From Hull's book when deriving coefficients of up and down movements, $u$ and $d$, of a stock price using binomial tree approach, at some point we get the following equation:

$$e^{\mu\Delta t}(u+d) - ud - e^{2\mu\Delta t} = \sigma^2\Delta t.$$

Then it is stated that from solving the above equation we obtain that $u = e^{\sigma\sqrt{\Delta t}}$ and $d= e^{-\sigma\sqrt{\Delta t}}$. It is also noted that we use Taylor's formula and throwing $\Delta t^2$ and higher terms:

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots.$$

Could you clarify how do we get to this result?

So far I get by using Taylor's formula:

$$e^{\mu\Delta t} \approx 1 + \mu\Delta t,$$ $$e^{2\mu\Delta t} \approx 1 + 2\mu\Delta t.$$

Then the above equation transforms to

$$(1+\mu\Delta t)(u+d) - ud - 1 - 2\mu\Delta t = \sigma^2\Delta t.$$

I am confused how to proceed from here. I tried to do some algebra but it gave no result. For instance, if we assume that $ud=1$ then we get

$$(1+\mu\Delta t)(u+d) - 2(1+\mu\Delta t) = \sigma^2\Delta t,$$ $$(1+\mu\Delta t)(u+d-2) = \sigma^2\Delta t$$ and $$u+d = \frac{\sigma^2\Delta t}{1+\mu\Delta t} + 2$$

Here I am stuck

  • $\begingroup$ note that there are over 30 binomial tree discretizations and this is just one of them, and it is really out of date. $\endgroup$
    – Mark Joshi
    Commented Apr 13, 2016 at 3:54

2 Answers 2


We assume that $u=e^x$ and $d = e^{-x}$. Note that \begin{align*} u &\approx 1+ x +\frac{x^2}{2}, \textrm{ and}\\ d &\approx 1- x +\frac{x^2}{2}. \end{align*} Substituting these into your last equation, \begin{align*} u+d = \frac{\sigma^2 \Delta t}{1+\mu\Delta t} + 2, \end{align*} we obtain that \begin{align*} x^2 \approx \frac{\sigma^2 \Delta t}{1+\mu\Delta t}. \end{align*} Note also that \begin{align*} \frac{1}{\sqrt{1+\mu\Delta t}} \approx 1-\frac{1}{2}\mu\Delta t+\frac{3}{8}(\mu\Delta t)^2. \end{align*} Consequently, \begin{align*} x & \approx \frac{\sigma \sqrt{\Delta t}}{\sqrt{1+\mu\Delta t}}\approx \sigma \sqrt{\Delta t} -\frac{1}{2}\mu\sigma (\Delta t)^{3/2} + \frac{3}{8}\sigma \mu^2 (\Delta t)^{5/2} \approx \sigma \sqrt{\Delta t}. \end{align*} That is, \begin{align*} u &= e^x =e^{\sigma \sqrt{\Delta t}},\\ d &= e^{-x} =e^{-\sigma \sqrt{\Delta t}}. \end{align*}

  • $\begingroup$ Thanks for your answer. Could you explain why $\frac{\sigma\sqrt{\Delta t}}{1 +\frac{1}{2}\mu\Delta t} \approx \sigma\sqrt{\Delta t}$? We assume that $\Delta t$ is small so that denominator is equal to one? Why then we assume $\sqrt{\Delta t}$ not too small so that numerator is not equal to zero? $\endgroup$
    – tosik
    Commented Apr 13, 2016 at 6:56
  • 2
    $\begingroup$ Remember that $\sqrt{x}$ tends slower towards zero than $x$ as $x \rightarrow 0 $. $\endgroup$
    – Quantuple
    Commented Apr 13, 2016 at 7:48
  • 1
    $\begingroup$ @IvanovNikita, please see my updates. $\endgroup$
    – Gordon
    Commented Apr 13, 2016 at 12:57
  • 1
    $\begingroup$ How do you obtain u and d without assuming that u = exp(x) and d = exp(-x)? At first glance it is not trivial to understand why we exponeniate only the left side of the equation. Or in other words how do you come to decide to equate u and d to exp (x) and exponeniate only the left side of the equation? $\endgroup$ Commented Aug 11, 2021 at 9:21

You can also get this result from Gauss-Hermite quadrature, and this is the subject of a recent paper in the Journal of Derivatives https://jod.pm-research.com/content/early/2021/04/02/jod.2021.1.130.short.

The price of an option can be written as the discounted integral of the payoff $\phi(S)$ times the transition probability density for a lognormal process: $$ C(S,T) = D(T)\int_0^\infty \phi(z)\exp\left(-\frac{\left(\ln\frac{z}{S} - \left(r-\frac{1}{2}\sigma^2\right)T\right)^2}{2\sigma^2T} ​\right)\frac{dz}{z\sqrt{2\pi\sigma^2T}}. $$ Making the substitution $y=\frac{\left(\ln\frac{z}{S} - \left(r-\frac{1}{2}\sigma^2\right)T\right)^2}{\sigma\sqrt{T}}$ leads to an integral amenable to Gauss-Hermite quadrature $$ C(S,T)=D(T)\int_{-\infty}^\infty \phi\left(Se^{\sigma\sqrt{T}y+\left(r-\frac{1}{2}\sigma^2\right)T}\right)e^{-\frac{y^2}{2}}\frac{dy}{\sqrt{2\pi}}. $$ Using second order Gauss-Hermite quadrature leads to $$ C(S,T) \approx \frac{D(T)}{2}\left[\phi\left(Se^{\sigma\sqrt{T} +\left(r-\frac{1}{2}\sigma^2\right)T}\right)+\phi\left(Se^{-\sigma\sqrt{T} +\left(r-\frac{1}{2}\sigma^2\right)T}\right) \right] $$

Now you may think this is very far from the answer, however two things need to be done. First discretize time into timesteps of constant variance by using the Chapman-Komolgorov equation, and second modify Gauss-Hermite quadrature to move the abscissa slightly off the roots of the third Hermite polynomial. The details are in the paper cited above. The result is $$ C(S,T) = D(T)\sum_{j=0}^{N}p^{j}(1-p)^{N-j} {N \choose j}f\left(xe^{(2j-N)\sigma\sqrt{\Delta t}}\right) $$ where $p=\left[\frac{1}{2} + \frac{\left(r-\frac{1}{2}\sigma^2\right)\Delta t}{2\sigma\sqrt{\Delta t}}\right]$. This is equivalent to the probabilities in the original paper by Cox, Ross and Rubinstein (bottom of page 252) Original CRR paper


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