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I have a little question and need some help with the notation. So, the question goes as follows:

A bond with a maturity of ten years that pays annual coupons of 8% has a price of \$90. A bond with a maturity of ten years and annual coupons of 4% has a price of \$80. What is the ten year zero rate?

I don't actually know what the ten-year zero rate is. I set up a system of equations with the information that is given: Is it just all about finding $y$?

\begin{align*} \$90 =& \sum_{i=1}^{10} e^{-yi} (0.08 Z) &+& e^{-y\cdot 10}Z\\ \$80 =& \sum_{i=1}^{10} e^{-yi} (0.04 Z) &+& e^{-y \cdot 10}Z\\ \Leftrightarrow \$10 =& \sum_{i=1}^{10} e^{-yi} (0.04 Z)\\ \Rightarrow \$70 =& e^{-y\cdot 10}Z \end{align*}

And is there any way to solve for the zero rate by hand? (This is the reason why I'm wondering; to solve this system, a calculator is necessary, but all the other homework problems were solvable by hand!)

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I'm especially curious now because, when I try to solve this system, I only get $y\approx -0.311$, and I don't think that $y$ should have a negative value here... – Marie. P. Sep 16 '12 at 21:37
up vote 2 down vote accepted

The way you are trying to solve these equations makes assumptions about the rates less than 10 years and therefore the shape of the yield curve. \$90 is the value of 8% coupons plus a 10-year zero-coupon bond. \$80 is the value of the 4% coupons plus a 10-year zero-coupon bond. 8% coupons are worth twice 4% coupons over the same period, regardless of the interest rates.

So 90 - 2*80 = value of 8% coupons - 2( value of 4% coupons) - 10-year zero-coupon bond


\begin{align*} \$70 &= e^{-10y} \cdot 100 \\ 0.7 &= e^{-10y} \\ \ln(0.7) &= -10 y \\ 0.35667 &= 10y \\ y &\approx 3.57 \% \quad\quad\\ \end{align*}

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Okay, so that confirms my initial thought, but how exactly do you get $\approx 3.57\%$, because I only get a value of $\approx -0.311$ when I type this into Mathematica as follows: $\verb{NSolve[{0.04*Z*Sum[Exp[-y*k], {k, 1, 10}] == 10,70 == Exp[-y*10]*Z},{y, Z}]}$ Does anyone see what I do wrong? – Marie. P. Sep 17 '12 at 12:56
There is no summation to solve. I added (corrected) computation. The only thing you need a calculator for is computing ln(0.7). – David Nehme Sep 17 '12 at 15:15
I forgot that we always know that $Z=100$. Thank you very much. – Marie. P. Sep 18 '12 at 7:08
How do you know Z = 100? ...... – Caroline Jun 2 at 14:47
@Caroline That's standard for any bond. – Chris Taylor Jun 3 at 7:35

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