Given the geometric Brownian motion, obtained with the Itô's Lemma: $$d \ln S_t = \left(\mu - \frac{\sigma^2}{2} \right) dt + \sigma d W_t$$ Where $\mu$ and $\sigma$ are constants, we have that: $$ \mathbb{E}_t[d \ln S_t] = \left(\mu - \frac{\sigma^2}{2} \right) dt $$ $$ \mathbb{V}_t[d \ln S_t] = \sigma^2 dt $$ If we call $\hat{\mu}$, the sample mean, and $\hat{\sigma}^2$, the sample covariance (of course divided by $n-1$) of the log-returns, we can estimate the population parameters by solving: $$ \left(\mu - \frac{\sigma^2}{2} \right) dt = \hat{\mu} $$ $$ \sigma^2 dt = \hat{\sigma}^2 $$ Obtaining: $$ \sigma^2=\frac{\hat{\sigma}^2}{dt} $$ $$ \mu = \frac{\hat{\mu}}{dt} + \frac{\sigma^2}{2} $$
Then my question is: is it possible that the actual sample mean $\hat{\mu}$ systematically gives better estimates of the population mean $\mu$ than the moments estimate? That is:
$$ \mu \simeq \hat{\mu} \neq \frac{\hat{\mu}}{dt} + \frac{\sigma^2}{2} $$
In order to check that I simulated $N$ stochastic processes ($S_{it}$) given some fixed $\mu_i$ and $\sigma_i$ for $t$ time periods, then I computed the log returns of the simulated process and then calculate the mean and standard deviation (both sample and with the moments method), like this (code is only for example):
S(1) = S0
for i = 2:T/dt
S(i) = S(i-1) + S(i-1) * mu * dt + S(i-1) * sigma * dW(i)
Where $dW_t$ is a vector with mean 0 and variance $\sqrt{dt}$.