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

Limiting Distribution of Scaled Random Walk

  1. Sep 27, 2012 #1
    Hi all,

    in text the formula for scaled random walk is:

    W^(n) (t) = (1/√n) M_nt

    in the example it says that:

    set t=0.25, n=100 and consider the set of possible values of W^(100) (0.25) = 1/10 M_25. This random variable is generated by 25 coin tosses, and since the unscaled random walk M_25 can take the value of any odd integer between -25 and 25. My first question is that why unscaled random walk takes only the odd integer?

    The scaled random walk W^(100) (0.25) can take any of the following values:
    -2.5, -2.3, -2.1,......,-0.3,-0.1,0.1,0.3,.......,2.1,2.3,2.5
    My second question is that why only this range?

    In order for W^(100) (0.25) to take the value 0.1, we must get 13 heads and 12 tails in the 25 coin tosses. The probability of this is

    P{W^(100) (0.25) = 0.1} = {(25!)/(13! *12!)} * (1/2)^25 = 0.1555

    by drawing a histogram bar centered at 0.1 with area 0.1555, since this bar has width 0.2, its height must be 0.1555/0.2 = 0.7775.
    My last question is that why 13 heads and 12 tails, and why we did the factorial part "{(25!)/(13! *12!)}"?

    Thanks in advance.
  2. jcsd
  3. Sep 27, 2012 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    Apparently the value of one realization of the process is the total number of excess heads in tossing 25 coins, which can't be an even number. (e.g. 4 more heads than tails would imply T + (T+4) = 25 so 2T = 21. )

    Were you thinking that the process involved tossing one coin per "step" and recording the accumulated number of heads after each step?
    Last edited: Sep 27, 2012
  4. Sep 27, 2012 #3
    Sorry, I still don't get your answer :(

    The text says that if head comes then move one step up (by taking +1 on y-axis), otherwise move one step down (by taking -1 on y-axis), and at each toss move 1 step forward on x-axis no matter what is the outcome. So, I guess that the answer to your question is YES!
  5. Sep 27, 2012 #4

    Stephen Tashi

    User Avatar
    Science Advisor

    You aren't being precise about what the text says because you aren't relating its words to the symbols your are using. If you quote exactly what the text says, perhaps someone can explain it.
  6. Sep 28, 2012 #5
    I have copied the original post from the text, the previous post in the text is as follow:

    To construct a symmetric random walk, we repeatedly toss a fair coin (p, the probability of H on each toss, and q=1-p, the probability of T on each toss, are both equal to 1/2). We denote the successive outcomes of the tosses by ω=ω_1ω_2ω_3...... In other words, ω is the infinite sequence of coin tosses, ω_n is the outcome of the nth toss. Let,

    X_j = {1 if ω_j=H , -1 if ω_j=T,

    and define M_0=0,

    M_K = j=1Ʃ^k X_j, k=1,2,3,...

    The process M_k, k=1,2,3,.... is a symmetric random walk. With each toss, is either steps up one unit or down one unit, and each of the two possibilities is equally likely.
  7. Sep 28, 2012 #6

    Stephen Tashi

    User Avatar
    Science Advisor

    According to that definition M_25 does count how much the number of heads in 25 tosses exceeds the number of tails. M_25 amounts to the random variable: (+1)(number of heads) + (-1) number of tails = (number of heads - number of tails) .

    The number of heads in 25 tosses can be an even number, such as 24, but how much this differs from the number of tails (which is 23 in that example) cannot be an even number.

    If I mind-read what your textbook is trying to do, I'd say it is taking a slow approach to introducing the Wiener process - or has it already done that?
  8. Sep 28, 2012 #7
    Another way to look at it, is that X_j = 2*B_j - 1 where the B_j are Bernoulli random variables. Plug this in and you get M_K = 2*N_k - k where N_k is a Binomial random variable. Once you're comfortable with this expression, the answers to your earlier questions should be fairly easy.
  9. Oct 31, 2012 #8
    Thank you sir
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook