Par Yield, Bond Yield and Zero Rate

Question

In the Hull's book, chapter 4.4, it says :

The par yield for a certain bond maturity is the coupon rate that causes the bond price to equal it's par value.

Then for this question (4.18) :

“When the zero curve is upward sloping, the zero rate for a particular maturity is greater than the par yield for that maturity. When the zero curve is downward sloping the reverse is true.” Explain why this is so.

the answer is :

The par yield is the yield on a coupon-bearing bond. The zero rate is the yield on a zero-coupon bond. When the yield curve is upward sloping, the yield on an N-year coupon-bearingbond is less than the yield on an N-year zero-coupon bond. This is because the coupons are discounted at a lower rate than the N-year rate and drag the yield down below this rate. Similarly, when the yield curve is downward sloping, the yield on an N-year coupon bearing bond is higher than the yield on an N-year zero-coupon bond.

• First, it's not very clear in my mind, is the par yield a coupon rate or a yield ?
• Secondly, i don't understand the answer, what does he mean by "drag the yield down below this rate" ? When he says "are discounted at a lower rate", of which rate does he talk about ? The bond's yield ?

Answer 1

Let's assume we have yearly cash flows, and let's focus on just two years - year 1 and year 2. Let $R_1$ and $R_2$ represent the zero rates of year 1 and year 2. So if you want to borrow for one year, you pay $R_1$ percent, and if you want to borrow for 2 years, you pay $R_2$ percent per year. So in an upward sloping scenario, these will look like this:

Now, you can easily convert these into the forward rates, so the first rate remains unchanged and the second rate would then be the interest rate for borrowing between year 1 and year 2, let's call it $R_{12}$ so the graph of the implied forward would be similar though the second point will be taller.

A 2-year maturity coupon bond will need to pay coupon rate of $R_1$ in the first year and a coupon rate of $R_{12}$ in the second year to be valued at par. But standard coupon bonds have fixed coupon rate (not variable by year), so the par coupon($c_2$, which is the par yield in the question wording), as you can see, will be some average of $R_1$ and $R_{12}$, which in an upward sloping environment will be lower than $R_2$. This $c_2$ is your 2-year par rate.

And you can extend the same logic to the third year and so on, to get the term structure of par yields, and if you plot it against the term structure of zero coupon, you see the c's will be lower than the R's (except for year 1 when they are equal), and this is what is meant.

Answer 2

This question came to me as well, the reasoning in the solution is not convincing, neither is the solution given by Magic.

In Magic's answer, the reasoning "$c_2$ is some average of $R_1$ and $R_{12}$, then lower than $R_2$" is incorrect since $R_{12}$ is bigger than $R_2$ in upward sloping.

We can prove it rigorously in the 2-year setting as below.

In continuous compounding setting, discrete par yield is computed by solving: $$ (100c)e^{-R_1}+(100c+100)e^{-2R_2}=100 $$ which gives: $$ c=\frac{1-e^{-2R_2}}{e^{-R_1}+e^{-2R_2}} $$ Next, convert c to continous par yield: $$c^*=ln(1+c)=ln\frac{1+e^{-R_1}}{e^{-R_1}+e^{-2R_2}}<ln\frac{1+e^{-R_2}}{e^{-R_2}+e^{-2R_2}}=ln(e^{R_2})=R_2$$

Answer 3

Just to test understanding of the reasoning:

$\frac{c}{1+R_{01}}+\frac{1+c}{\left(1+R_{01}\right)\left(1+R_{12}\right)}=1$

Multiply through by 1+c:

$c\frac{1+c}{1+R_{01}}+\frac{\left(1+c\right)^2}{\left(1+R_{01}\right)\left(1+R_{12}\right)}=1+c$

Now $\frac{1+c}{1+R_{01}}>1$ as per the reasoning in the previous answer because the curve is upward sloping, which means:

$\frac{\left(1+c\right)^2}{\left(1+R_{01}\right)\left(1+R_{12}\right)}<1$

which by definition means:

$\frac{\left(1+c\right)^2}{\left(1+R_{2}\right)^2}<1$

Answer 4

In discrete and 2-year setting, the question can be formulated to: given $R_1<R_2$, show $R_1<c<R_2$ from $$\frac{c}{1+R_1}+\frac{1+c}{(1+R_2)^2}=1.$$

This can be proven by contradiction:

1) assume $c<=R_1$, then $$1=\frac{c}{1+R_1}+\frac{1+c}{(1+R_2)^2}<\frac{R_1}{1+R_1}+\frac{1+R_1}{(1+R_1)^2}=1$$ 2) assume $c>=R_2$, then $$1=\frac{c}{1+R_1}+\frac{1+c}{(1+R_2)^2}>\frac{c}{1+c}+\frac{1+c}{(1+c)^2}=1$$