I am a highschool student with a small but decent knowledge of stocks. For a maths project, I'm investigating the probabilities and correlations of stocks.

I came across this formula for calculating the periodic daily return as a tool to help calculate historical volatility of a stock:

$$ \text{Periodic Daily Return} = \ln(\frac{\text{Today's Stock Price}}{\text{Yesterday's Stock Price}}) $$

I have a basic understanding of $e$ (in that it is the continuous compounding interest of $\$1$ for one period of time at a rate of $ 100\%$), and hence I understand how e can be manipulated to get the above formula, and how it has advantages over a simple interest calculation.

I'm curious as to whether $\log_{10} $ or even $\log_b$ can be used to the same effect, and if not, why $\ln$ is more advantageous.

Any help would be much appreciated, cheers!

  • $\begingroup$ I think the equation is $P(t)=P(0)\cdot e^{\delta \cdot t}$, For $t=1$ we get the daily return. Here we can take the logarithmus naturalis only. $$\delta=\ln\left(\frac{P(1)}{P(0)}\right)$$ $\endgroup$ – callculus May 18 at 23:43

It's not a big difference as $\log x = \ln x / \ln 10$ - changing logarithm base is just multiplication of it's value by constant. In terms of percents, $\ln 10$ is period you need to get tenfold increase at continuous 100% rate (and $\ln e = 1$ is period you need to get an increase in $e$ times).

You can use $\log$, but then you will need to divide result by $\ln 10$ if you want to use it instead of $\ln$. Depending on further formulas you will use the return in, it can make them a bit simpler or a bit more complex.

$e$ is good base for logarithm for several reasons - for example derivative $\ln'(x) = \frac{1}{x}$, and any other base will require to replace $1$ in numerator with something else.

  • $\begingroup$ Oh wow, that multiply by a constant rule is very interesting, thanks for that mihaild (and Raskolnikov for the important edit). Apart from ease-of-use, are there any other benefits that you know of for using base e rather than base 10? $\endgroup$ – K King May 21 at 6:59
  • $\begingroup$ I can't imagine anything but convenience for using function $f(x)$ instead of $a\cdot f(x)$ - we can always divide by $a$ all occurences of $f$. $\endgroup$ – mihaild May 21 at 8:58

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