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It appears that in a 1D random walk the expected RMS distance from the origin is some positive value with any positive number of steps (N):

sqrt(N)/2

However, in this same walk, the expected deviation of the fraction of steps to any one direction tends towards zero. 

1/(2 * sqrt(N))

I cannot intuitively see why the former doesn't tend towards zero as well if the average steps to the left cancels with the steps to the right.

Does anyone have an intuitive explanation for this?

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3 Answers 3

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I think I intuitively see this more clearly now.

If one draws two parallel lines A and B with lengths equal to coin flips where heads adds a cm to A and tails adds a cm to B, then the more flips one takes, the larger the difference in lengths between A and B. This tends towards a positive value, much like the RMS distance from the origin.

However, if one takes the fraction of that difference (A-B)/(A+B) over expected difference, which is 0.5 for balanced flipping, then that fraction tends towards zero with more flips.

The excess tends to increase whereas the proportion of the excess over the total steps tends towards zero.

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I assume you have perhaps already seen the maths but only want to get a better intuitive feel for this situation.

The distances from the origin will of course take many values, some big and some small. The effect of the squaring step in the RMS calculation will be to greatly weight matters towards the larger distances before you take the mean.

To see the effect in a simple context try finding the mean and the RMS of a few simple sets of numbers and see how these averages behave when some numbers are much larger than others

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  • $\begingroup$ Thanks. I actually meant to ask why RMS wouldn't tend towards zero rather than some positive value that increases with steps. I think I just now managed to see why, but am a bit ashamed b/c it might seem too obvious to others here like yourself that my question may be confusing. $\endgroup$
    – imagineerThat
    Commented Sep 4, 2019 at 23:53
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One way it helps me to think about this is in terms of a coin toss. The process of an $n$-step discrete random walk in 1d is the same flipping $2n$ couns, getting $h$ heads, and taking $2*(h-n)$. If I flip a fair coin 10 million times, how likely is it I'll get exactly 5 million heads? Very unlikely, right? And if I flip 10 billion coins, it's even less likely I'll get exactly 5 billion heads. So, the coins make it obvious to me at least that the distribution gets more spread out in terms of the actual number of heads you get, so the expected RMS gets bigger, and indeed as big as you want, which isn't maybe that hard to guess since it's intuitively going up.

For the expected proportion though, the coins make it clear too. Because, if I flip 10 million fair coins, even though it's very unlikely I get exactly 5 million heads, it's also pretty clearly unlikely that 51% of them will come up heads. And if I flip 10 billion coins it's even less likely. So, intuitively, the expected percent difference is going to zero, and that's again equivalent to the expected variance in the proportion of steps in either direction in your random walk going to zero.

By the way, if you're familiar with the Demoivre-Laplace theorem, this can pretty well be rigorized in a somewhat more intuitive way, since the distributions are widening, so the expected RMS is increasing, but on taking proportions we contract the number line so that they're shortening for proportions, since your contracting by more and more, which is pretty much a totally rigorizable argument.

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  • $\begingroup$ These are nice and simple explanations. I can intuitively see the distribution of heads widening with more tosses, making the RMS larger. Thank you! $\endgroup$
    – imagineerThat
    Commented Sep 5, 2019 at 0:07

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