Given that

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

I would like to show that

$u=e^{\sigma\sqrt{\Delta t}}$

I know I must somehow use Taylor's approximation $e^x = 1 + x + \frac{x^2}{2}+...$ and ignore terms of $\Delta t$ higher than 1, but I can't seem to get to the value of $u$. Can someone show me this derivation?

  • $\begingroup$ To those interested the question asks to explain the derivation of the $u$ term in The chapter on Binomial Trees (chapter 17.1 in my edition) in Hull's Options Futures and other Derivatives $\endgroup$ – Probilitator Mar 20 '14 at 19:43

I see your porblem - Hull unfortunately does not explain the reasoning behind the approach.

The hint the books gives is correct. Using Taylor series $e^x$ can be written as $e^x = 1 + x + \frac{x^2}{2}+...$. Hull also incoporates a dividend rate $q$ but we can disregard it here.

$p$ is given by $p=\frac{e^{r\Delta }-d}{u-d}$. We also have $u=\frac{1}{d}$. So to complete our setup wie primarily just need to find a propper $u$ that satisfies equation $(*)$ $$e^{r\Delta t}(u+d)-ud-e^{2r\Delta t} = \sigma^2\Delta t $$ One can assume that $u$ will be some function of $\Delta t$ and thus write $u(\Delta t)$. Furthermore we do not need $u(\Delta t)$ to solve $(*)$ for huge $\Delta t$. If we assume that the function has some taylor approximation we can just work with the truncated taylors some for it will approximate $u(\Delta t)$ well enough for small $\Delta t$.

So we set $u(\Delta t) = e^{-\sigma\sqrt{\Delta t}}$

Now obviosly this choice does not satifsy equation $(*)$. Still we would use the result if it's second order taylor approximation will do the job (thus fullfill the equation quite well for smaller $\Delta t$) - remember we can use a taylor expansion to approximate the function - this is the somewhat simplified statement of Taylor's Theorem

The second order Taylor Sum for $e^t$ is given by $e^t \approx 1+t+0.5t^2$. Inserting $\sigma \sqrt{\Delta t}$ for $t$ gives $e^{\sigma \sqrt{\Delta t}}\approx 1+\sigma \sqrt{\Delta t}+0.5\sigma^2 \Delta t$. And thus $u(\Delta t) \approx 1+\sigma \sqrt{\Delta t}+0.5\sigma^2 \Delta t$ and $d(\Delta t) \approx 1-\sigma \sqrt{\Delta t}+0.5\sigma^2 \Delta t$ for small enough $\Delta t$.

Using the same tecnique we approximate the $e^{r\Delta t},e^{2 r\Delta t}$ terms by their first order taylor sums and get $e^{r\Delta t}=1+r \Delta t$ and $e^{2 r\Delta t}=1+2r \Delta t$.

If you substite the terms in equation $(*)$ by the approximations derived here and kill/ignore all the terms containing $(\Delta t)^2$ you will get the desired result.


$$(1+r\Delta t)(1+\sigma \sqrt{\Delta t}+0.5\sigma^2 \Delta t+1-\sigma \sqrt{\Delta t}+0.5\sigma^2 \Delta t)-(1+\sigma \sqrt{\Delta t}+0.5\sigma^2 \Delta t)(1-\sigma \sqrt{\Delta t}+0.5\sigma^2 \Delta t)-(1+2r\Delta t) = \sigma^2\Delta t $$

Simplify (thus just carry out the multiplication) whenver you encounter a term containg $(\Delta t)^2$ (e.g. $\sigma^4(\Delta t)^2$) just set it to zero.

Edit on the background for the choice of $u$

By using the relation $d=1/u$ one can simplify equation $(*)$ to $$ u^2-\frac{1+e^{2r\Delta t}+\sigma^2\Delta t}{e^{r\Delta t}}u+1=0 $$

Setting $A=0.5\frac{1+e^{2r\Delta t}+\sigma^2\Delta t}{e^{r\Delta t}}=0.5(e^{-r\Delta t}+e^{r\Delta t}+\sigma^2\Delta t e^{r\Delta t})$ one arrives at the quadratic equation $u^2-2Au+1=0$

Solving this equation gives $u=A+\sqrt{A^2-1}, d=A-\sqrt{A^2-1}$ Now if one inserts the actual formula for $A$ into this equations, substitutes $e^{r \Delta t},e^{-r \Delta t}$ with $1+r \Delta t,1-r \Delta t$, simplifies and then neglects all the terms containing $(\Delta t)^2$ or higher one arrives at $$ u=1+\sigma \sqrt{\Delta t}+0.5\sigma^2 \Delta t $$ This is the second order Taylor approximation of $e^{ \sigma \sqrt{\Delta t}}$

  • $\begingroup$ How do you know u is a function of $\Delta t$ and not, say, a function of $\Delta$ and $r$? And how do you "guess" to set $u(\Delta t) = e^{-\sigma \sqrt{\Delta t}}$? $\endgroup$ – mathjacks Mar 21 '14 at 0:55
  • $\begingroup$ I meant to say "a function of $\Delta t$" not just $\Delta$. $\endgroup$ – mathjacks Mar 21 '14 at 2:04
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    $\begingroup$ Regarding the choice of $u$ I will add something to the answer later today. $\Delta t$ ist actually mainly a "notational" thing. It means a time step rather than a poin in time. So if we had a monthly binomial grid $\Delta t=1/12$. $\endgroup$ – Probilitator Mar 21 '14 at 8:05
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    $\begingroup$ Great thank you. Looking forward to your addition to the answer. It's been incredibly helpful so far. $\endgroup$ – mathjacks Mar 21 '14 at 12:54
  • $\begingroup$ I added an explanation on the choice of $u$. I can also recommend Modeling Derivatives in C++ (eu.wiley.com/WileyCDA/WileyTitle/productCd-0471654647.html). Chapter 3 explains very well the backgrounds of the binomial tree model. There you will also find several choices for $u$ explained in more detail than in Hull - cheers $\endgroup$ – Probilitator Mar 21 '14 at 22:15

Not all binomial trees take $u=e^{\sigma\sqrt{\Delta t}}$. Thinking of the binomial tree as a discrete approximation (on a grid) to a continuous process, it makes sense that a variety of choices for where to place grid points will work.

For a listing of a few different choices of $u$, see the Tian Tree settings and others. From this Sitmo page you can see, for example, that Jarrow and Rudd take

$$ u = e^{(r-q-\frac12 \sigma^2)\Delta t + \sigma\sqrt{\Delta t}} $$

  • $\begingroup$ I'm actually just curious about how to derive the value of u that I listed given the formula that I have. From Hull, I don't get a clear picture of how that derivation happens and I imagine it's a matter of simple mathematics? $\endgroup$ – mathjacks Mar 20 '14 at 18:20

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