Question on yield curve fitting from Wilmott on Quant Finance p.529

Question

My last question is related.

At the top of p. 529, it says,

"From the Taylor series expansion for $Z$ we find that the yield to maturity is given by

$$-\frac{log Z(r,t;T)}{(T-t)}\approx-a+\left(\frac{1}{2}a^{2}-b\right)(T-t)+\left(ab-c-\frac{1}{3}a^{3}\right)(T-t)^{2}+\dots$$

for short times to maturity."

We know that we derive yield to maturity from the inverse of the zero coupon bond equation

$$Z(r,t;T)=e^{-r(T-t)}$$

by taking logs and dividing by $(T-t)$ and multiplying both sides by $-1$.

If we plug the solutions for $a(r)$, $b(r)$, and $c(r)$ into the series expansion of $Z$ from p.528 we have that $Z$ is equal to

$$ Z\approx-r(T-t)+(\frac{1}{2}r^2-\frac{1}{2}(u-\lambda w))(T-t)^2+... $$

But the solution provided on p. 529 shows that no such substitutions for the values of a, b, and c have yet been made.

Taking logs, dividing by $(T-t)$, and multiplying both sides by $-1$ on our equation does not give us

$$-\frac{log Z(r,t;T)}{(T-t)}\approx-a+\left(\frac{1}{2}a^{2}-b\right)(T-t)+\left(ab-c-\frac{1}{3}a^{3}\right)(T-t)^{2}+\dots$$

and neither does starting with

$$Z\approx 1+a(r)(T-t)+b(r)(T-t)^2+c(r)(T-t)^3$$

and taking logs, etc..

So, obviously I am starting from the wrong place. Can you help me see which $Z$ to start with? I don't believe it is the computations that I am having trouble with, I am having trouble seeing the plan, as Polya would say.

Thanks in advance.

Answer 1

As @michipilli said, if

• $Z = 1+ as + bs^2 + cs^3$ (where I have substituted $T-t$ by $s$ for ease of notation and also suppressed the dependencies of $a$, $b$ and $c$) and
• $\log (1+\zeta) = \zeta - \frac{1}{2}\zeta^2 + \frac{1}{3}\zeta^3 + ...$ then,

\begin{align*} \log Z &= (as + bs^2 + cs^3) - \frac{1}{2}(as + bs^2 + cs^3)^2 + \frac{1}{3}(as + bs^2 + cs^3)^3 + ... \end{align*} and \begin{align*} -\frac{\log Z}{s} &= -\frac{1}{s}(as + bs^2 + cs^3) +\frac{1}{2s}(as + bs^2 + cs^3)^2 - \frac{1}{3s}(as + bs^2 + cs^3)^3 + ... \end{align*} Now,

• $(as + bs^2 + cs^3)^2 = a^2s^2 + b^2s^4 + 2acs^4 + 2abs^3 + o(s^5)$
• $(as + bs^2 + cs^3)^3 = a^3s^3 + 3a^2bs^4 + o(s^5)$

Substituting back we get \begin{align*} -\frac{\log Z}{s} &= -a - bs - cs^2 + \frac{a^2s}{2} + \frac{b^2s^3}{2} + acs^3 + abs^2 - \frac{a^3s^2}{3} - a^2bs^3 + o(s^3)\\ &= -a + \left(\frac{a^2}{2} - b\right)s + \left(ab -c -\frac{a^3}{3}\right)s^2 + o(s^3) \end{align*}

which is what you desire.

Answer 2

Is the author taking logs (and dividing by (T-t) etc) of our previous Z expansion from the previous page?

He does, as you will see if you try to do the computation. What did you prevent to find this out by yourself? (I am trying to be constructive.)

Mathematically, it doesn't add up to what the author provides as the answer. What am I missing here?

The sequel of the book probably tells.