I would like to understand Geometric Brownian Motion. In my lecture it was introduced as follows: Let $\mathcal T = \{ 0 , \delta t, 2 \delta t, \dots, n \delta t = T\}$ be a discrete time grid. $(S_t)_{t \in \mathcal T}$ follows discrete-time geometric Brownian motion if its logarithm $Z_t = \log (S_t)$ $$ Z_{t+\delta t} = Z_t + (\mu − \frac 1 2 \sigma^2) \delta t + \sigma \sqrt{\delta t} * \epsilon_t $$ for all $t \in \mathcal T$ where the $\epsilon_t$ are independent standard normally distributed random variables.
Now I thought about the distribution of $Z_t$ because I already heard that $S_t$ should have log-normal distribution, therefore $Z_t$ should be normally distributed. I found the following equation: $$ Z_{n \delta t} = Z_0 + n (\mu - \frac 1 2 \sigma^2) \delta t + \sigma \sqrt{\delta t} * \sum_{i = 0}^{n-1} \epsilon_i $$ where $\sum_{i=0}^{n-1} \epsilon_i \sim \mathcal N(0,n)$. Because $Z_0$ should be deterministic (not random) I thought about $Z_{n \delta t} \sim \mathcal N(Z_0 + n(\mu - \frac 1 2 \sigma^2) \delta t , \sigma^2 n \delta t )$. But this seems to be a little bit wrong. However, I found someone claiming $Z_t \sim \mathcal N(\mu t , \sigma^2 t)$. So what did I do wrong?
EDIT: If we define $t = n * \delta t$, then we have $Z_t \sim \mathcal N(Z_0 + t(\mu + \frac 1 2 \sigma^2), t\sigma^2)$. So the variance seems to be okay, but concerning the expected value - there are way too many terms.
The reason why I think about the distribution: I want to calcute the 5th percentile of a GBM. I already did this in simulations with generating many paths and plot the 5th percentile. But there should be a way to express the percentile in terms of $\Phi^{-1}$, where $\Phi$ is the cdf of a standard normal distribution.
I hope that you understand what I want to do and I would be very grateful for any kind of help!
PS: We didn't talk about Wiener Process or Brownian Motion or SDE, so the definition above is the only thing I know. And in general I have some trouble of understanding what we are doing :)