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I am struggling to understand "force of interest" in simple interest. I understand that "force of interest" can be interpreted as compound interest with infinitely small time interval. That is, the new interest generated any infinitely short moment is added to the value.

However, as simple interest means that the interest has a linear continuous relationship with time, I am not sure how to understand "force of interest" in simple interest. For example, how can I interpret the equation $\delta _t = \frac{i}{1+it}$ realistically in simple interest?

Valia
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3 Answers3

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You seem to have taken the force of interest to be $\dfrac{a'(t)}{a(t)}$ where $a(t)$ is the accumulated amount at time $t$

So your expression $\delta _t = \dfrac{i}{1+it}$ seems to be correct for simple interest with an interest rate of $i$, since $a(t)=a(0) (1+it)$ and $a'(t) = a(0)i$

In this case, $\delta_t$ is the momentary compounding rate, and is a decreasing function of $t$:

  • when $t=0$ and the starting amount is $a(0)$, it is $\delta_0=i$,
  • while for example when $t=\frac1i$ and the accumulated amount is $2a(0)$, it is $\delta_{1/i}=\frac12 i$.
Henry
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  • Thanks for answering! I am still not sure why $\delta _t$ is a momentary compounding rate that can exist in a simple interest. – Valia May 28 '17 at 23:25
  • @Akroma: do you understand that $\delta_t$ varies with $t$ when the simple interest rate is constant, because the accumulated amount increases while the interest (in money terms) is constant? If not, try the comments at https://math.stackexchange.com/questions/1020120/force-of-interest-for-simple-interest – Henry May 28 '17 at 23:30
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Perhaps it is easier to analyze the problem mathematically: given an accumulation function $a(t)$, a function $\delta(t)=\frac{\mathbb{d}\ln(a(t))}{\mathbb{d}t}$ may exist. If $a(t)$ is discrete, then $\delta(t)$ is not defined at the jumps. Similarly, if $a(t)$ is continuous and differentiable, then $\delta(t)$ is continuous.

For simple discrete interest conversion period, $\delta(t)$ is not defined at the jumps.

For simple continuous interest conversion period, $\delta(t)=\frac{i}{1+it}$.

For compound discrete interest conversion period, $\delta(t)$ is not defined at the jumps.

For compound continuous interest conversion period, $\delta(t)=k$ (constant).

There is much confusion in the literature on whether the function $\delta(t)=\frac{\mathbb{d}\ln(a(t))}{\mathbb{d}t}$ itself, or whether the $\delta(t)=k$ for the compound continuous interest conversion period case only, is called the force of interest.

Reference for differentiability of function: Link

Nam Tran
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late to the party but; combining the two answers above clarifies beautifully.

  1. simple interest implies that as time increases, the degree interest has on accumulated value decreases, because AV increases with t(linear continuous relationship) but it increases continuously by the same amount. a larger quantity of time applies a proportionate level of change as a lesser amount. so the force((i feel like this should be “momentum“) of interest has to decrease as time increases such that the amount(monetary, which i feel should be where “force” comes into play) of change is constant.

  2. although interest is simple(product of length of period), the period in question is not discrete, even if it is an instant(momentary). and if the interest were applied at discrete intervals, the case would then become that the force of interest could only exist during interest respective moments of application.

Teddy La Fleche
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