Are Random Walks with Different Step Sizes Identical in Brownian Motion Limit?

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SUMMARY

The discussion centers on the comparison of two random walks in the context of Brownian motion. One walk consists of N steps with a step size of 1, while the other has \(\alpha^2 N\) steps with a step size of \(\frac{1}{\alpha}\). Both walks exhibit the same mean deviation of \(\sqrt{N}\). The inquiry focuses on whether these two random walks appear identical in the Brownian motion limit as N approaches infinity.

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sjweinberg
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Consider a random walk (in any dimension) with N steps and a step size of 1. Take a real number \alpha > 0 and consider another random walk which takes \alpha^2 N steps but wil step size \frac{1}{\alpha}.

I immediately noticed that the mean deviation after the full walk in both cases is the same: \frac{1}{\alpha} \sqrt{\alpha^2 N} = \sqrt{N}. However, I'm curious to what extent these two random walks look identical. If were to take a Brownian motion type limit (where N becomes large and the step size 1 is thought of as being small), would the walks look identical?

Thanks in advance to any masters of statistics.
 
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