Expanding the Short-Maturity Yield Curve Bond Pricing Equation
Summary
This discussion examines a power series expansion for a zero-coupon bond price near maturity and how it is substituted into a bond pricing partial differential equation. The answer explains that the displayed groupings arise because the drift and volatility functions are expanded in time around maturity. Since volatility depends on both the short rate and time, its squared value receives a first-order time expansion; the drift term is treated similarly. This accounts for terms involving time derivatives of those coefficients.
The response also confirms two suspected typographical errors in the source expression: the rate derivative grouping should carry the maturity-time powers shown by the series, and the final bond-price term should use the second-order coefficient rather than the third. The exchange gives an algebraic explanation and corrections, but no numerical fit or empirical yield-curve evidence. The expansion is local to short maturities and relies on the stated pricing equation and coefficient smoothness sufficient for the Taylor expansions.
Key ideas
- The coefficient groupings come from expanding time-dependent drift and volatility terms around maturity.
- Expanding squared volatility produces a first-order time correction involving its time derivative.
- The drift coefficient receives a similar Taylor correction around maturity.
- The stated rate-derivative term requires maturity-time powers consistent with the series expansion.
- The final pricing term should use the second-order series coefficient identified in the response.
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Full text
# Yield curve fitting example in Wilmott on Quant Finance p.528
# Yield curve fitting example in Wilmott on Quant Finance p.528
In Wilmott on Quantitative Finance Vol. 2, p. 528, Section 31.4.2, is given a power series expansion for a zero coupon bond
$$Z(r,t;T)=1+a(r)(T-t)+b(r)(T-t)^2+c(r)(T-t)^3+\dots$$
then it says to substitute this into the bond pricing equation, which is, of course
$$Z_t+\frac{1}{2}w^2Z_{rr}+(u+\lambda w)Z_r-rZ=0$$
the result of which is given as
$$ -a-2b(T-t)-3c(T-t)^{2}+\frac{1}{2}\left(w^{2}-2(T-t)w\frac{\partial w}{\partial t}\right)\left(\left(T-t\right)\frac{\partial^{2}a}{\partial r^{2}}+(T-t)^{2}\frac{\partial^{2}b}{\partial r^{2}}\right) $$ $$ \text{ }+\left(\left(u-\lambda w)-(T-t\right)\frac{\partial\left(u-\lambda w\right)}{\partial t}\right)(T-t)\left(\frac{da}{dr}+\left(T-t\right)^{2}\frac{db}{dr}\right)-r\left(1+a(T-t)+c(T-t)^{2}\right)+\dots=0 $$
My conundrum, for starters, is that I don't know how the first parenthical grouping in each of
$$\left(w^{2}-2(T-t)w\frac{\partial w}{\partial t}\right)\left(\left(T-t\right)\frac{\partial^{2}a}{\partial r^{2}}+(T-t)^{2}\frac{\partial^{2}b}{\partial r^{2}}\right)$$
and
$$\left(\left(u-\lambda w)-(T-t\right)\frac{\partial\left(u-\lambda w\right)}{\partial t}\right)(T-t)\left(\frac{da}{dr}+\left(T-t\right)^{2}\frac{db}{dr}\right)$$
terms came to be (and the latter line, it seems like errata for the partial in $r$, shouldn't it be $\left((T-t)\frac{da}{dr}+\left(T-t\right)^{2}\frac{db}{dr}\right)$ not preceded by the $(T-t)$ term (because when it distributes it will give the wrong power for $(T-t)$ on the $\frac{db}{dr}$ term?
When I compute partial derivatives with respect to $t$, $r$, and $rr$ of $Z$ I don't get the same result for the partials of $r$ and $rr$. Obviously, some steps are missing. I do get the same first three terms as that given above, due to $Z_t$, but after that it diverges from the answer given until the $-rZ$ term at the end of the bond pricing equation (though, again, I do think it should have been a $b(T-t)^2$ instead of $c(T-t)^2$ in that last term.
I know this is asking a bit much perhaps, but any help is appreciated. Thanks in advance.
## Answer by Theja Tulabandhula (score 6, accepted)
https://quant.stackexchange.com/a/9132
Your observations are pretty much correct.
The groupings are because of the fine print "Note how I have expanded the drift and volatility terms at $t = T$; in the above these are evaluated at $r$ and $T$." on the same page (p.528).
Basically, $w$ is a function of both $r$ and $t$. Since we want to use $w(r,T)$ instead of $w(r,t)$, we taylor expand $w(r,t)$ around $w(r,T)$ with respect to $t$. The same is true for functions of $w(r,t).$ Thus,
\begin{align} w(r,t)^2 &= w(r,T)^2 + (t-T)\frac{\partial w(r,T)^2}{\partial t}\\ & = w(r,T)^2 -2 (T-t)w(r,T)\frac{\partial w(r,T)}{\partial t} \end{align}
The same is true for $u(r,t) - \lambda w(r,t)$ where now $u$ may also be a function of $r$ and $t$: \begin{align} u - \lambda w(r,t) &= (u(r,T) - \lambda w(r,T)) + (t-T)\frac{\partial (u(r,T) - \lambda w(r,T))}{\partial t}\\ &= (u(r,T) - \lambda w(r,T)) - (T-t)\frac{\partial (u(r,T) - \lambda w(r,T))}{\partial t} \end{align}
I hope this solves your conundrum.
### Other points
- Yes there is an errata: it should be $\left((T-t)\frac{da}{dr}+\left(T-t\right)^{2}\frac{db}{dr}\right)$.
- It should have been a $b(T-t)^2$ instead of $c(T-t)^2$ in the last term as you point out at the end of your question.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.