Deriving Risk-Neutral Probability in a Forward Binomial Tree
Summary
The document derives a simplified expression for the risk-neutral up-move probability in a standard forward binomial tree. The tree uses up and down factors built from the risk-free rate, dividend yield, volatility, and period length. Starting from the usual no-arbitrage probability formula, the derivation factors out the common drift term, leaving a ratio involving exponentials of volatility and the square root of the time step.
To obtain the textbook form, the answer multiplies the numerator and denominator by a matching factor and simplifies, yielding the reciprocal of one plus the positive volatility exponential. This is an algebraic clarification of an identity under the stated definitions, rather than a broader treatment of option valuation or tree construction. It does not discuss parameter estimation, convergence, or practical limitations of the binomial model.
Key ideas
- The risk-neutral probability follows from matching the expected one-step growth factor to the risk-free growth factor adjusted for dividends.
- Factoring out the common drift exponential reduces the probability expression to a volatility-based ratio.
- Multiplying by a suitable exponential factor simplifies the ratio to the stated reciprocal form.
- The derivation assumes the up and down factors defined for the standard forward tree.
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Full text
# How to derive the formula for risk-neutral probability for a Standard Binomial Tree (Forward Tree)
# How to derive the formula for risk-neutral probability for a Standard Binomial Tree (Forward Tree)
Consider a standard binomial tree. Let $u = e^{(r - \delta)h + \sigma\sqrt{h}}$ and $d = e^{(r - \delta)h - \sigma\sqrt{h}},$ where $\delta$ is the continuously compounded dividend yield, $h$ is the length of one period in a binomial model, and $\sigma$ is volatility.
I am told in my textbook that the risk-neutral probability $p*$ is given by:
$$p^* = \frac{e^{(r - \delta)h} - d}{u - d} = \frac{1}{1 + e^{\sigma\sqrt{h}}}.$$
I tried deriving the second equality as follows:
$\begin{align}\frac{e^{(r - \delta)h} - d}{u - d} &= \frac{e^{(r - \delta)h} - e^{(r - \delta)h - \sigma\sqrt{h}}}{e^{(r - \delta)h + \sigma\sqrt{h}} - e^{(r - \delta)h - \sigma\sqrt{h}}}\\ &= \frac{e^{(r - \delta)h}(1 - e^{-\sigma\sqrt{h}})}{e^{(r - \delta)h}(e^{\sigma\sqrt{h}} - e^{-\sigma\sqrt{h}})}\\ &= \frac{1 - e^{-\sigma\sqrt{h}}}{e^{\sigma\sqrt{h}} - e^{-\sigma\sqrt{h}}}\end{align}.$
Now at this point I am stuck and I'm unsure if it's either something algebraic I am not seeing, or if there is some property of forward trees that we can use to reach the conclusion.
## Answer by KarolisR (score 2, accepted)
https://quant.stackexchange.com/a/30172
Here's one algebraic way to derive it: $$ \frac{(1 - e^{-\sigma\sqrt{h}})(1 + e^{\sigma\sqrt{h}})}{(e^{\sigma\sqrt{h}} - e^{-\sigma\sqrt{h}})(1 + e^{\sigma\sqrt{h}})} = \frac{1 - e^{-\sigma\sqrt{h}} + e^{\sigma\sqrt{h}} - 1}{(e^{\sigma\sqrt{h}} - e^{-\sigma\sqrt{h}})(1 + e^{\sigma\sqrt{h}})} = \frac{e^{\sigma\sqrt{h}} - e^{-\sigma\sqrt{h}}}{(e^{\sigma\sqrt{h}} - e^{-\sigma\sqrt{h}})(1 + e^{\sigma\sqrt{h}})} = \frac{1}{1 + e^{\sigma\sqrt{h}}} $$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.