# Help with expansion of Continuously Compounded Interest Equation

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• Oct 4th 2012, 02:52 PM
georgop
Help with expansion of Continuously Compounded Interest Equation
Hello-

Was hoping for some insight into a solution for a specific issue...

I'm trying to estimate the earned value of a set of assets, where the known variables are the starting and ending values, and time.

At first, I thought the formula for continuously compounding interest (solving for interest rate) would be the ticket:

r = ln(A/P)^1/t

where:
A= final value,
P = starting value
t = time
r = interest rate

But then I realized that I am not dealing with a closed system-- i.e., the starting value is being incremented continuously during time period t... for example:

Start with 100 units; end with 1,000 units. Each unit creates interest. However, over time period t (say one year), 2 units were added to A every day. What was the interest earned on all the units deposited?

So I suppose I'm looking for some sort of series to describe this activity?

Forgive me if I'm not making sense... I figure there's a formula for this that would make life easy.. :)

Thank you all in advance.
• Oct 4th 2012, 03:01 PM
johnsomeone
Re: Help with expansion of Continuously Compounded Interest Equation
Do I understand:?
100 widgets on day 1. They'll earn interest for 365 days.
Then 3 widgets added on day 2. These will only earn interest for 364 days.
Then 3 more widgets added on day 3. These will only earn interest for 363 days.
Etc.
In the continuous approximation, f(t) = the number of items there at time t, which is increasing (you aren't removing items). All items are having their interest compounded instantaneously.
Is that your scenario?
Or do you need the number of items to be discrete? That means do you need it to look like 100, 103, 106, ..., or can it be described as f(t) = 100 + 3t.
• Oct 4th 2012, 03:12 PM
georgop
Re: Help with expansion of Continuously Compounded Interest Equation
Quote:

"In the continuous approximation, f(t) = the number of items there at time t, which is increasing (you aren't removing items). All items are having their interest compounded instantaneously."
Yes, this is what I need for right now...
• Oct 6th 2012, 05:29 AM
johnsomeone
Re: Help with expansion of Continuously Compounded Interest Equation
I'll tell you what I got (derived using Riemann sums in two different ways). It's definitely wrong, but also defintiely close to being right, as I'll explain. I'll then tell you what's a *guess* at correcting it, based on looking at step functions.

Principal at time t given by f(t). Interest compounded continuously at rate r.

Wrong answer: Total at time $T = re^{rT} \int_{0}^{T} f(\tilde{t})e^{-r\tilde{t}} d\tilde{t}$

It's wrong because for constant functions like $f(\tilde{t}) = A$, the answer it gives is $Ae^{rT} - A$ instead of $Ae^{rT}$.

I get similar incorrect answers when using step functions - an extra term at the end.

Guess at correct answer: Total at time $T = f(T) + re^{rT} \int_{0}^{T} f(\tilde{t})e^{-r\tilde{t}} d\tilde{t}$

The reason I'm guessing that's correct is that it works on step functions (which in later math become the basis for defining the integral), and also it's what showed up in one of my initial derivations that I thought was mistaken. Please don't go setting someone's portfolio based on such weak reasoning - but it seems like it should be correct. Especially, as I said, given that integrals are defined as limits of step functions, and this is correct on simple step functions.