stochastic calculus - Geometric Brownian Motion: percentage returns vs log-returns

In classical calculus, we know that the limit of percentage return (ie $dS/S$) equals that of the log return (ie. $dln(S)$ ).

With uncertainty, we rely on Ito Lemma to draw a relationship between the two:

\begin{equation*} dS = \mu S dt + \sigma Sdz \end{equation*}

and

\begin{equation*} dln(S) = (\mu - \sigma^2/2) dt + \sigma dz \end{equation*}

I understand the mathematics behind but I would like to know more about the intuition, mainly

with uncertainties, when we "switch" from percentage return to log return, why do we have a smaller drift $(\mu - \sigma^2/2)$? Is there any intuition or financial sense behind?

Moreover, when we discretize the process, can we draw the same relationship and say something like \begin{equation*} \Delta S = \mu S \Delta t + \sigma S \Delta z \end{equation*}

and \begin{equation*} \Delta ln(S) = (\mu - \sigma^2/2) \Delta t + \sigma \Delta z \end{equation*}

Thank you in advance.

Answer

The percentage return over the infinitesimal interval $[t, t+dt]$ is given by \begin{align*} \frac{S_{t+dt} - S_t}{S_t} \approx \mu dt + \sigma \sqrt{dt} \xi, \end{align*} where $\xi$ is a standard normal random variable. On the log-return, note that, for $x$ sufficiently small, \begin{align*} \ln (1+x) \approx x -\frac{x^2}{2}, \end{align*} then, by ignoring the higher order terms (relative to $dt$), \begin{align*} \ln \frac{S_{t+dt}}{S_t} &= \ln \left(1+ \frac{S_{t+dt} - S_t}{S_t} \right)\\ &\approx \frac{S_{t+dt} - S_t}{S_t} -\frac{1}{2} \left( \frac{S_{t+dt} - S_t}{S_t}\right)^2\\ &\approx \mu dt + \sigma \sqrt{dt} \xi -\frac{1}{2} \left(\mu dt + \sigma \sqrt{dt} \xi\right)^2\\ &\approx \mu dt + \sigma \sqrt{dt} \xi -\frac{1}{2}\sigma^2\xi^2 dt\\ &\approx \left(\mu - \sigma^2/2 \right)dt + \sigma \sqrt{dt} \xi. \end{align*} Here, we assume that \begin{align*} \xi^2 \approx E(\xi^2) = 1. \end{align*}