Page 1 of 1

Simple Compounding, why?

Posted: December 10th, 2013, 7:47 pm
by Cuchulainn
Take a 1-period (length dt) approximation to exp(-r dt), i.e. the popular1 / (1 + r dt)Questions:1. Why this? Is there a financial reasoning or is it just numerical convenience?2. If we had a more accurate and more stable approximation, is that 'financially acceptable'?3. What do you think of this approximation to exp(-r dt)?(1 - r dt) / (1 + r dt)

Simple Compounding, why?

Posted: December 10th, 2013, 8:05 pm
by Traden4Alpha
Isn't it just historical path dependency? Given principal, rate, and time (P, r, dt), the simplest formula for interest to be paid on an account is P*r*dt which is then added to P. When I was little, we got monthly compounding of simple interest and we were darn glad of it, too! Then, in school, we learned that the exponential form was almost magical mathematical formula that results from getting interest-on-the-interest-on-the-interest-on-the-... and it seemed almost sneaky or subversive. P.S. I'd avoid division and more complicated maths!

Simple Compounding, why?

Posted: December 10th, 2013, 8:16 pm
by Cuchulainn
No.Both forms are simple rational functions. So, what is the compelling rationale (e.g. convention, QED), then fair enough. BTW these rational functions are in a yield curve fitting routine. For zero coupon bond.

Simple Compounding, why?

Posted: December 10th, 2013, 8:32 pm
by Traden4Alpha
QuoteOriginally posted by: CuchulainnBoth forms are simple rational functions. So, what is the compelling rationale (e.g. convention, QED), then fair enough.I'm not joking about division. Moreover, the simpler method can be done on a four-function calculator (no stack or memory) and the other cannot. In a pre-computer, pre-numerate environment, only addition and multiplication were possible (and even the multiplication might be handled by an interest rate table).That's not to say a better approximation won't be appreciated. It's just a rational explanation for why rational functions weren't acceptable historically.What about Taylor series? I think we have a plain English explanation of it somewhere here.

Simple Compounding, why?

Posted: December 11th, 2013, 2:14 am
by DavidJN
The simple (i.e. non-compounded) method is not an approximation, it is an economically equivalent convention that is used in the marketplace for many financial instruments with maturities under one year. You would not use the same numerical value for the interest rate in both equations. Changing your notation ever so slightly, denote rs the simple rate and rc the continuous compound rate.1/(1 + rs dt) = exp(-rc dt)If you know one of the two rates you can solve for the other. Different numerical rates using different mathematical conventions that yield the same discount factor. Equivalent transforms.Few, if any financial contracts use continuous compounding, they instead either use simple compounding or discrete compounding (e.g. semi-annual). Continuous compounding is a theorist's way of looking at things, a mathematical and maybe even programming convenience at times.

Simple Compounding, why?

Posted: December 11th, 2013, 8:31 am
by ppauper
QuoteOriginally posted by: Cuchulainn3. What do you think of this approximation to exp(-r dt)?(1 - r dt) / (1 + r dt)not much[$] exp(-r\,dt))\approx 1 - r\,dt[$][$](1 - r\,dt) / (1 + r\,dt)\approx 1 - 2r\,dt[$]

Simple Compounding, why?

Posted: December 11th, 2013, 10:49 am
by Cuchulainn
QuoteOriginally posted by: ppauperQuoteOriginally posted by: Cuchulainn3. What do you think of this approximation to exp(-r dt)?(1 - r dt) / (1 + r dt)not much[$] exp(-r\,dt))\approx 1 - r\,dt[$][$](1 - r\,dt) / (1 + r\,dt)\approx 1 - 2r\,dt[$]Both forms are special cases of Pade(p,q) rational approximants, P(0,1) O(dt^2) and P(1,1) O(dt^3), respectively. (for PDE/FDM they correspond to fully implicit and Crank Nicolson, respectively, well established).PadeP(1,1) is more accurate and robust than P(0,1) which is nice but if simple compounding is agreed on financially then the results that P(1,1) produces will be more accurate and at the same time "wrong"? (?). A different value will be calibrated for the SDE.This goes back to the remarks of DavidJN concerning the numerical Vs programming Vs financial interpretation of this formula. // I put it in Student because #questions > #answers (speaking for myself)

Simple Compounding, why?

Posted: December 11th, 2013, 1:22 pm
by DavidJN
You are over thinking this to a staggaring degree. I repeat, this is not an approximation, just a transformed way (i.e. a particular convention out of many) of getting to the same present value factor.I also repeat that the exp(-r dt) form (i.e. continuously compound interest rate) is not something found in actual financial contracts but rather in textbooks. There is nothing stopping you from using that form (or any other form) of computing present values so long as you adjust the rate accordingly.

Simple Compounding, why?

Posted: December 11th, 2013, 3:00 pm
by cemil
QuoteOriginally posted by: CuchulainnTake a 1-period (length dt) approximation to exp(-r dt), i.e. the popular1 / (1 + r dt)Questions:1. Why this? Is there a financial reasoning or is it just numerical convenience?2. If we had a more accurate and more stable approximation, is that 'financially acceptable'?3. What do you think of this approximation to exp(-r dt)?(1 - r dt) / (1 + r dt)Imagine you have 1euro with rd (discrete rate) for dt period, so at time dt you will have 1x(1+rd dt)Imagine you have 1euro with rc (continuous rate) for dt period, so at time dt you will have 1xexp(rc dt)If we suppose we have the same money at the date dt, we can write 1x(1+rd dt)=1xexp(rc dt)<=> exp(-rc dt)= 1/(1+rd dt)So it isn't an approximation.

Simple Compounding, why?

Posted: December 11th, 2013, 4:47 pm
by Cuchulainn
Clear. Thanks all.

Simple Compounding, why?

Posted: December 11th, 2013, 6:22 pm
by ppauper
QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: ppauperQuoteOriginally posted by: Cuchulainn3. What do you think of this approximation to exp(-r dt)?(1 - r dt) / (1 + r dt)not much[$] exp(-r\,dt))\approx 1 - r\,dt[$][$](1 - r\,dt) / (1 + r\,dt)\approx 1 - 2r\,dt[$]Both forms are special cases of Pade(p,q) rational approximants, P(0,1) O(dt^2) and P(1,1) O(dt^3), respectively. (for PDE/FDM they correspond to fully implicit and Crank Nicolson, respectively, well established).PadeP(1,1) is more accurate and robust than P(0,1) which is nice but if simple compounding is agreed on financially then the results that P(1,1) produces will be more accurate and at the same time "wrong"? (?). A different value will be calibrated for the SDE.This goes back to the remarks of DavidJN concerning the numerical Vs programming Vs financial interpretation of this formula. // I put it in Student because #questions > #answers (speaking for myself) you might want to put a factor of 1/2 in one....[$] exp(-r\,dt))\approx 1 - r\,dt[$][$](1 - r\,dt/2) / (1 + r\,dt2)\approx 1 - r\,dt[$][$](1 -\alpha r\,dt) / (1 + (1-\alpha) r\,dt)\approx 1 - r\,dt[$]

Simple Compounding, why?

Posted: December 11th, 2013, 8:17 pm
by AVt
I *guess* (like T4A) that it is historical, in German "Zins" and "Zehent"may even have common roots - may be you dig in Etymology and Bible texts:https://de.wikipedia.org/wiki/Zinsverbo ... stamentYou may find that "Zins für Speise" (~ in the middle of the screen) can notmean rates, it means payment for something (here: food for a 'brother').So I guess it started from simple rules.And they still exists in laws: for example the german law "BGB" uses it (andforbids compounding) for some kinds of debts. So I would say, it is the other way round: compounding naturally leads toto 'exponential' growth, using exp or similar (when it became 'handy' for the folks in Finance).

Simple Compounding, why?

Posted: December 12th, 2013, 8:36 am
by Cuchulainn
QuoteOriginally posted by: ppauperQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: ppauperQuoteOriginally posted by: Cuchulainn3. What do you think of this approximation to exp(-r dt)?(1 - r dt) / (1 + r dt)not much[$] exp(-r\,dt))\approx 1 - r\,dt[$][$](1 - r\,dt) / (1 + r\,dt)\approx 1 - 2r\,dt[$]Both forms are special cases of Pade(p,q) rational approximants, P(0,1) O(dt^2) and P(1,1) O(dt^3), respectively. (for PDE/FDM they correspond to fully implicit and Crank Nicolson, respectively, well established).P(1,1) is more accurate and robust than P(0,1) which is nice but if simple compounding is agreed on financially then the results that P(1,1) produces will be more accurate and at the same time "wrong"? (?). A different value will be calibrated for the SDE.This goes back to the remarks of DavidJN concerning the numerical Vs programming Vs financial interpretation of this formula. // I put it in Student because #questions > #answers (speaking for myself) you might want to put a factor of 1/2 in one....[$] exp(-r\,dt))\approx 1 - r\,dt[$][$](1 - r\,dt/2) / (1 + r\,dt/2)\approx 1 - r\,dt[$] // edit DD[$](1 -\alpha r\,dt) / (1 + (1-\alpha) r\,dt)\approx 1 - r\,dt[$]You are correct. I wrote it down too fast. In fact, exp(- z) is handily written as (2 - z)/(2 + z)// I put a '/' into your formulaVarga's classic book "Matrix Iterative Analysis" has a nice table for Pade(p,q) page 266.