Difference Between Random Walks

Question

I am considering 2 formulations of random walks that both supposedly converge to Brownian motion. Let $\mu >0$ be the drift and $\sigma>0$ be the standard deviation (volatility). Consider a discretization with $t_0 = 0$ and $t_n = nh$ for some small step size $0<h\ll1$. Further define a binomial random variable $$B(\pm1,p) = \begin{cases} 1, \text{ with probability } p \\ -1, \text{ with probability } 1- p\end{cases}$$ then fix $x_0$ and consider \begin{align} x_{t_{n+1}} = x_{(n+1)h} &= x_{nh} + \mu h + \sigma \sqrt{h}B\big(\pm1,\frac{1}{2}\big) \\ x_{t_{n+1}} = x_{(n+1)h} &= x_{nh} + \sigma \sqrt{h}B\big(\pm1, \frac{1}{2}(1 + \frac{\mu}{\sigma}\sqrt{h})\big) \end{align} I know that both of the converge to Brownian motion $BM(x_0,\mu,\sigma^2)$ in the limit that $h\rightarrow 0$. I am just wondering what the difference is between the formulations or can we somehow scale the binomial distribution so that they look more similar?.

Answer 1

You have two walks that are biased so that the particle tends to move to the right. The first walk is biased by adding a constant $\mu h$ to a symmetric Bernoulli distribution. The second walk is biased by making the Bernoulli distribution asymmetric.

In either case the particle is expected to move $\mu h$ units to the right during one step of the random walk. So it seems to me the distributions already are scaled in order to make this true.

Have you done any simulations or convolutions to see what the distribution of states looks like after a large number of steps? How dissimilar were the two results?