# Deriving $u$ and $d$ coefficients using binomial tree approach

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

• note that there are over 30 binomial tree discretizations and this is just one of them, and it is really out of date. – Mark Joshi Apr 13 '16 at 3:54

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*}
• 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? – tosik Apr 13 '16 at 6:56
• Remember that $\sqrt{x}$ tends slower towards zero than $x$ as $x \rightarrow 0$. – Quantuple Apr 13 '16 at 7:48