Average displacement based on probability .

Click For Summary

Discussion Overview

The discussion revolves around the concept of average displacement, particularly when the average value is zero. Participants explore the implications of a zero average displacement in terms of probability, variation, and the possibility of calculating root-mean-square values. The conversation touches on theoretical aspects and examples related to random walks and displacement in one-dimensional motion.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions whether an average displacement of zero implies equal chances of being on either side of the initial position.
  • Another participant argues that a zero average does not necessarily indicate equal displacement chances, citing examples where larger displacements can skew the average.
  • A participant provides a specific example with values to illustrate how a single large negative displacement can influence the average to be zero despite multiple smaller positive displacements.
  • There is curiosity about the derivation of root-mean-square values when the average displacement is zero, with some suggesting it is still possible due to variation in results.
  • One participant expresses confusion about the implications of a zero average and seeks clarification on the influence of larger displacements on the average.
  • Another participant mentions random walks and the challenge of incorporating time into the analysis of displacement.

Areas of Agreement / Disagreement

Participants express differing views on the implications of a zero average displacement, with some agreeing that it does not imply equal chances of displacement while others provide counterexamples. The discussion remains unresolved regarding the relationship between average displacement and the underlying probabilities.

Contextual Notes

Participants reference specific examples and scenarios that highlight the complexity of average displacement calculations, but there are unresolved assumptions about the definitions and implications of average values in this context.

terp.asessed
Messages
126
Reaction score
3
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?
 
Physics news on Phys.org
terp.asessed said:
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.
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.
Oh, I am also curious as to if it be still possible to derive root-mean-square even if displacement <x> = 0?
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.
 
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.
 
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 anyone of the 1's.
 
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?
 
Right
 
Thank you! I think I am understanding better!
 
  • Like
Likes   Reactions: FactChecker
terp.asessed said:
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?

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. .
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K