Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Random walk probability

  1. Sep 2, 2011 #1
    Hi guys,

    I was reading about random walks and i encountered one step of a proof which i dont know how to derive in a mathematically rigorous way.

    the problem is in the attached file and S is a random walk with X_i as increments, X_i =

    I know that intuitively we can switch the indices to obtain the second equation from the first but how do we prove it rigorously?

    EDIT: btw, im just looking for hints, not the entire solution. i think one of the possible hints is that the X_i's are i.i.d. but i cant think of a way to use this

    Attached Files:

    Last edited: Sep 2, 2011
  2. jcsd
  3. Sep 2, 2011 #2


    User Avatar
    Science Advisor

    Since all X_i have the same distribution renumbering the indices makes no difference.
  4. Sep 3, 2011 #3
    hi mathman, thanks btw! so there's no rigorous proof for this?
  5. Sep 3, 2011 #4


    User Avatar
    Science Advisor

    I don't what you need to make it rigorous.
  6. Sep 3, 2011 #5
    Mathman's lemma: (X1,X2) has the same distribution as (X2,X1).

    Proof: P[X1<=x1,X2<=x2] = P[X1<=x1]P[X2<=x2] = P[X2<=x1]P[X1<=x2] = P[X2<=x1,X1<=x2]
  7. Sep 4, 2011 #6
    You started with the assumption that the Xi's were iid. Part of the definition of iid is that they are identical - that is, every marginal probability statement for one variable can be exchanged for any probability statement about another. The other bard of the definition of iid is that they are independent. This fact allows us to extend the above from marginal probability statements to any arbitrary joint probability statement.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook