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I am reading a book that is talking about continuous random walk. It first starts with defining one dimensional discrete random walk as starting at point 0 and move to either to the right or left at the rate of $1$ per unit of time.

Then, it said that if we instead move $\sqrt{h}$ per $h$ units of time, and took the limit as $h$->$0$, we would have continuous random walk. It will then converge into brownian motion.

My questions are: 1) Why are me moving $\sqrt{h}$ instead of $h$ per $h$ unit of time? Can someone show me the calculation? 2) Before taking the limit to $0$, how do we proof that the random variable is indeed binomial? I undestand that from CLT, it will converge into normal rv but, I am baffled on how to proof that it is indeed binomial. I am considering on proofing it through it's moment but I wonder if there is a simpler way.

mohlee
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  1. You move by $\sqrt{h}$ so that the variance of a step is proportional to $h$. This ensures that the variance after a fixed finite time is some fixed finite number. If you move by asymptotically less than $\sqrt{h}$ then the variance at a fixed finite time would go to zero as $h$ went to zero; if you move by asymptotically more then the variance would blow up as $h$ went to zero. Notice that the mean is kept zero (if it were not, then the mean would dominate over the fluctuations as $h$ went to zero) so the variance is the next thing up to be kept bounded.
  2. The number of steps to, say, the right after $N$ steps is a binomial random variable with parameters $N$ and $1/2$. Call it $R$. The number of steps to the left is $N-R$ so the displacement is $R-(N-R)=2R-N$, in units of whatever the step size is. It isn't a binomial random variable, strictly speaking, but its distribution looks very similar. The CLT applies regardless (because the mean is kept zero, the total variance is kept fixed, the summands are independent, and the number of summands is sent to $\infty$). Now the discrete random walk at a time $t=Nh$ is $\sqrt{h}(2R-N)=\frac{\sqrt{t}}{\sqrt{N}}(2R-N)$. You may recognize this as a "z-score" now.

By the way, the situation for an asymmetric random walk is different, because there is an overall trend in one direction. In this case the step size must be proportional to $h$, not $\sqrt{h}$, in order for the continuum limit to be of order 1 (i.e. neither blowing up nor going to zero). In this case CLT still applies, but not to leading order: the leading order term is a deterministic "transport" term.

Ian
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