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Variance in Random Walk

  1. Oct 30, 2009 #1
    Hi,

    I know that the expectation E(Sn) for a one-dimensional simple random walk is zero. But what about the variance?

    I read in http://en.wikipedia.org/wiki/Random_walk#One-dimensional_random_walk" that the variance should be E(Sn2) = n.

    Why is that? Can anyone prove it?

    Thank you very much!
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Oct 30, 2009 #2
    Just write down the definition of [itex]S_n[/itex] and you will be able to answer your question yourself.
     
  4. Oct 31, 2009 #3
    Var(Sn) = E(Sn2) = E(Z12 + Z22 + Z32 + ... + Zn2) =* E(Z12) + E(Z22) + ... + E(Zn2) = 1 + 1 + ... + 1 (n times) = n

    *variables are independent and uncorrelated

    Is this correct then?
     
  5. Oct 31, 2009 #4
    This is almost correct. [itex]S_n[/itex] is defined to be [itex]Z_1+\ldots +Z_n[/itex], where the [itex]Z_i[/itex] are independent (or at least uncorrelated) with mean zero and variance one. It follows that
    [tex]
    S_n^2 = \sum_{i,j=1}^n{Z_i Z_j}
    [/tex]
    and not, as you wrote,
    [tex]
    S_n^2 = \sum_{i=1}^n{Z_i^2}
    [/tex]

    However, using independence of the [itex]Z_i[/itex] you can still do a similar computation to prove [tex]\mathbb{E}\left[S_n^2\right]=n[/tex].
     
  6. Oct 31, 2009 #5
    Thank you!
     
  7. Oct 31, 2009 #6
    You're welcome:smile:
     
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