Please teach me how to prove this random walk problem....

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The discussion revolves around proving the mean position in a random walk problem, specifically addressing cases of equal and varying step lengths. Participants explore how to demonstrate that the mean position after a series of steps returns to the starting point, emphasizing the importance of symmetry in the calculations. They suggest breaking down longer steps into smaller equal-length steps to simplify the proof. There is a debate on the necessity of complex mathematics versus straightforward reasoning to arrive at the conclusion. Ultimately, the conversation highlights the need for clarity in mapping the problem's conditions to the proof's requirements.
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Homework Statement
prove that the mean position after a given number of steps is the starting position
Relevant Equations
random walk
246418

I think I can prove equal length version problem, But I am confusing in this case,,,

Please help me!
 
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If you understand the case of 1 path length, consider the case of two allowed path lengths. If you can do that I believe the generalization will be evident.
 
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Ok, I tried it like this.
Is it right??
1562850042417.png
 
OK, So how do you show the single length case to give zero..
 
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You know the mean of equal-distance lengths is zero. Now take the hint given: you can make those equal-length steps as small as you like!
 
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Thanks your reply.
If I make the length of one step 0, the mean of length is 0.
##k→0,<k>=0##
So,,,,,I don't know what does this mean :<
 
arcTomato said:
Problem Statement: prove that the mean position after a given number of steps is the starting position
Relevant Equations: random walk

I think I can prove equal length version problem
First things first. Make good on your "I think I can"...and show me
Then we will worry about the way to generalize
 
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hutchphd said:
First things first. Make good on your "I think I can"...and show me

246458

Is it wrong this photo??I thought this can prove equal length version.
If this is wrong, I have to withdraw my statement. ;;
 
arcTomato said:
View attachment 246458
Is it wrong this photo??I thought this can prove equal length version.
If this is wrong, I have to withdraw my statement. ;;
Your formula for <k> is correct but how do I know it sums to zero?
 
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  • #10
##P(n,k)=P(-n,k)##
So, the pair of numerator are canceled and the one having ##k=0## is uncancelled. this shows the sum is zero, right??
 
  • #11
arcTomato said:
##P(n,k)=P(-n,k)##
So, the pair of numerator are canceled and the one having ##k=0## is uncancelled. this shows the sum is zero, right??
Exactly. You would formally write that as
##ΣkP(n,k)=ΣkP(n,k)-ΣkP(-n,k)##
##=k Σ(P(n,k)-P(-n,k))=0 ##,, by symmetry
and you would explicitly delimit the sums (I'm too lazy )

I think you see that the multi-length sum can be manipulated simply to produce the same result
 
  • #12
You don't need any fancy math.
Say you have two steps of different lengths x1 and x2. How about dividing each of those into smaller, equal-length steps of x0 each? So x1 + x2 = a*x0 + b*x0 where a and b are different positive integers. Now you have a series .of equal-length steps. And you know that equal-length steps of any number give a zero mean, right?
 
  • #13
rude man said:
You don't need any fancy math.
Say you have two steps of different lengths x1 and x2. How about dividing each of those into smaller, equal-length steps of x0 each? So x1 + x2 = a*x0 + b*x0 where a and b are different positive integers. Now you have a series .of equal-length steps. And you know that equal-length steps of any number give a zero mean, right?
Why would you do this? The other result is two lines and it is patently correct. And the "fancy math" is the common language of probability which is the pedagogic purpose here.
 
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  • #14
I think I got it
How can I write the calculation process?

##<k>=<k_0+k_1+k_2,,,,>=<k_0>+<k_1>+<k_2>+,,,,,=0##
right??
 
  • #15
arcTomato said:
I think I got it
How can I write the calculation process?

##<k>=<k_0+k_1+k_2,,,,>=<k_0>+<k_1>+<k_2>+,,,,,=0##
right??
Absolutely.
I will let you worry with the formalism for a while...work it from the end backwards if it helps. Or meet in the middle.
 
  • #16
I appreciate for your help!
Tell me again if there is a problem I do not understand!
 
  • #17
rude man said:
You don't need any fancy math.
Say you have two steps of different lengths x1 and x2. How about dividing each of those into smaller, equal-length steps of x0 each? So x1 + x2 = a*x0 + b*x0 where a and b are different positive integers. Now you have a series .of equal-length steps. And you know that equal-length steps of any number give a zero mean, right?
I don't see how that results in a proof. What is the mapping between this and the original problem? E.g. how does a sequence of n steps translate, and where might you end up?
Secondly, it gets a little messy if the two lengths have an irrational ratio.
Thirdly, although I do not get the point of the hint, you seem to have misread it. It says, a succession of random walks, wherein each random walk has a single step length (implying that different walks in the sequence can have different step lengths).
 
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