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 & 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 $\frac{1}{L}$?