# 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 $$y_{1}, then $$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.

• What do you mean by ‘kth period spot rate’ pls?
– dm63
Commented Mar 5, 2022 at 16:32
• It looks like you have defined it as the kth period forward rate actually
– dm63
Commented Mar 5, 2022 at 17:44
• @dm63 S(k) is the yield to maturity of a $k$-period zero-coupon bond. Commented Mar 6, 2022 at 0:14

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 $$y_1 and $$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.
• 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.