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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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  • $\begingroup$ 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. $\endgroup$ – Ilya Mar 30 '11 at 6:29
  • $\begingroup$ Here's the wikipedia article: en.wikipedia.org/wiki/Binomial_options_pricing_model $\endgroup$ – user98 Mar 30 '11 at 20:14
  • $\begingroup$ Getting a prob > 1 (or < 0) means that you've chosen your lattice spacing in a way that makes the algorithm numerically unstable. $\endgroup$ – Foster Boondoggle Dec 20 '11 at 3:54
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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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  • $\begingroup$ Where did you find this formula? Citation needed. $\endgroup$ – chrisaycock Mar 31 '11 at 15:57
  • $\begingroup$ It comes from the definition of the risk-neutrality. In fact in CRR model for the riskless income $1+r$ is used rather then $e^{r\Delta t}$. $\endgroup$ – Ilya Mar 31 '11 at 16:37
  • $\begingroup$ @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 $nColumn\geq 20 000$. $\endgroup$ – Ilya Apr 4 '11 at 6:16
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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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  • $\begingroup$ For small enough time steps it will. The binomial tree is just an approximation, and you can't really call $p$ a genuine probability. $\endgroup$ – quant_dev Mar 29 '11 at 21:37

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