Difference between PSD estimate and variance of DFT In Bartlett's PSD estimate one averages over L segments of the squared DFT coefficients. From wikipedia I found this formula

$$ \textrm{PSD}(k) = \frac{1}{L}\sum_{l=1}^{L} \frac{1}{M} \lvert X^{[l]}(k)\rvert^2 \tag{1}$$
where $X^{[l]}$ denotes the DFT of the $l^{\rm th}$ segment.

With my (not so standard) DFT Definition

$$
X(k) = \frac{1}{\sqrt{N}} \sum_{n=1}^{N}x[n]e^{-j\omega_kn},  \textrm{with } \omega_k = \frac{2\pi k}{N} \tag{2}
$$

Bartlett's method results in

$$\textrm{PSD}(k) = \frac{1}{L}\sum_{l=1}^{L} \frac{1}{M} \lvert\sqrt{M}\cdot X^{[l]}(k)\rvert^2 = \frac{1}{L}\sum_{l=1}^{L} \lvert X^{[l]}(k)\rvert^2 \tag{3}$$

A similar formula is given in my textbook for the (unbiased) sample variance of the DFT coefficients of a periodic signal:

$$ \hat{\sigma}_x^2(k) = \frac{1}{L-1}\sum_{l=1}^{L} \lvert X^{[l]}(k) - \hat{X}(k)\rvert^2 \tag4$$

with sample mean $\hat{X}(k) = \frac{1}{L} \sum_{l=1}^{L} X^{[l]}(k)$

Assuming that $\hat{X}=0$, equations $(3)$ and $(4)$  seem to be biased &amp; unbiased estimates of the variance $\sigma_X^2(k)$.

Questions:


Always assuming  $\hat{X}=0$, is it generally correct  to interpret the PSD as an biased estimate of the DFT variance? If not, can someone  explain  the difference, please?
Would Bartlett's method become an unbiased estimator for the DFT variance if $(3)$  was scaled by $\frac{1}{L-1}$ instead of $\frac{1}{L}$?
So, as usual, I start by pointing out what the PSD actually is:

The power spectral density is the Fourier Transform of the autocorrelation function of a wide-sense stationary random process.


  Always assuming  $\hat{X}=0$, is it generally correct  to interpret the PSD as an biased estimate of the DFT variance? If not, can someone  explain  the difference, please?


The Wiener-Chintschin¹ theorem points out that this happens to coincide with the expectation value of the absolute of the Fourier transform of a signal.  In other words, to answer your first question:

The PSD is a stochastic (hence, "invisible") property of a random process. The DFT can be used to estimate that; not the other way around!

 

To answer your second question, a bit of juggling the term of variance is required:


  Would Bartlett's method become an unbiased estimator for the DFT variance if (3) was scaled by $\frac{1}{L-1}$ instead of $\frac{1}{L}$?


Variance is formally defined as the expectation of the square of the deviation of any observation to the mean:

$$Var(X) = \mathbf E\left[\left(X-\mu\right)^2\right]$$

Now, applying Parseval's theorem, the power in a time and frequency domain should be identical. Considering a white random process, where the covariance of two samples is 0, it's easy to see that variance is kind-of-linear:

$$Var\left(\sum\limits_{i=1}^L X_i\right) = \sum\limits_{i=1}^N Var(X_i)$$

Considering the special case of L=1, it becomes clear that $(4)$ isn't applicable here – you kind of get the standard deviation of the process, but not the variance.


¹ there's more latin-lettered spellings to Chintschin than there's letters in in his cyrillic name...