- #1
sjweinberg
- 6
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Consider a random walk (in any dimension) with [itex]N[/itex] steps and a step size of 1. Take a real number [itex]\alpha > 0[/itex] and consider another random walk which takes [itex]\alpha^2 N[/itex] steps but wil step size [itex]\frac{1}{\alpha}[/itex].
I immediately noticed that the mean deviation after the full walk in both cases is the same: [itex]\frac{1}{\alpha} \sqrt{\alpha^2 N} = \sqrt{N}[/itex]. 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.
I immediately noticed that the mean deviation after the full walk in both cases is the same: [itex]\frac{1}{\alpha} \sqrt{\alpha^2 N} = \sqrt{N}[/itex]. 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.