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Choosing a Discount Factor for Fractional-Year Periods

Article Quant Q&A · Author: Eva Dahlbeck

Summary

The discussion compares two discount factors for a zero-coupon rate held constant over whole years plus a fractional year. One approach compounds the rate continuously across the full elapsed term in discrete compounding form; the other compounds for the whole years, then applies simple interest to the remaining fraction. The answer expresses both formulas and explains that they coincide when the fractional period reaches a full year.

The choice depends on the convention intended for that partial period: use the simple-interest form when the fraction is treated with simple interest, or the compound-interest form when compounding continues through the fraction. The response observes that for a fraction between zero and one, the two values will be close, but gives no numerical comparison. In practice, the governing market convention and rate definition determine which factor is appropriate; proximity does not make the conventions interchangeable.

Key ideas

  • A fractional-year discount factor depends on whether the partial period uses simple or compound interest.
  • The simple-interest form applies a linear rate adjustment to the fractional year.
  • The compound-interest form raises the accumulation factor to the fractional-year exponent.
  • The two expressions agree when the fractional period is one full year.
  • The applicable convention, including day count, determines which expression to use.

Tags

Full text
# Discount factor


# Discount factor












Suppose we have :

$r$ - zero coupon rate, constant over time,

$n$ - a number of years (an integer),

$\theta$ - a fraction of a year $(\theta < 1)$ , calculated with the relevant day count convention.

Which discount factor is the correct one ?

(I'm inclined to think it's $\beta_{2}$ because over a period of time of less than a year, it's a simple interest that is computed).

$\beta_{1} = \frac{1}{(1+r)^{(n+\theta)}}$

$\beta_{2} = \frac{1}{(1+r)^n}\cdot \frac{1}{(1 + r\theta)}$

## Answer by Pandaaaaaaa (score 2)

https://quant.stackexchange.com/a/25166

You have

$\beta_1=\frac{1}{(1+r)^n}\frac{1}{(1+r)^\theta}$

and

$\beta_2=\frac{1}{(1+r)^n}\frac{1}{(1+\theta r)}$.

Both are equal when $\theta=1$. If you consider simple interest then go for $\beta_2$. If you would like compound interest within fraction of year then pick $\beta_1$.

However, because $\theta$ is between 0 and 1 then values $\beta$'s won't be that different.

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.