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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.

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  • $\begingroup$ What do you mean by ‘kth period spot rate’ pls? $\endgroup$
    – dm63
    Mar 5 at 16:32
  • $\begingroup$ It looks like you have defined it as the kth period forward rate actually $\endgroup$
    – dm63
    Mar 5 at 17:44
  • $\begingroup$ @dm63 S(k) is the yield to maturity of a $k$-period zero-coupon bond. $\endgroup$
    – cici30725
    Mar 6 at 0:14

1 Answer 1

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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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  • $\begingroup$ also one can see that $y_k$ is a complex average of the $S_i$ , for i=1 to k, so $y_k<S_k$ follows if you can show that $S_i$ is increasing. $\endgroup$
    – dm63
    Mar 7 at 16:22

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