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Covariance function property

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  1. Oct 25, 2014 #1
    Hi all. My task is to prove the property of covariance function:

    ##(r(n)-r(m))^2≤2r(0)(r(0-r(n-m)))##

    My solution:

    ##1) (r(n)-r(m))^2=r(n)^2-2r(n)r(m)+r(m)^2##
    ##2) 2r(0)(r(0)-r(n-m)))=2r(0)^2-2r(0)r(n-m)##

    From covariance function properties I know that ##2r(0)^2≥r(n)^2+r(m)^2##
    So now I just need to prove that ##r(0)r(n-m)≤r(n)r(m)##
    But dont know how to do that. Any thoughts? I'm not sure, maybe my way of solution is bad and I need to find other one. Any help would be appreciate.
     
  2. jcsd
  3. Oct 25, 2014 #2

    Ray Vickson

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    What is the function ##r(n)##; that is, how is it defined, and what is its relation to anything?
     
  4. Oct 26, 2014 #3
    If I understand you I'll try to answer.
    ##r(n)## is covariance function of stationary process. ##r(n)## show the covariance between two random variables and ##n## is operator of distance.
    ##r(n)=EX_nX_0-EX_nEX_0##
    Because process X is stationary, then ##EX_nX_0-EX_nEX_0=EX_{n+h}X_h-EX_{n+h}EX_h##
    And some properties:
    ##r(0)≥0##
    ##r(h)=r(-h)##
    ##|r(h)|≤r(0)##
    But it does not help me to solve the problem :(
     
  5. Oct 26, 2014 #4

    Ray Vickson

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    OK, so you have a stationary process in which
    [tex] \text{Cov}(X_{n+k},X_n) = r(k), k=0,1,2, \ldots . [/tex]
    In that case, the notation in the question still does not make sense: the RHS involves ##r(0-r(n-m))= -r(n-m)##, and I cannot fathom this. If we stick to your definition, this means
    [tex] \text{Cov}(X_k,X_{k-(-r(n-m))}),[/tex]
    and there is no reason for a time like ##t = k -(-r(n-m)) = k + r(n-m)## to be integer-valued.

    Did your problem statement have a typo in it? Did you really mean ##r(0) - r(n-m)?##
     
    Last edited: Oct 26, 2014
  6. Oct 26, 2014 #5
    Oh my God. I made a mistake :) In the first line it should be ##r(0)*(r(0)−r(n−m))##
     
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