# Average displacement based on probability .

1. Sep 26, 2014

### terp.asessed

I am just curious, if I happen to get average value of displacement x = 0 (as in <x> = 0), does it mean it is zero because there are equal chances of being right or left of the initial position (assuming a point moves horizontally only, to right or left of the initial postion = 0)? Then, even if I want to know how long it takes for a point to take from one place to another, would <x> be of no use in answering a question as to how long a point usually takes?

Oh, I am also curious as to if it be still possible to derive root-mean-square even if displacement <x> = 0?

2. Sep 26, 2014

### FactChecker

No. There can be one displacement of -1,000 and it would take one thousand displacements of 1 to average out to 0. So larger displacements in one direction have more effect on the average than small displacements.
Assuming <x> still means average (a non-standard notation), then the answer is yes. An average of zero does not mean that there is no variation in the results. There is still a root mean square of results minus average.

3. Sep 26, 2014

### terp.asessed

Could you please clarify the following words you typed? I'm sorry, but I still have trouble understanding. Then, what's the point of <x> =0 (yep, it means average) if it's possible?:

No. There can be one displacement of -1,000 and it would take one thousand displacements of 1 to average out to 0. So larger displacements in one direction have more effect on the average than small displacements.

4. Sep 26, 2014

### FactChecker

In the example I mentioned, there are 10001 values. One value of -1000 and a thousand values of 1. So the average is (-1000 + 1 + 1 + ...+1)/1001, where there are a thousand 1s. That is 0/1001 = 0. So the one large displacement in the negative direction had much more influence on the average than any one of the 1's.

5. Sep 26, 2014

### terp.asessed

Hmmm. Interesting. But, if, in another example, a person walks to both sides equally, which means that there are equal displacement in both negative and positive, still that would make average to 0. Still, it means that <x> does not necessarily mean there are equal chances of displacement to both sides, right?

6. Sep 26, 2014

### FactChecker

Right

7. Sep 26, 2014

### terp.asessed

Thank you!!!!!!!!! I think I am understanding better!

8. Oct 1, 2014

### WWGD

I think this has to see with Random Walks on the Real line, tho with Random Walks you can only figure out the probability of being in a certain spot/coordinate, but I don't know how to take time into account. .

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