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This is a follow up question to this

I found two different ways to "adjust for inflation" on a daily basis given an annual rate (without doing daily compounding). Both yield the same result at the end of the timeframe however I can't determine if there is any error in the interpolated result.

$$r: \text{annual rate}$$ $$t: \text{time in days}$$ $$years: \text{total time in years, constant}$$ $$PV: \text{Present Value, constant}$$ $$FV(t): \text{Future Value for a given time } t \text{ (in days)}$$

For example, if $years = 2$ then $t \in [0, 730]$

Plots below are generated using $r=0.50$ and $PV=1000$

  1. The first approach is using Giskard's proposal (see related anwer) and let the rate fixed and let the time vary:

$$x = \sqrt[365]{1 + r} - 1$$ $$FV(t) = PV(1 + x)^t$$

This produced the following plot for 1 year: First Approach

And this for 2 years: enter image description here

  1. The second solution fixes the time and changes the interest rate based on time:

$$x(t) = \frac{r}{365} * \frac{t}{365}$$ $$FV(t) = PV(1 + x(t))^{years}$$

This produced the following plot for 1 year: enter image description here

And this for 2 years:

enter image description here


I think the second approach is incorrect because when computing 2 years, the value for one year changed from $666.67$ to $\approx 640$ whereas in the first method the original value is kept but I cannot justify why this happens.

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    $\begingroup$ You did not take my suggestion to explain the variables to heart. What is $t$, and what is $years$ in this formula: $$FV(t) = PV(1 + x(t))^{years}?$$ Is $years$ simply $t/365$? $\endgroup$
    – Giskard
    Jan 10, 2022 at 20:59
  • $\begingroup$ If so, why do you think these two formulas should come to the same conclusion? $\endgroup$
    – Giskard
    Jan 10, 2022 at 21:01
  • $\begingroup$ $years$ is a constant (integer) and $t$ is a variable that varies from 0 to $365$*$years$. The idea is that in the plot the x-axis shows $t$ and $years$ is the upper-boundary for $t$. Maybe the confusion is that I do not want to calculate the FV for a given $t$ but rather I want the FV for all possible values less than $365$*$years$. To represent that "all possible values" I made FV a function of $t$ which is the precise day $\endgroup$ Jan 10, 2022 at 23:46
  • $\begingroup$ I just thought that since the final value is the same in both methods, the daily interpolation might have some meaning in each case. but I don't know exactly what that meaning would be and what would be its interpretation $\endgroup$ Jan 11, 2022 at 0:14

1 Answer 1

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Your graphs do not seem to match your explanation of the formulas.


Let us look at the formulas $$x(t) = \frac{r}{365} \cdot \frac{t}{365}$$ $$FV(t) = PV(1 + x(t))^{years}$$

at the end of the period, that is where $t = 365 \cdot years$.

$$x(365 \cdot years) = \frac{r}{365} \cdot \frac{365 \cdot years}{365} = \frac{r}{365} \cdot years$$ $$FV(365 \cdot years) = PV(1 + x(365 \cdot years))^{years} = PV\left(1 + \frac{r}{365}\cdot years\right)^{years}.$$

For $years = 1$ this would mean an $FV$ of $PV\left(1 + \frac{r}{365}\right)$, much smaller than the expected $PV\left(1 + r\right)$.


Perhaps you meant to write $x(t) = \frac{r}{365} \cdot t?$

For $years = 1$ and small $r$ values this results a decent but imperfect approximation of the result yielded by daily compounding formula with $x = \sqrt[365]{1 + r} - 1$.

However for $years = 10$ we have $$FV(365 \cdot 2) = PV\left(1 + r \cdot 10\right)^{10}.$$

The compounding is going on for 10 years as expected, but the yearly interest rate has also increased tenfold, which is probably not intentional.


Instead of experimenting with your own formulas for $FV$, I recommend reading up on this is a well-explored subject, as there are several formulas for discounting, including one for continuous time, and it is proven that the smaller time period you choose the closer you get to this.

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  • $\begingroup$ You were right, the actual formula was $x(t)= \frac{r}{365} \cdot t$. I just rushed and didn't pay enough attention to the details. I will have a look at the wikipedia page and post a new question if there is anything else. Thank you Giskard for taking the time for answering and explaining everything $\endgroup$ Jan 11, 2022 at 22:56

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