Which one is larger, the continuously compunded return of an investment, or its annually compounded return?
I think the second one. For the same investment, these different yields will end up giving the same return. One is multiplied continuously, one only once. That means that the latter must be greater than the former so that it cancels out at the end. No?
Akukuu is correct; the R_cc will be smaller than the R_ac for the reason listed above.
The larger the number of subperiods in the period, the greater is the compounded return. I would say the continuously compounded is greater than the annually. Let’s try an example: initial value of investment: 1000 grows to 1100 at the end of the first year grows to 1500 at the end of the second year drops to 1300 at the end of the third year compound annual growth: (Ending value/Beginning value)^(1/number of years)-1=(1300/1000)^(1/3)-1=9.139% continuous compounding: ln(Ending value/Beginning value)=26.236%
It is counter intuitive, but here is how it works: Rc : return compounded on a continuous basis Ra: return compounded on an anuual basis Rc = Ln (1+Ra), Ln = natural log If Ra=10%, then Rc = 9.53%. yet, I still don’t see how this is true!
Also trying to understand the relationship here. 5% annual return compounded semi-annually (1+.025)^2 -1 = 5.0625% 5% compounded continuously e^.05 - 1 = 5.1271% Then, Initial Value Beginning = 1000 Initial Value Time1 = 2000 Return = 2000/1000 -1 = 100% Contrast to: ln (2000/1000) = 69.31%
Compounding is interest on the principal plus re-invested interest. The more your compound, in effect, the higher the return as interest is being reinvested more often or continously.
Here is what is meant. You bought a stock for $10, and sold it a year later for $15, so your annual compounded return is 50% (there is only 1 year, so there is no effect from compunding). For this *same* stock, what is the continuously compounded rate? $10 *e^r = $15, solving for r we get: e^r = 1.5 r = Ln (1.5) = 40.5%, which is less than the annually compounded return.
There is another way to prove that the annually compounded rate is higher that the continuously compounded rate: Using BaII+ : 2nd ICONV (number 2) Arrow down Eff = effective rate = 50 (following the example given by dreary) Enter Arrow down C/Y = compouding periods / year, since we want the continuously coumpeded rate, just plug a high number ie. 50000 Enter arrow down Nom CPT ==> 40.5%
However, do not confuse that with this: Clerk: Sir, we offer a 10% return. Would you like that to be compunded annually or continuously? You: Well, folks at AF said that the continuously compounded rate is less than the annually compounded rate, so I’ll go with annual compounding. Clerk: What a guy!
That’s not the continuously compounded. That’s the APR (nominal) compared to EAY(effective). Dreary, I guess this the the exception: if you have only 1 period, than yes, the annual is greater than the continuously compounded. Otherwise, no.
map1, here it is with 3 years: You bought a stock for $10, and sold it 3 years later for $15, so your annual compounded return is 14.47%. For this *same* stock, what is the continuously compounded rate? $10 *e^3r = $15, solving for r we get: e^3r = 1.5 3r = Ln (1.5), r =13.51%, which is less than the annually compounded return.
Check the table 1 on top of page 182 of Volume 1 CFAi text.
map1 Wrote: ------------------------------------------------------- > The larger the number of subperiods in the period, > the greater is the compounded return. I would say > the continuously compounded is greater than the > annually. > > Let’s try an example: > > initial value of investment: 1000 > grows to 1100 at the end of the first year > grows to 1500 at the end of the second year > drops to 1300 at the end of the third year > > compound annual growth: (Ending value/Beginning > value)^(1/number of > years)-1=(1300/1000)^(1/3)-1=9.139% > > continuous compounding: ln(Ending value/Beginning > value)=26.236% Map, I think your calculation is wrong in this example To make calculation simple, I use holding period return (HPR) as bridge compound annual rate = (1+HPR) ^1/3 -1 = (1300/1000)^1/3 -1 =0.0914 Rcc = Ln [(1+HPR)^1/3] = Ln 1.0914 = 0.0875 annual compound > Rcc, it applies generally.
forgot the N, uh?
for the 10 annual return to 9.53 compounded continuous … amount obtained by compounding at 10% (a greater rate) is equivalent to amount obtained by compounding at 9.53% (a smaller rate) simply cuz the compounding intervals are far too many. thus one may say continous compounding at the ARITHMETICALLY SAME RATES will produce the HIGHER returns than for annual compounding as for Dreary there… I’d like to think the Clerk was indeed WRONG in asking that. Compounded continuously, 10% would give much more dollar returns. Pls do correct me if im wrong.
The clerk is not necessarily ethical.