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Relating Coupon-Bond Yields to Spot Rates on an Upward Curve

Article Quant Q&A · Author: cici30725

Summary

The document considers a fixed-income claim about coupon bonds: when yields to maturity increase with maturity, the maturity-matched spot rate is at least as high as the bond’s yield to maturity. It explains that a coupon bond’s price can be represented either by discounting its cash flows at one yield to maturity or by discounting each cash flow at its corresponding spot rate. The questioner notes that this equality alone did not yield a proof.

The answer emphasizes including the principal redemption payment in the price. It establishes equality of the one-period spot rate and yield, then compares the two-period bond’s cash flows under the two discounting representations to infer a higher two-period spot rate. The argument suggests induction for longer maturities but does not provide that full proof. The reasoning also relies on the stated upward ordering and the bond cash-flow setup; broader curve shapes and conventions are not covered.

Key ideas

  • A coupon bond can be priced using one yield to maturity or a set of maturity-specific spot rates.
  • The principal redemption payment must be included among the bond cash flows.
  • For a one-period bond, the spot rate equals its yield to maturity.
  • The answer uses a two-period comparison to suggest an induction proof, but does not complete it.

Tags

Full text
# Spot rate dominates the yield to maturity if the yield curve is normal


# Spot rate dominates the yield to maturity if the yield curve is normal












> Let $y_{k}$ denote the yield-to-maturity of a $k$-period coupon bond. Let $S(k)$ denote the $k$-th period spot rate. If $y_{1}<y_{2}<y_{3}<\cdots$, then $S(k)\geq y_{k}$ for all $k\in \mathbb{N}$.

I approached this problem by first considering the price of a $k$-period bond can be priced as both $\sum_{i=1}^{k} \frac{C}{(1+y_{k})^{i}}$ and $\sum_{i=1}^{k} \frac{C}{(1+S(i))^{i}}$, where $C$ is the cash flow for a period. Thus $\sum_{i=1}^{k} \frac{C}{(1+y_{k})^{i}} = \sum_{i=1}^{k} \frac{C}{(1+S(i))^{i}}$.I tried to use induction and some other ways, but I can't seem to prove that $S(k)\geq y_{k}$.

Any help is appreciated.

## Answer by Kurt G. (score 1)

https://quant.stackexchange.com/a/70086

OK: $S_k$ is the yield to maturity of a $k$-period zero coupon bond. I would however include the redemption term into the bond price.

For $k=1$ $$ P=\frac{C}{1+y_1}+\frac{1}{1+y_1}=\frac{C}{1+S_1}+\frac{1}{1+S_1} $$ This implies $S_1=y_1\,.$ For $k=2\,,$ $$ P=\frac{C}{1+y_2}+\frac{C}{(1+y_2)^2}+\frac{1}{(1+y_2)^2}=\frac{C}{1+S_1}+\frac{C}{(1+S_2)^2}+\frac{1}{(1+S_2)^2}\,. $$ Since $y_1<y_2$ and $S_1=y_1$ we see that $$ \frac{C}{1+y_1}+\frac{C}{(1+y_2)^2}+\frac{1}{(1+y_2)^2}>\frac{C}{1+S_1}+\frac{C}{(1+S_2)^2}+\frac{1}{(1+S_2)^2}\,, $$ or $$ \frac{C}{(1+y_2)^2}+\frac{1}{(1+y_2)^2}>\frac{C}{(1+S_2)^2}+\frac{1}{(1+S_2)^2}\,. $$ This clearly implies $S_2>y_2\,.$ This suggests that $S_k\ge y_k\,\,\forall k$ can be proved by induction.

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