<p>It is an approximation, but it is appropriate for <em>small</em> values (i.e. returns close to zero, what exactly <em>small</em> means depends on the precision you want to achieve) and it comes from the first order Taylor expansion of logarithmic function.</p>
<p>The correct way to compute a return, calling <span class="math-container">$P_t$</span> and <span class="math-container">$P_{t-1}$</span> final and initial price, is:</p>
<p><span class="math-container">$$
R_t=\frac{P_t-P_{t-1}}{P_{t-1}}
$$</span></p>
<p>Now, let's consider <span class="math-container">$x=\frac{P_t}{P_{t-1}}$</span>.</p>
<p>The first order Taylor expansion of <span class="math-container">$f(x)$</span> around <span class="math-container">$x_0$</span> is:</p>
<p><span class="math-container">$$
f(x)\approx f(x_0)+f'(x_0)\left(x-x_0\right)
$$</span></p>
<p>By substituting <span class="math-container">$f(x)$</span> with <span class="math-container">$ln(x)$</span>, considering that the derivative of <span class="math-container">$ln(x)$</span> is <span class="math-container">$\frac{1}{x}$</span> and taking <span class="math-container">$x_0=1$</span>, you get:
<span class="math-container">$$
ln(x)\approx ln(1)+\left(x-1\right)
$$</span></p>
<p><span class="math-container">$$
ln(x)\approx x-1
$$</span></p>
<p>Substituting:</p>
<p><span class="math-container">$$
ln(\frac{P_t}{P_{t-1}})\approx \frac{P_t}{P_{t-1}}-1
$$</span></p>
<p><span class="math-container">$$
ln(\frac{P_t}{P_{t-1}})\approx \frac{P_t-P_{t-1}}{P_{t-1}}=R_t
$$</span></p>
<p>Which completes the proof.</p>

CC BY-SA 4.0
2023-07-03T18:48:49.757Z
And how can I apply this to my case where $$P_t = Close_t$$ and $$P_{t-1} = Open_t$$. Also isn't $$P_t - P_{t-1} =  ΔP$$
BlankerHans
CC BY-SA 4.0
2023-07-03T22:09:31.067Z
I am not sure I understood the problem, what do you mean with "Also isnt' ..."?
Don
CC BY-SA 4.0
2023-07-03T22:50:35.393Z
I meant: Isn't the following true? Sorry :D
BlankerHans
CC BY-SA 4.0
2023-07-04T06:39:53.830Z
This answer seems to address a different question than the one asked.
Richard Hardy
CC BY-SA 4.0
2023-07-04T07:25:21.967Z
@RichardHardy can you explain why it is not an answer? He asked if its representation of returns was appropriate and a way to show it in a mathamatical correct way.
Don
CC BY-SA 4.0
2023-07-04T07:25:26.170Z
@BlankerHans Yes, it is true, to apply to your case just substitute "P_t" with "Close_t" and "P_t-1" with "Open_t".
Don
CC BY-SA 4.0
2023-07-04T11:46:59.767Z
@Don I am unsure about it because the time index is little bit messed up isn't it? close_t = p_t-1 seems not right.
BlankerHans
CC BY-SA 4.0
2023-07-04T13:30:44.797Z
@Don, your answer focuses on log- vs. % returns, but the question is about intra-day vs. daily returns. Your later comment goes in the right direction, I think.
Richard Hardy
CC BY-SA 4.0
2023-07-04T13:55:10.633Z
@BlankerHans Time index shouldn't be messed up if you consider "Δt" as 24 hours. If closure and opening prices are taken at the same hour, just consider "close_(t-1)=opening_t", (Bitcoin is negotiated 24h/day).
Don
CC BY-SA 4.0
2023-07-03T16:31:53.193Z
2023-07-03T16:31:53.193Z
Don
<p>So I have time series data on the bitcoin price. For each day there is an open price and a close price. In many papers they calculae the return like this:
<span class="math-container">$$ R_t = \Delta P  / P_{t-1} = ln(P_t/P_{t-1}) $$</span>
This is more like a day on day return.</p>
<p>I calculated it like this:
<span class="math-container">$$ R_t = ln( Close_t / Open_t ) $$</span>
I would call it an intra-day return. Is this an appropriate way to do it and how can I formulate this in a mathematical correct way?</p>

<time-series><price><capital-returns><bitcoin>
How to formulate intra-day return?
BlankerHans
# How to formulate intra-day return?

(Asked by: username_0 on 2023-07-03)

So I have time series data on the bitcoin price. For each day there is an open price and a close price. In many papers they calculae the return like this:
$$ R\_t = \Delta P / P\_{t-1} = ln(P\_t/P\_{t-1}) $$
This is more like a day on day return.

I calculated it like this:
$$ R\_t = ln( Close\_t / Open\_t ) $$
I would call it an intra-day return. Is this an appropriate way to do it and how can I formulate this in a mathematical correct way?

**Tags:** time-series, price, capital-returns, bitcoin


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## 1 answers

### Answer by username_1 on 2023-07-03. Score: 1

It is an approximation, but it is appropriate for *small* values (i.e. returns close to zero, what exactly *small* means depends on the precision you want to achieve) and it comes from the first order Taylor expansion of logarithmic function.

The correct way to compute a return, calling $P\_t$ and $P\_{t-1}$ final and initial price, is:

$$
R\_t=\frac{P\_t-P\_{t-1}}{P\_{t-1}}
$$

Now, let's consider $x=\frac{P\_t}{P\_{t-1}}$.

The first order Taylor expansion of $f(x)$ around $x\_0$ is:

$$
f(x)\approx f(x\_0)+f'(x\_0)\left(x-x\_0\right)
$$

By substituting $f(x)$ with $ln(x)$, considering that the derivative of $ln(x)$ is $\frac{1}{x}$ and taking $x\_0=1$, you get:
$$
ln(x)\approx ln(1)+\left(x-1\right)
$$

$$
ln(x)\approx x-1
$$

Substituting:

$$
ln(\frac{P\_t}{P\_{t-1}})\approx \frac{P\_t}{P\_{t-1}}-1
$$

$$
ln(\frac{P\_t}{P\_{t-1}})\approx \frac{P\_t-P\_{t-1}}{P\_{t-1}}=R\_t
$$

Which completes the proof.

> **Comments:**
> **username_0**: And how can I apply this to my case where $$P_t = Close_t$$ and $$P_{t-1} = Open_t$$. Also isn't $$P_t - P_{t-1} =  ΔP$$  
> **username_1**: I am not sure I understood the problem, what do you mean with "Also isnt' ..."?  
> **username_0**: I meant: Isn't the following true? Sorry :D  
> **username_2**: This answer seems to address a different question than the one asked.  
> **username_1**: @username_2 can you explain why it is not an answer? He asked if its representation of returns was appropriate and a way to show it in a mathamatical correct way.  
> **username_1**: @username_0 Yes, it is true, to apply to your case just substitute "P_t" with "Close_t" and "P_t-1" with "Open_t".  
> **username_0**: @username_1 I am unsure about it because the time index is little bit messed up isn't it? close_t = p_t-1 seems not right.  
> **username_2**: @username_1, your answer focuses on log- vs. % returns, but the question is about intra-day vs. daily returns. Your later comment goes in the right direction, I think.  
> **username_1**: @username_0 Time index shouldn't be messed up if you consider "Δt" as 24 hours. If closure and opening prices are taken at the same hour, just consider "close_(t-1)=opening_t", (Bitcoin is negotiated 24h/day).  

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