Hello All
I know that for white noise integrating PSD results in the variance. I know the prove of that statement.
however I do not the why integrating PSD of any random noise other than white noise results in variance.
Thank you
Mija
Because E(X_{n+k}X_n) is the k-th Fourier coefficient of the PSD (by the definition) and E(X^2) should be the 0-th coefficient of the PSD. Probably you have to assume that E(X)=0 (as usually) and therefore Var(X)=E(X^2).
Dear Mina,
A relatively simple explanation for your question can be found in books/lectures on stationary random processes (of both - the discrete and continuous time). As an example one can start with the notes by Henri P. Gavin (just an example among maaany other sources):
http://people.duke.edu/~hpgavin/cee541/PowerSpectra.pdf
where the idea based on the Parseval equality is presented. However, complete proofs require a little bit of Functional Analysis. Consult also the following chapters/lectures/presentations
http://www.le.ac.uk/users/dsgp1/COURSES/MESOMET/ECMETXT/11mesmet.pdf
http://www.cas.usf.edu/~cconnor/geolsoc/html/chapter11.pdf
http://www.stat.berkeley.edu/~bartlett/courses/153-fall2010/lectures/23.pdf
In the above treatises the general noise is by defintion any stationary process X_t in the wide sense with finite variance, constant mean and the covariance dependent on the difference of instants only, i.e. Cov( X_t, X_s) = R(t-s). All relations between the process and the spectral density are expressed in terms of the moments of order 2 at most.
Good wishes
Dear Mija
The inverse Fourier transform of the psd (power spectral density) is equal to the autocorrelation function of the random process. So, we can easily see that the autocorrelation function at lag zero is equal to the integral of the psd. on the other hand, the autocorrelation function at lag zero is equal to the power of the random process. Note that if the random process is zero mean, then its power is equal to its variance.
Regards
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