# Quant Question

Can somebody please explain the difference between: Continuous compounding: e^r - 1 = EAR and Continuously compounded rate of return: rcc = ln ( 1 + HPR ) Thanks, best of luck.

Use 10% as HPY e^.1 - 1 = 10.517% ln(1+.10) - 1 = 9.531% At t=0 I make a \$100 investment and earn 9.531% continuous compounding. At t=1 I will have \$110.00. ----> PV * e^(ln(1+HPY)) At t=0 I make a \$100 investment and earn 10% continuous compounding. At t=1 I will have \$110.51 ------> PV * e^HPY The relationship is e^(ln(1+HPY)-1) = 1+HPY

Thanks…I am still somewhat unclear as to the conceptual meaning of this?

(1+i/n)^n can be used to find yearly, monthly, qtly, daily, hourly, … effective rates. as n–> infinity (aka continuous compounding [charging interest every nanosecond]) (1+i/n)^n approaches “e”. e is useful because it answers the question: what is this annual rate equivalent to if I break it up into pieces (infinite pieces). 10% annually is not as good as 2.5% quarterly, and 2.5% qtly is not as good as .00027397 daily, and so on. For quarterly equivalent you still need to use (1+i/n)^n, but for “nanosecond” equivalent you can by-pass that formula and use e. e^.10 = 10.5170918, but quarterly is (1+.1/4)^4 which is 10.381128%. Now ln(1+rate) answers the question: if bank A is giving me less of a rate on my savings than bank B, at what point would it make sense to use bank A? Bank B gives me 10% annually, but bank B gives me 9.531% compounded continuously…which should I use? They are the same as you will end up with 1.1x your initial investment. ln(1+.10) tells you the ‘break-even’ point. If this doesn’t help, use this http://www.cmss.monash.edu.au/Events/Meeting08/Talks/WorkshopBjorkNotes.pdf