Dear list,
I had some confusion regarding what function too use in order too relate
results from spec.pgram() too a chi-square distribution. The documentation
indicates that the PSD estimate can be approximated by a chi-square
distribution with 2 degrees of freedom, but I am having trouble figuring out
how to do it in R, and figuring out what specifically that statement in the
documentation means. I have very little exposure to distribution functions
in R.
I have started with a signal that is simply a sine function, with a visible
periodicity within the nyquist range.
>a <- sin(2*pi *(0:127)/16)
>plot(a, type="l")
I then get the raw power spectrum using, for instance:
>PSD <- spec.pgram(a, spans=NULL, detrend=F)
Next I did a search for information on chi-square.
>help.search("chi square")
. provided a list of potentials, of which Chisquare() seemed to fit because
it mentioned the distribution.
>?Chisquare
. provided the documentation, which shows four functions and their
descriptions. I've assumed that I need too use dchisq() for my purposes, so
that the fitted distribution would be: [assuming the power returned by
spec.pgram() ARE regarded as quantiles.]
>plot(dchisq(PSD$spec, df=2))
. but the values are not between (0,1).
>plot(PSD$freq, pchisq(PSD$spec, df=2, lower.tail=F))
. looks better because values range from 0-1.
Please clarify how to fit the PSD estimate to a chi-square distribution and
any transforms that may be needed to get the quantiles from the PSD. Is what
I tried what the documentation 'had in mind' when it says "df: The
distribution of the spectral density estimate can be approximated by a chi
square distribution with 'df' degrees of freedom." ? If so, what is the
appropriate function call too use (pchisq(), dchisq(), or qchisq())? If not,
what function should I consider?
Thanks in advance,
Keith C.
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