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I have discussion with my colleague on why a general assumption $$ud=1$$ in binomial tree option pricing model would be necessary?

I take it a simplification of the problem, otherwise, there will be more intermediate nodes in the tree, which will be hard to compute. While my colleague insists that since the underlying is often lognormal with 0 mean, $ud=1$ were a fair assumption.

I think his explanation is acceptable. The question, is there any other reason to make $ud=1$ assumption?

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  • $\begingroup$ I think there is also an important distinction between theory and modeling to remember. Modeling sometimes may have assumptions that are not necessarily realistic but mathematically simple so this may be the case here $\endgroup$
    – Kamster
    Commented Jan 20, 2015 at 2:20

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you don't need $ud=1.$ In fact, there are now about 30 binomial trees which converge to Black--Scholes in the large step limit. Most of them do not have $ud=1.$ All you need is

$$ d < e^{r \Delta t} < u $$

The tree recombines provided $u$ and $d$ don't change from step to step.

See my book More Mathematical Finance for a comprehensive review and classification of binomial trees.

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  • $\begingroup$ Led $d=-1000<e^*<u=1,1$. The tree does not converge? $\endgroup$
    – emcor
    Commented Jan 20, 2015 at 1:26
  • $\begingroup$ binomial trees are approximations to the BS model. $u$ and $d$ will be functions of the BS parameters and the time-step size. eg $$ u = e^{(r-0.5\sigma^2) \Delta t + \sigma \sqrt{\Delta t}}$$, $$ d = e^{(r-0.5\sigma^2) \Delta t -\sigma \sqrt{\Delta t}}$$ $\endgroup$
    – Mark Joshi
    Commented Jan 20, 2015 at 4:14
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The condition

$$ud=1\text{, or equivalently }u=1/d$$

is necessary to ensure convergence of the Binomial tree's mean $\mu$ and standard deviation $\sigma$ to nonfinite values when $n$ (number of steps) goes to infinity.

Cox-Rubinstein-Ross showed in their famous paper, that to achieve this, we must have:

$$u=e^{\sigma\sqrt{t/n}}\text{, }d=e^{-\sigma\sqrt{t/n}}$$ respectively, which holds exactly: $$ud=1.$$ We can aswell choose $\mu$ and $\sigma$ to fit $u$ and $d$ respectively, so $ud=1$ is the actual condition.

Paper excerpt: enter image description here

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    $\begingroup$ CRR's condition ud=1 leads to a recombinant tree, but binomial trees need not be recombinant, they are just much easier to calculate when they are. $\endgroup$
    – experquisite
    Commented Sep 30, 2014 at 21:46
  • $\begingroup$ @experquisite The paper says the tree is only convergent under this parametrization. $\endgroup$
    – emcor
    Commented Sep 30, 2014 at 23:12
  • $\begingroup$ I'm pretty sure the condition for it to be convergent in the way you describe is just that the step size must be ~ $\sqrt{t}$. If you start assuming a log-normal process and constant volatility, that's when you start to get boxed into particular parameterizations. $\endgroup$
    – experquisite
    Commented Oct 1, 2014 at 0:23
  • $\begingroup$ researchgate.net/publication/… seems to be regarding non-recombining binomial trees for options pricing, but I haven't read the paper. $\endgroup$
    – experquisite
    Commented Oct 1, 2014 at 0:24
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    $\begingroup$ Right, my point is that there exist binomial models for non-recombining trees where your $ud=1$ condition doesn't hold, which still converge to a price as $\Delta t \rightarrow 0$. $\endgroup$
    – experquisite
    Commented Oct 1, 2014 at 16:58

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