# 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)?

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

$$\exp \left( \frac{10}{365} \times \frac{0.222994}{100} \right) = 1.000061096$$

• (Albeit I don't think any actual Libors are quoted 30/360) Nov 9 '20 at 8:53
• True. I just wanted to keep the maths simple and get 0.25 rather than have 91/360. I might change it.
– Dom
Nov 9 '20 at 11:25

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 here