<p><strong>Your graphs do not seem to match your explanation of the formulas.</strong></p>
<hr />
<p>Let us look at the formulas
<span class="math-container">$$x(t) = \frac{r}{365} \cdot \frac{t}{365}$$</span>
<span class="math-container">$$FV(t) = PV(1 + x(t))^{years}$$</span></p>
<p>at the end of the period, that is where <span class="math-container">$t = 365 \cdot years$</span>.</p>
<p><span class="math-container">$$x(365 \cdot years) = \frac{r}{365} \cdot \frac{365 \cdot years}{365} = \frac{r}{365} \cdot years$$</span>
<span class="math-container">$$FV(365 \cdot years) = PV(1 + x(365 \cdot years))^{years} = PV\left(1 + \frac{r}{365}\cdot years\right)^{years}.$$</span></p>
<p>For <span class="math-container">$years = 1$</span> this would mean an <span class="math-container">$FV$</span> of <span class="math-container">$PV\left(1 + \frac{r}{365}\right)$</span>, much smaller than the expected <span class="math-container">$PV\left(1 + r\right)$</span>.</p>
<hr />
<p>Perhaps you meant to write <span class="math-container">$x(t) = \frac{r}{365} \cdot t?$</span></p>
<p>For <span class="math-container">$years = 1$</span> and small <span class="math-container">$r$</span> values this results a decent but imperfect approximation of the result yielded by daily compounding formula with <span class="math-container">$x = \sqrt[365]{1 + r} - 1$</span>.</p>
<p>However for <span class="math-container">$years = 10$</span> we have
<span class="math-container">$$FV(365 \cdot 2) = PV\left(1 + r \cdot 10\right)^{10}.$$</span></p>
<p>The compounding is going on for 10 years as expected, but the yearly interest rate has also increased <em>tenfold</em>, which is probably not intentional.</p>
<hr />
<p>Instead of experimenting with your own formulas for <span class="math-container">$FV$</span>, I recommend reading up on this is a <a href="https://en.wikipedia.org/wiki/Compound_interest#Calculation" rel="nofollow noreferrer">well-explored subject, as there are several formulas for discounting, including one for continuous time</a>, and it is proven that the smaller time period you choose the closer you get to this.</p>

CC BY-SA 4.0
2022-01-11T22:56:05.963Z
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
Ezequiel Castaño
CC BY-SA 4.0
2022-01-11T04:25:30.537Z
2022-01-11T07:06:10.190Z
2022-01-11T07:06:10.190Z
Giskard
<p>This is a follow up question to <a href="https://economics.stackexchange.com/q/50016/40138">this</a></p>
<p>I found two different ways to &quot;adjust for inflation&quot; 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.</p>
<p><span class="math-container">$$r: \text{annual rate}$$</span>
<span class="math-container">$$t: \text{time in days}$$</span>
<span class="math-container">$$years: \text{total time in years, constant}$$</span>
<span class="math-container">$$PV: \text{Present Value, constant}$$</span>
<span class="math-container">$$FV(t): \text{Future Value for a given time } t \text{ (in days)}$$</span></p>
<p>For example, if <span class="math-container">$years = 2$</span> then <span class="math-container">$t \in [0, 730]$</span></p>
<p>Plots below are generated using <span class="math-container">$r=0.50$</span> and <span class="math-container">$PV=1000$</span></p>
<ol>
<li>The first approach is using Giskard's proposal (see related anwer) and let the rate fixed and let the time vary:</li>
</ol>
<p><span class="math-container">$$x = \sqrt[365]{1 + r} - 1$$</span>
<span class="math-container">$$FV(t) = PV(1 + x)^t$$</span></p>
<p>This produced the following plot for 1 year:
<a href="https://i.sstatic.net/05eaK.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/05eaK.png" alt="First Approach" /></a></p>
<p>And this for 2 years:
<a href="https://i.sstatic.net/HSm2U.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/HSm2U.png" alt="enter image description here" /></a></p>
<ol start="2">
<li>The second solution fixes the time and changes the interest rate based on time:</li>
</ol>
<p><span class="math-container">$$x(t) = \frac{r}{365} * \frac{t}{365}$$</span>
<span class="math-container">$$FV(t) = PV(1 + x(t))^{years}$$</span></p>
<p>This produced the following plot for 1 year:
<a href="https://i.sstatic.net/gtMgg.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/gtMgg.png" alt="enter image description here" /></a></p>
<p>And this for 2 years:</p>
<p><a href="https://i.sstatic.net/ly82n.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ly82n.png" alt="enter image description here" /></a></p>
<hr />
<p>I think the second approach is incorrect because when computing 2 years, the value for one year changed from <span class="math-container">$666.67$</span> to <span class="math-container">$\approx 640$</span> whereas in the first method the original value is kept but I cannot justify why this happens.</p>

CC BY-SA 4.0
2022-01-10T20:59:27.947Z
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 $t/365$?
Giskard
CC BY-SA 4.0
2022-01-10T21:01:29.230Z
If so, why do you think these two formulas should come to the same conclusion?
Giskard
CC BY-SA 4.0
2022-01-10T23:46:26.863Z
$years$ 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
Ezequiel Castaño
CC BY-SA 4.0
2022-01-11T00:14:10.097Z
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
Ezequiel Castaño
<inflation><nonexpert>
Difference between two different ways of adjusting for inflation daily given annual rate
Ezequiel Castaño
# Difference between two different ways of adjusting for inflation daily given annual rate

(Asked by: username_0 on 2022-01-10)

This is a follow up question to [this](https://economics.stackexchange.com/q/50016/40138)

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 username_1'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](https://i.sstatic.net/05eaK.png)](https://i.sstatic.net/05eaK.png)

And this for 2 years:
[![enter image description here](https://i.sstatic.net/HSm2U.png)](https://i.sstatic.net/HSm2U.png)

2. 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](https://i.sstatic.net/gtMgg.png)](https://i.sstatic.net/gtMgg.png)

And this for 2 years:

[![enter image description here](https://i.sstatic.net/ly82n.png)](https://i.sstatic.net/ly82n.png)

---

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.

**Tags:** inflation, nonexpert

> **Comments:**
> **username_1**: 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 $t/365$?  
> **username_1**: If so, why do you think these two formulas should come to the same conclusion?  
> **username_0**: $years$ 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  
> **username_0**: 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  

---

## 1 answers

### Answer by username_1 on 2022-01-11. Score: 2 | Accepted answer

**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](https://en.wikipedia.org/wiki/Compound_interest#Calculation), and it is proven that the smaller time period you choose the closer you get to this.

> **Comments:**
> **username_0**: 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 username_1 for taking the time for answering and explaining everything  

---

