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

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 $p = \frac{1}{2} \left( 1+\mu \sqrt{\Delta t} \right)$.