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

Question

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 1

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.