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I would like to understand the time-dependence of the average distance from the start-point in 1D Brownian motion. It's said the expected distance in Brownian motion is 0, which I would call the average end-position, including (-) signs. But here I am interested in the average distance using only (+) signs!

It's said the expected "spread" is √t (p,q .. probability for left,right, t.. time). Unfortunately I am not sure if "spread" is what I am looking for. For this reason I would like to ask in the form of this specific example, so that misunderstandings are prevented:

Imagine an object randomly moving either left or right, with P(left)=0.25 and P(right)=0.75! Each "time" it moves by length 1. I would like to understand how to calculate the average distance from the start point after e.g. 100 times.

I know how to calculate the average distance based on the Binomial distribution:

P(48 x left, 52 x right) x2 + P(53 x left, 47 x right) x3 + .. every outcome-probability x distance .. you get the idea!

So is there a formula to calculate this "average distance"? I would especially be interested in the time-dependence.

KaPy3141
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  • You possibly want the average of a Brownian excursion. – Paul Nov 18 '19 at 11:03
  • You want something like $\int_0^T |B'(t)|dt$? Or rather, $\int_0^T |B(t)|dt/T$ ? The former measures the sum of the sizes of all the little steps the BM took up to time $T$, the latter the time-average distance from the start position? – kimchi lover Nov 18 '19 at 11:49
  • @kimchilover: I am not sure if either of both... I actually meant to observe infinite particles: Each single particle diffuses somewhere and you measure the distance from start (always positive). What is the average distance dependent on time for all particles? – KaPy3141 Nov 18 '19 at 11:57

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It sounds, from comments, that you want, for fixed time $t$, the expected value of the absolute value of where a Brownian motion particle is, assuming it started at $0$ at time $t=0$. That is, you want $E[ |B(t) |]$.

(As if you released a zillion BM particles at $0$ at time $0$, and at some later $t$ each sends in a report telling how far from $0$ it is just then, and you take the average of all those reported distances.)

Since $B(t)\sim N(0,t)$, you want $\sqrt t$ times the expected value of the absolute value of standard normal random variable: $$ E[|B(t)|] = \frac {2\sqrt t} {\sqrt{2\pi}} \int_0^\infty x e^{-x^2/2}dx = \sqrt{\frac {2t} \pi}.$$

kimchi lover
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  • Yes, that is what I meant, thanks so much! But is this formula adjustable for different p and q (probability for left,right in 1D)? For example is p=1 and q=0, then the time-dependence obviously is 100% linear. (No Brownian motion) – KaPy3141 Nov 18 '19 at 12:23
  • The answer I gave is for (basically) the gaussian approximation to the symmetric discrete random walk. The exact formula is messy, and messier if $p\ne q$. You are asking, I think, for $E[|X-np|]$ where $X\sim Bi(n,p,q)$. – kimchi lover Nov 18 '19 at 12:31
  • @kimchilover I assume the integral is in $\mathrm{d}x$? Also, what is the difference between this and square root of mean squared displacement? Mean squared displacement gives $\sqrt{2Dt}$. – ck1987pd Feb 04 '21 at 11:44
  • @C.Koca Of course, and thanks for spotting it. Since fixed. The difference is between "mean absolute deviation" and "root mean square deviation". Since $x^2$ and $\sqrt x$ are non-linear functions, you cannot pass them thru integral signs as you may with linear functions. – kimchi lover Feb 04 '21 at 12:39
  • @kimchilover So, shouldn't I get 1/2 if $(1/\sqrt{4\pi t})\int_0^\sqrt{2t/\pi}e^{-x^2/2t}\mathrm{d}x$? but I don't. Where do I go wrong. Did I gravely misunderstand something? – ck1987pd Feb 04 '21 at 13:41
  • @C.Koca I think you did. I'm busy now, and don't have time to figure what you are doing. Better would be for you to post a new question, with background including what problem you think you are solving, and how you think it should be solved, and exactly what the point of confusion or contradiction is. – kimchi lover Feb 04 '21 at 13:52