here is my question:
You have a binomial process where \begin{equation} Z_j \end{equation}
is a random variable taking the value of one with probability p and the value of minus one otherwise.
We have a stock price the log of which follows the following process: \begin{equation} logS_t=logS_0+\mu(T)+\sigma\sqrt(T/N)\ \sum_{j=1}^{N} Z_j \end{equation}
We want to find out what happens as N goes to infinity
The risk-neutral probability for the up move is given by the following equation:
\begin{equation} q=\frac{e^{(r-\mu)dt}-e^{-\sigma\sqrt{dt}}}{e^{\sigma\sqrt{dt}}-e^{-\sigma\sqrt{dt}}} \end{equation}
which after expansion equals:
\begin{equation} q= 1/2(1-\sqrt{dt}(\frac{\mu+0.5\sigma^2-r}{\sigma})) \end{equation}
So now \begin{equation} Z_j \end{equation}
instead of zero and 1 has approximately:
\begin{equation} mean= \sqrt{dt}(\frac{r-\mu-0.5\sigma^2}{\sigma}) \end{equation}
and
\begin{equation} variance= 1-dt(\frac{r-\mu-0.5\sigma^2}{\sigma})^2 \end{equation}
Now what I don't understand is how from this follows that the mean of logSt is:
\begin{equation} logS_0+(r-0.5\sigma^2)T \end{equation}
and the variance of logSt is:
\begin{equation} \sigma^2T \end{equation}