# Variance of periodogram estimate of the power spectrum

I have been reading chapter 13.4. ("Power Spectrum Estimation Using the FFT") of the Numerical Recipies Book.

Some things related to the expectation value of the "periodogram estimate of the power spectrum" became not clear to me though.

Background

Suppose we have an equally-spaced $N$-point sample $c_0 ... c_{N-1}$ of the discrete signal $c(t)$ and its Fourier transform $$C_k = \sum_{j=0}^{N−1} = c_j \exp{ 2πijk/N}$$ $$k = 0, . . . , N − 1$$ Then we can define the "periodogram estimate of the power spectrum" $P(f_k)$ as $$P(f_k) = \frac{1}{N^2} [{|C_k|^2 + |C_{N−k}|^2]}$$

with $k = 1, 2, . . . , \frac{N}{2}− 1$ and $f_k = \frac{k}{N\Delta}$ for $k = 0, 1, . . . ,\frac{N}{2}$ ($\Delta$ is the sampling interval).

So far everything is clear.

Now the book says

"... the variance of the periodogram estimate at a frequency $f_k$ is always equal to the square of its expectation value at that frequency. In other words, the standard deviation is always 100 percent of the value, independent of N!"

My questions

1. What is the expectation value of the periodogram estimate at a frequency $f_k$? Why (can you show me how to derive it)?
2. The expectation value doesn't change with $N$, so changing the length of the input doesn't improve the power spectrum estimation. But if I have a non-integral number of periods in my sample, more samples (longer sampling time) means less weight on the end of the signal (where there is a disconnect) so less smearing of frequency components measured due to the disconnect between the beginning and the end of the signal. So in this case a higher $N$ means a better estimation? Where am I wrong?
3. If the SD is always 100 percent of the expectation value, doesn't that make this simple periodogram estimation (using just a single DFT of $N$ samples) pretty much useless? Should one always opt for Welch's or Bartlett's method e.g.?

Thank you very much in advance!

• 3. yes-- this is the motivation to introduce Welch's method, which uses averaging to exchange frequency resolution for lower variance. – nibot Sep 19 '12 at 18:55

$\ var\{{I_{xx}}\} = S^2_{xx}e^{(j\Omega)}.[1 + (sinN\Omega / Nsin\Omega)^2]$
As $\ N \rightarrow \infty$ denominator of the second part in the parenthesis goes to infinity and the division goes to zero because of the N but summing with 1 prevents all equation to become zero. So variance converges to the square of the PSD, not the zero.