Why a Par-Coupon Bond Has a Conversion Factor of One

Summary

The note verifies that a bond whose coupon rate equals its yield has a conversion factor of one in the stated setup. It expresses the bond’s price as the present value of its coupon payments and principal, then divides by par value. With coupon and yield both at six percent, the discounted principal is the reciprocal of the yield factor raised to maturity, while the coupon stream forms a geometric sum.

The geometric sum simplifies to one minus the discounted-principal term. Adding that amount to the discounted principal gives one, establishing the result for any stated maturity N. This is an algebraic illustration of a bond-pricing identity rather than a discussion of futures conversion-factor conventions in broader market practice. It assumes the coupon frequency, discounting, and yield conventions implicit in the formula; different payment schedules or quoting rules would require corresponding adjustments.

Key ideas

• When coupon equals yield, the bond is priced at par under the stated assumptions.
• The coupon cash flows form a geometric series when discounted at the yield.
• The coupon present value and principal present value sum to par value.
• The proof relies on the payment and discounting conventions encoded in the formula.

Tags

# Conversion factor for bond with coupon=yield


# Conversion factor for bond with coupon=yield












Please illustrate that a bond with maturity N years that has coupon equal to its yield is associated with the conversion factor of 1.

I do this by writing out $$\frac1{100} \left( \sum_{t=1}^N \left[ \frac{100 (0.06)}{1.06^t} \right]+\frac{100}{1.06^N} \right)$$ but I do not get that this = 1.

I use the formula:

$$\sum_{k=m}^n a^k = \begin{cases}\frac{a^{n+1} - a^m}{a-1}, \quad &a \neq 1\\n-m+1, \quad &a=1\end{cases}$$

## Answer by jacob (score 2, accepted)

https://quant.stackexchange.com/a/20729

We are given a bond with Coupon = Yield = $6 \%$ and Maturity $N$. We want to check that the conversion factor = 1, in other words that $$\frac1{100} \left( \frac{100}{1.06^N} + \sum_{t=1}^N \frac{100 \cdot 0.06}{1.06^t} \right) = 1 $$ or equivalently $$ \frac{1}{1.06^N} + \sum_{t=1}^N \frac{0.06}{1.06^t} = 1. $$

The first term is $\frac{1}{1.06^N} = \left( \frac{1}{1.06} \right)^N.$ The second term can be simplified using the formula for geometric sums $$\sum_{t=1}^N \frac{0.06}{1.06^t} = 1 - \left( \frac{1}{1.06} \right)^N.$$ Now we add the first term and the second term and see that $$\left( \frac{1}{1.06} \right)^N \quad + \quad 1 - \left( \frac{1}{1.06} \right)^N = 1. $$

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