No-Arbitrage Conditions for Binomial Risk-Neutral Probabilities
Summary
The document explains why the risk-neutral probability in a binomial option tree can fall outside the zero-to-one range. In the risk-neutral formula, the probability depends on the up and down factors and the risk-free growth over one time step. For it to represent a valid probability in an arbitrage-free model, the risk-free growth factor must lie strictly between the down and up factors. If this condition fails, the formula produces an invalid value; reducing the time step may restore the required ordering for a given parameterization.
The answer connects the condition to the chosen up and down factors and gives a time-step bound for its stated setup. A second response also recommends smaller steps and points out that the supplied volatility and rate may be atypical. The numerical bound is model-dependent, so it should not be applied without checking the tree’s definitions and units. The probability formula is valid only when the no-arbitrage ordering holds; an out-of-range result signals a problem with the inputs or discretization.
Key ideas
- The binomial risk-neutral probability is valid only when the risk-free growth factor lies between the down and up factors.
- A probability outside the zero-to-one range indicates that the tree parameters violate the required no-arbitrage condition.
- Reducing the time step can help satisfy the condition, depending on the chosen parameterization.
- The appropriate time-step bound depends on the model’s definitions and input units.
Tags
Full text
# Risk neutral probability in binomial lattice option coming greater than 1...what's wrong?
# Risk neutral probability in binomial lattice option coming greater than 1...what's wrong?
I am substituting reasonable values in the below fomula (like r=0.12, T=20, nColumn=16, sigma=0.004)...why is probability coming out to be greater than 1? Any help? Thanks!
```
del_T=T./nColumn; % where n is the number of columns in binomial lattice
u=exp(sigma.*sqrt(del_T));
d=1./u;
p=(exp(r.*del_T)-d)./(u-d); % risk neutral probability
```
## Answer by SBF (score 2, accepted)
https://quant.stackexchange.com/a/873
Yeah, I've found this formula. So you just need to put $$ \Delta t < \frac{\log{u}}{r}. $$
Edited: To avoid arbitrage one should have $0<d<1+r<u$ - (Shreve, Stochastic Calculus for Finance I), or in you case $0<d(\Delta t)<\mathrm{e}^{r\Delta t}<u(\Delta t)$. Only under this condition your formula $$ p = \frac{\mathrm{e}^{r\Delta t}-d}{u-d} $$ is valid and the probability will be less than $1$ and greater than $0$ - in fact I told you the same from the beginning. Using the formula for $u(\Delta t)$ we have that for a time step $$ \Delta t < \frac{\sigma^2}{r^2}. $$
It's strange that this conditions are not presented in wikipedia. Moreover they abuse notation for $u(\Delta t)$ and $d(\Delta t)$ using there $t$ rather than $\Delta t$.
## Answer by quant_dev (score 0)
https://quant.stackexchange.com/a/844
Simple: decrease the time step. The binomial tree is just an approximation, and you can't really call $p$ a genuine probability.
Your parameters are also rather far from "typical". I would choose:
$r = 0.05$ (still higher than the current risk-free rate)
$\sigma = 0.2$ (more typical volatility value)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.