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The relationship between annual discrete and continuous compounding interest rates is given as:

$$1+r_d = e^{r_c}$$

My question is what are the properties of the difference between $r_d$ and $r_c$?

For example, it should hold $r_d>r_c$ because more compounding should have lower interest to arrive at the same value. Can you show this mathematically?

I am not sure what other properties could exist?

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    $\begingroup$ From the convexity of the exponential function, we have $e^x\geq 1+x$ for all $x\in\mathbb{R}$. Thus, $1+r_d=e^{r_c}\geq 1+r_c$ which implies $r_d\geq r_c$. $\endgroup$ – Kevin Jul 27 at 13:27
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Let's add a time variable to extend to non-annual periods $$1 + r_d t = e^{r_c t}$$

The taylor expansion of exponential is \begin{align} e^{r_c t} &= \sum_{n=0}^\infty {\frac {(r_c t)^n} {n!}}\\ &= 1 + r_c t + {\frac 1 2}(r_c t)^2 + \cdots \end{align}

so by equating the two equations, we see that $$r_d = r_c + {\frac 1 2}r_c^2 t + O(t^2)$$

Two things we can see from this:

  1. $r_d > r_c$
  2. for $t$ small, the two rates are almost the same. As $t$ gets bigger, and the rates are no longer the same
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  • $\begingroup$ Are you assuming $r_c<1$ for your first conclusion? Why is the error term $O(t^2)$ when the expansion continues with $r^3$? Sry it may be obvious but I`m not Mathematician. $\endgroup$ – emcor Jul 28 at 14:34
  • $\begingroup$ I think it should be $(1+r_d)^t$ to add time? $\endgroup$ – emcor Jul 28 at 14:36
  • $\begingroup$ Certainly I am assuming $r \ll 1$, risk-free rates are typically in the single digit percentages and have probably never approached 100% in all of history. I divided both sides through by a factor of $t$ above which is why the lower on $t$ is lower than $r$. As for the compounding... yes potentially, I'd assumed this question was about simple vs. compounding rates but was that a misinterpretation? For $t=1$ the analysis is the same. $\endgroup$ – StackG Jul 29 at 0:25

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