Difference between PSD estimate and variance of DFT

Question

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 $\frac{1}{L}$?

Answer 1

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!

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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.

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