Obtaining the drift of a Wiener process formed from a random walk

Question

I'm trying to understand how the equation for Geometric Brownian Motion is formed from a random walk. I'm following the book 'Statistics of Financial Markets' but I'm struggling to follow how the drift is found from the expected value. I understand that for a random walk $ \left \{ X_n; n \geq 0 \right \}$, $ E(X_t) = n(2p-1) \cdot \Delta x$ where $p$ is the probability of increasing the random variable by $\Delta x$ and that for $t = n \Delta t$

$$E(X_t) = (2p-1) \cdot t \frac{\Delta x}{\Delta t} $$

however the book claims that for

$$ \Delta t \rightarrow 0, \Delta x = \sqrt{\Delta t}, p = \frac{1}{2} \left( 1+\mu \sqrt{\Delta t} \right)$$ we obtain that $\forall t$ $E(X_t) \rightarrow \mu t $.

How is this result obtained? Earlier in the book it suggests that these values may have been chosen to prevent the Variance from converging to $0$, however I don't understand why $\Delta x = \sqrt{\Delta t} $ and in particular why we have chosen $p = \frac{1}{2} \left( 1+\mu \sqrt{\Delta t} \right)$.

Answer 1

Here Δx must be in the sense of deviation and as a measure of deviation √Δt suits well, as it's the standart deviation of the increments. As this is not a symmetrical process, it has a drift (upper or lower), meaning that the probability of going in any direction is skewed. The bigger the time interval between two points, the bigger the skewness of X's path from it's mean. μ should be representing the skewness of the distribution, if X has an upper drift, then the probability of going up is bigger, and the "biggerness" is determined by μ itself. I'm guessing, that p was chosen this way just to get to the mean result. Because the real distribution of the increments is normal and can not be a linear function of the mean.