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Without being educated in high Mathematics, I randomly claim that the relation of expected distance to time (spread = time^(0.5)) is based on the probability of a random direction to lead to an increase of distance. This probability is 50%!

Now, assuming this is correct, how would the the formula for estimating the spread look if we doubled that probability? (Doubled in an independent sense: Given a random motion would lead to a decrease in distance distance from the start, this randomness is applied a second time and the motion is only executed if both random events point inwards. =would decrease the distance)

I assume the formula would change to: spread = time^(3/4)

So my 2 questions are:

  1. Is the premise of this question any valid?
  2. How would the time dependence change, given these assumptions?
KaPy3141
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    There are two things that one should keep separate here. First: what is the expected position to be at after time $t$? This is $0$ for the symmetric case (by symmetry). What is non-zero is the "spread" in the distribution of walkers which grows as $\sqrt{t}$. This is what we usually say is the "distance" it has gone. A well-studied random walk similar to what you describe is the case where you go right with a probabillity $p$ and left with a probabillity $q$. The expected position after $N$ steps is now $\propto t(p-q)$ while the expected spread is $\sqrt{pq t}$. It stays a square root. – Winther Nov 15 '19 at 15:14
  • Thanks a lot, I totally confused it! I think I get it now! – KaPy3141 Nov 15 '19 at 15:27
  • @Winther: This is exactly the answer I wanted! You should have posted it as an answer! – KaPy3141 Nov 15 '19 at 15:35

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