# How to formulate intra-day return?

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?

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.

• And how can I apply this to my case where $$P_t = Close_t$$ and $$P_{t-1} = Open_t$$. Also isn't $$P_t - P_{t-1} = ΔP$$ Jul 3, 2023 at 18:48
• I am not sure I understood the problem, what do you mean with "Also isnt' ..."?
– Don
Jul 3, 2023 at 22:09
• I meant: Isn't the following true? Sorry :D Jul 3, 2023 at 22:50
• This answer seems to address a different question than the one asked. Jul 4, 2023 at 6:39
• @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
Jul 4, 2023 at 7:25