Deriving Brownian Drift from a Scaled Random Walk
Summary
The document asks how a biased random walk converges to a process with nonzero drift as its time step shrinks. It starts from the expected displacement after repeated up-or-down steps, then considers steps of size proportional to the square root of the time increment and an upward-step probability slightly above one half. Under that scaling, the expected displacement approaches a quantity proportional to elapsed time.
The response gives an intuitive explanation: square-root scaling preserves the scale of random fluctuations, while the probability bias represents directional drift. It does not carefully derive the limit or explain the precise parameter scaling; instead, it speculates that the probability was chosen to produce the stated mean. The discussion therefore introduces the intuition but leaves a rigorous derivation and the corresponding variance convergence to the reader.
Key ideas
- Scaling step size with the square root of the time increment preserves nonvanishing random fluctuations in the limit.
- A small imbalance in up and down probabilities creates directional drift.
- The stated probability adjustment is intended to make expected displacement grow linearly with time.
- The response is intuitive and does not provide a rigorous convergence derivation.
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# Obtaining the drift of a Wiener process formed from a random walk
# 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 $p = \frac{1}{2} \left( 1+\mu \sqrt{\Delta t} \right)$.
## Answer by iNarek94 (score 0, accepted)
https://quant.stackexchange.com/a/21399
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.