binomial option pricing model - problem with risk-neutral probability












I have a little problem: in the binomial option pricing model, the price of a european derivative security $V_{n}$ satisfies: $V_{n}=[1/(1+r)]*[\tilde{p}*optionUp +\tilde{q}*optionDown]$ where: $\tilde{p}=\frac{e^{r*\Delta T} -d}{u-d}$ But when I read the article "option pricing model" on Wikipedia(http://en.wikipedia.org/wiki/Binomial_options_pricing_model), the $\tilde{p}$ of Wikipedia's $\textbf{algorithm}$ is slightly different: $\tilde{p}=\frac{(ue^{-r*\Delta T} -1)*u}{u^2-1}$ (I take q=0) I try to compare these 2 forms but they are not equal... why ??? Thanks ! :)

## Answer by Gordon (score 1, accepted)

https://quant.stackexchange.com/a/17118

In the link you provided, by noting the construction of array p[], p0 and p1 are respectively the discounted $\texttt{down}$ and $\texttt{up}$ probabilities. Since $d=\frac{1}{u}$, then \begin{align*} p0 &= e^{-r \Delta T}\, \frac{u-e^{(r-q)\Delta T}}{u-d}\\ &= \frac{\big(u\,e^{-r \Delta T} -e^{-q\Delta T}\big)u }{u^2-1}, \end{align*} and \begin{align*} p1 &= e^{-r \Delta T}\,\big(1-p0\, e^{r \Delta T}\big)\\ &=e^{-r \Delta T} - p0. \end{align*} Note that $p1\, e^{r \Delta T}$ and $p0\, e^{r \Delta T}$ are respectively the $\texttt{up}$ and $\texttt{down}$ probabilities.

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