node31 Smoothing STAT 510 - Applied Time Series Analysis

node31 Smoothing STAT 510 - Applied Time Series Analysis -...

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This is Google's cache of http://onlinecourses.science.psu.edu/stat510/node/31 . It is a snapshot of the page as it appeared on 22 Aug 2010 03:50:30 GMT. The current page could have changed in the meantime. Learn more Text-only version STAT 510 - Applied Time Series Analysis ANGEL Department of Statistics Eberly College of Science Home // Section 3: Spectral Domain Models Smoothing Submitted by gfj100 on Sun, 03/28/2010 - 16:37 Quick Review . .. Let's quickly review what was done before. We looked at the discrete Fourier transform of the data where ω j = j / n when j / n ≤ 0.5 and ω j = j / n- 1 when j / n > 0.5 for j = 0, . .. , n -1. Now, these can be interpreted as the coefficients of a regression. where Now, for a particular frequency This is hardly surprising since
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as an approximation of the spectral density. This is definitely an approximation, but we can make this more concrete by saying that Another interesting thing to note is that the d c j ) and d c k ) for different j and k are nearly independent -- and are even less dependent for larger n . If we use a slightly different notation -- for n data points, we perform the periodogram at the ω j . Let's write this as ω j : n . Imagine that we get more and more data ( n → ∞) and that ω j : n → ω. It's not hard to imagine the that: E [ I j : n )] → f (ω) That's great, but there is a problem. One important thing that we do in statistics is to find confidence intervals for our estimators. However, for the periodogram, the problem is that the number of parameters that we are fitting (the f j : n )) is growing at the same rate as the data. More data means more frequencies where we want to estimate the spectral density. Asymptotically, This gives a confidence interval of Note that the width of the confidence interval never shrinks -- regardless of the size of the sample. This contrasts with a similar result for sample variance, where Note, that the mean of a is n - 1 and variance is 2( n - 1). If we divide by n - 1 we see the variance of S 2 is shrinking. This does not occur above for the periodogram. Another problem is this -- what we really want to do is to have a good representation of f (ω) which is presumably a continuous function in ω and we want to obtain this continuous function from discrete observations{in this case the periodogram I j ) which is a function across the discrete values ω j . We need to find a way to fix this.
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node31 Smoothing STAT 510 - Applied Time Series Analysis -...

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