Converting Simple LIBOR Quotes to Continuous Compounding
Summary
This Q&A clarifies that a three-month LIBOR quote is an annualized simple rate, rather than a continuously compounded rate. Interest accrual uses the applicable day-count convention; the example assumes Actual/360 for the LIBOR accrual period. To express that accrual as an equivalent continuously compounded rate on an Actual/365 basis, it equates the simple growth factor over the period with an exponential growth factor and solves for the continuous rate. That converted rate can then be applied to a shorter interval using the selected basis.
The answer illustrates the conversion with a low quoted rate and a sample period, observing that under those conditions the simple and continuous rates are close. A second response emphasizes that neither proposed exponent denominator alone captures the convention exactly: the rate is annualized, but simple compounding and specific day counts determine the accrual factor. The numerical result depends on the assumed period length and day-count bases, so these should be checked for the instrument and market in question.
Key ideas
- LIBOR is quoted as an annualized simple rate rather than a continuously compounded rate.
- The accrual factor depends on the relevant day-count convention.
- An equivalent continuous rate can be found by matching growth factors over the same period.
- The converted rate can be applied to a shorter interval using its chosen time basis.
- The numerical conversion depends on the assumed accrual length and day-count conventions.
Tags
Full text
# Is 3-month LIBOR quoted in annual terms?
# Is 3-month LIBOR quoted in annual terms?
If 3 month LIBOR is 0.22% and I want to find the interest rate with continuous compounding for 10 days do I multiply my principal by `e^(10*r/90)` or `e^(10*r/365)`?
## Answer by Dom (score 4)
https://quant.stackexchange.com/a/59225
Libor interest payments are determined using simple annualised compounding rather than continuous compounding (using time in units of years). To find the equivalent interest rate with continuous compounding, you would solve for $r_c$ where the Libor rate is for 91 days and I assume the Libor has an Actual/360 basis convention. To calculate the elapsed time for the continuous compounding, I decide to use Actual/365. This gives:
$\exp \left( \frac{91}{365} \times r_c \right) = \left( 1 + \frac{91}{360} \times \frac{0.22}{100} \right)$
So
$r_c = \frac{365}{91} \times \ln \left( 1 + \frac{91}{360} \times \frac{0.22}{100} \right)$
which gives
$r_c = 0.222994\%$
The low level of interest rates and the short time period mean that the Libor rate and continuously compounded rate are almost identical.
You can then apply this for 10 days (once again using Actual/365 for calculating times) to get a growth factor of
$\exp \left( \frac{10}{365} \times \frac{0.222994}{100} \right) = 1.000061096$
## Answer by piterbarg (score 3)
https://quant.stackexchange.com/a/59221
well, strictly speaking neither, but the second answer gets you closer to the truth, as Libor is indeed quoted in annual terms. However it is not quoted as continuously-compounded but as simply compounded. Eg in your example a 3M compounding factor would be $1 + (1/4) \times 0.22\%$ where the factor 1/4 is also somewhat approximate as in reality specific day counting rules are used. See eg hereShown 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.