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Solving a Coupon Bond Yield with a Change of Variables

Article Quant Q&A · Author: mindandfields
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Summary

The document shows how to solve for the continuously compounded yield of a two-year coupon bond when its price is expressed as the discounted value of the coupon payments and principal. The suggested substitution sets the exponential discount factor equal to a new variable, turning the pricing equation into a quadratic. After solving for the positive root, the yield is recovered by taking the negative natural logarithm.

The explanation gives an algebraic route for a bond priced at 98 with a 3.5 coupon and 100 face value, but it does not work through the quadratic or report a yield. The approach applies when the cash flows and discounting convention match the stated equation; other coupon schedules or compounding conventions require a different setup. The document also does not discuss yield selection beyond noting that the discount factor must be positive.

Key ideas

  • Set the exponential discount factor to a new positive variable to simplify the bond equation.
  • The resulting equation is quadratic in the discount factor.
  • Convert the positive solution back to yield with the negative natural logarithm.
  • The method depends on using the cash flows and continuous compounding assumed in the equation.

Tags

Fixed incomeStatistics
Full text
# Calculating coupon yield and continous compounding


# Calculating coupon yield and continous compounding












I need to calculate the yield of a 2 year Coupon Bond. Price = 98, Coupon = 3.5, N = 100.

Now when I try to solve this, I arrive at the equation: $$ 98 = 3,5*e^{-y}+103,5*e^{-2*y} $$ But I can't figure out how to solve this equation for the yield y. I tried with Wolfram Alpha but don't know how to interpret the result. Is my math just too rusty or is my approach wrong?

## Answer by ir7 (score 3, accepted)

https://quant.stackexchange.com/a/55478

Hint: Let $$z = \mathrm{e}^{-y} $$

That way you get a quadratic equation in $z$ (note that $z$ is positive) and then you can get back to $y$ using:

$$ y = -\ln (z) $$

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.