# 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
````
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could you please five a reference to this formula in the literature? It seems that this is an approximation of probability in the sense, that it can be $<1$ only under some conditions, not always. – Ilya Mar 30 '11 at 6:29
– S_H Mar 30 '11 at 20:14
Getting a prob > 1 (or < 0) means that you've chosen your lattice spacing in a way that makes the algorithm numerically unstable. – Foster Boondoggle Dec 20 '11 at 3:54

## 2 Answers

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 $\Delta t$.

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Where did you find this formula? Citation needed. – chrisaycock♦ Mar 31 '11 at 15:57
It comes from the definition of the risk-neutrality. In fact in CRR model for the riskless income $1+r$ is used rather then $e^{r\Delta t}$. – Ilya Mar 31 '11 at 16:37
The first condition doesn't do any good either. Usually `u~[1.001,1.5]`, and as I said in comment to `quant_dev` that by putting `nColumn=1000`, which gives `del_T=0.02`, then you can calculate that your given condition is satisfied, however the `p` is still greater than 1. – S_H Apr 1 '11 at 18:08
@H_S: Using your parameters you will get that if $nColumn = 1000$ then $\frac{\log u}{r} = 0.0047$ - so $del_T = 0.02$ does not satisfy the first condition. Please do you calculations properly before writing that something does not work. First condition is right but not convenient since by changing $del_T$ you also change $u$, therefore I also provide the second condition. If you apply it you will see that $nColumn\geq 20 000$. – Ilya Apr 4 '11 at 6:16

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)

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It doesn't. Even if I make `nColumn=1000` and `T=20`, still `p=4.1291`. – S_H Mar 29 '11 at 21:21
For small enough time steps it will. The binomial tree is just an approximation, and you can't really call $p$ a genuine probability. – quant_dev Mar 29 '11 at 21:37
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