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.
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: $$ E[|B(t)|] = \frac {2\sqrt t} {\sqrt{2\pi}} \int_0^\infty x e^{-x^2/2}dx = \sqrt{\frac {2t} \pi}.$$