Continuously Compounded Returns

I’m in a financial “math” class that I really shouldn’t have taken (it’s in the business school) and our lecture notes the other day included some incredibly basic properties about interest calculations.  Anyway, I told one of my friends that it was super fucking easy to derive the formula for continuous interest by taking the limit as the number of compounding periods approaches infinity.  Upon further review, I stand by the fact that it is conceptually easy to do this, but there are some computational issues that one could run into – for example, I had to use a relatively simple substitution in order to avoid using L’Hopital’s rule or something else messy like the definition of a difference quotient, etc.  But, regardless, below is the derivation, starting with the definition of the effective annual rate for an interest rate, r, compounded n periods per year for t years.  Suppose there are n compounding periods per year and r is the interest rate.

Screen shot 2013-09-17 at 2.46.18 AM

How do we show that the effective annual rate under continuously compounded interest (i.e. effective interest with arbitrarily large n) is er – 1?  We need to show the equation below, which equates the limit of the EAR as the compounding periods approach infinity to the formula we claimed represents the continuously compounded rate (Exp(rt)-1):

Screen shot 2013-09-17 at 2.45.36 AM

If I were being tested on this in a math class for whatever reason, I might not show the continuously compounded interest rate this way.  This is more of an intuitive justification than a real proof.  To see a real proof, you can check out notes from Wharton here – it’s not really any “harder” per se, but some of the steps might be less obvious.  For example, the proof involves taking the natural logarithm of both sides of the equation (as shown below) which serves a purpose, though it may not seem like it at first.

Screen shot 2013-09-17 at 2.59.49 AM

This was done in order to take advantage of a useful property of logarithms: the log of a product is the sum of the logs.  Using this, the right hand side is split up into a sum and the first term no longer depends on the limit because it doesn’t have an n in it.  You’re left with a sort of messy equation that involves logs on both sides (it’s implicitly defined), so to rectify that you just need to remember that e and the natural log are inverses, so e raised to the power ln is 1 and the ln of e is 1.

Note: just so you don’t sound like an asshole, e is Euler’s number, and it is pronounced like “oiler”.

Another note:  There are tons of definitions of e, but the one I think of first when I think of e is the one below:

Screen shot 2013-09-17 at 3.06.44 AM

In words, e is the positive real number such that the integral from 1 to e of the function 1/t is equal to 1.  Equivalently, the area of the integral of 1/t from 1 to e is 1.  ln(x) crosses the x-axis when x=1 (since ln(1)=0), so the area in between the function ln(x) and the x-axis from its x-intercept to x = e is 1.  Obviously, ln(x) takes on the value 1 when x=e since ln(e)=1.  I’m not good at making fancy graphs online, but below is a heinous picture of the point I’m trying to convey.  Sorry Euler 😦

Screen shot 2013-09-17 at 3.16.06 AM

About schapshow

Math & Statistics graduate who likes gymnastics, 90s alternative music, and statistical modeling. View all posts by schapshow

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