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Limit of Sample ACF for X_t = cos(wt)

  1. Feb 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Show that the sample acf at lag h for n observations of xt = cos(wt) converges to cos(wh) as n tends to infinity.

    2. Relevant equations
    The sample acf at lag h is defined as the sample autocovariance function at lag h divided by the sample autocovariance function at lag 0, where the sample autocovariance function at lag h is defined as:

    acvf(h) = [itex]\sum[/itex]t=1n-|h|(xt+|h| - [itex]\overline{x}[/itex])(xt - [itex]\overline{x}[/itex])

    Here, [itex]\overline{x}[/itex] is the sample mean of the n observations.


    3. The attempt at a solution
    A previous question asked to show something similar for xt = a + bt, so what I did there was just substitute everything in (everything works out nicely because there are formulas for summing integers and squares from 1 to n).

    I attempted to do the same:

    xt - [itex]\overline{x}[/itex] = cos(wt) - 1/n [itex]\sum[/itex]s=1n cos(ws)


    xt+h - [itex]\overline{x}[/itex] = cos(wt + wh) - 1/n [itex]\sum[/itex]s=1n cos(ws)

    So then the numerator of acvf(h) becomes:
    [itex]\sum[/itex]t=1n-|h|(cos(wt)cos(wt+wh) - 1/n cos(wt+wh)[itex]\sum[/itex]s=1n cos(ws) - 1/n cos(wt) [itex]\sum[/itex]s=1n cos(ws) + 1/n2 [itex]\sum[/itex]s=1n [itex]\sum[/itex]r=1n cos(ws)cos(wr))

    And the denominator of acvf(h) becomes:
    [itex]\sum[/itex]t=1ncos(wt)cos(wt) - 1/n [itex]\sum[/itex]s=1n [itex]\sum[/itex]r=1n cos(ws)cos(wr)

    And at this point, I have no idea how to evaluate the cosine sums to find the limit as n goes to infinity of the fraction. I attempted to convert the cosines to exponentials by Euler's formula but I ended up with more terms that were even messier.

    Any help would be appreciated, even just some starting point or confirmation that some particular method works, and I'll try to grind through it again. Thanks.
     
  2. jcsd
  3. Feb 12, 2014 #2
    UPDATE: I solved the problem using exponentials (a bit tedious but it worked). This thread can be closed now.
     
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