Why Cramér spectral representation and not DTFT for stochastic process

Question

In a lot of time-series analysis references I find (written by mathematicians or statisticians rather than engineers), I find the following signal decomposition for a stochastic process, termed the "Cramér representation" (e.g. eqn 8.11 of this reference): $$ X[n] = \int_{\langle 2\pi \rangle} e^{-j\omega n} d Z(\omega) $$

The factor $dZ(\omega)$ is referred to as a spectral increment. I found another reference (ref, eqn 77) that said that the spectral increments are orthogonal (w.r.t. the expectation operator) if the process is stationary.

Compare this to the inverse discrete-time Fourier Transform (IDTFT), non-normalized, angular frequency convention (eqn 4.2.28 of Proakis & Manolakis, Digital Signal Processing, 4th ed): $$ X[n] = \frac{1}{2\pi} \int_{\langle 2\pi \rangle} e^{j\omega n} X(\omega) d\omega $$

With the exception of trivial differences in convention (minus sign in the exponent, normalization factor), the two representations appear to be the same. Ignoring the minus sign convention on $\omega$ for now, I am tempted to just conclude: $$ dZ(\omega) = \frac{1}{2\pi} X(\omega) d\omega $$ but I suspect there is a deeper mathematical reason why this would be wrong and that the statistics literature uses spectral increments instead.

Why do statisticians prefer the Cramér representation? Are there any computational or theoretical advantages to using it?

Does it have something to do with the convergence (or existence) of some type of integral? Or some issue relating to the fact that $X[n]$ is explicitly a stochastic process in the Cramér representation whereas the DTFT might rely on the signal being deterministic.

I wonder this because engineering education (at least mine was this way) tends to abuse notation or gloss over certain mathematical difficulties because those nuances wouldn't matter for the situations in which an engineer would be using said mathematical tools. For instance, as an undergrad I never had to learn what a Lebesgue integral was, even though I was implicitly computing Lebesgue integrals in my probability course.

Answer 1

I will introduce some terminology and intuition that will be helpful when reading other references. It will be neither complete nor completely rigorous.

---

The measures that we first encounter in real analysis assign sizes (non-negative real numbers) to measurable subsets of $\mathbb{R}$; Lebesgue measure is the measure that agrees with the intuition we build in calculus (the measure of the interval $[a,b]$ is $b-a$, etc).

$Z$ is a measure, but it is a stochastic measure†. It does not assign numbers to measurable subsets of $[0,2\pi]$. Rather, it assigns a random variable to each such subset: \begin{equation} X_A = \int_{A}dZ(\omega). \end{equation} The convergence of the integral on the right-hand side is an issue that I would rather not try to explain (to you or to myself).

In particular, the $Z$ used for WSS processes is an orthogonal stochastic measure. One result is that random variables assigned to non-overlapping sets are independent of one another.

If $A$ is a Lebesgue-measurable set, then $Z(A)$ is a random variable, and the expectation of $\left|Z(A)\right|^2$ is \begin{equation} \mathsf{E}[\left|Z(A)\right|^2] = \textrm{Lebesgue measure of $A$}. \end{equation} Hence, Lebesgue measure is "under the hood" even if we stick with the notation $dZ(\omega)$.

Just as we can use Lebesgue measure to integrate functions over subsets of $\mathbb{R}$, we can use $Z$ to integrate functions over subsets of $[0,2\pi]$ (such as all of $[0,2\pi]$).

---

Let $\mu$ be Lebesgue measure on $\mathbb{R}$, and let $\nu$ be another measure on $\mathbb{R}$. $\nu$ is said to be absolutely continuous with respect to Lebesgue measure if there is a function $f$ such that $d\nu = fd\mu$, or the measure $\nu(A)$ of $A$ is equal to \begin{equation} \nu(A) = \int_{A}f(x)d\mu(x). \end{equation} The function $f$ is called the Radon–Nikodym derivative of $\nu$ with respect to $\mu$.

Not all measures are absolutely continuous with respect to Lebesgue measure. The example most familiar to electrical engineers is Dirac measure. Lebesgue measure assigns measure zero to any set consisting of a single point, and a measure that is absolutely continuous with respect to Lebesgue measure must do the same. But the Dirac measure $\delta_0$ assigns measure 1 to the set $\{0\}$ and to any set that contains $0$. Since $\delta_0$ is not absolutely continuous with respect to Lebesgue measure, $d\delta_0$ cannot be written as $fd\mu$.

There are also more exotic measures that are not absolutely continuous with respect to the Lebesgue measure.

---

I have found no evidence of the notion of absolute continuity of stochastic measures.

EDIT: While theoretical results about spectral representations of WSS processes are crucial for applications, the $dZ$ notation may be off-putting and perhaps even doubt-inducing. I suspect that writing $Y(\omega)d\omega$ for $dZ(\omega)$ is a useful abuse of notation that allows the user to manipulate symbols as though some analogue of the Radon-Nikodym derivative existed. Rigor can be added after the fact.

Note that rigor might arrive decades after the fact. Plenty of ideas seem to work just fine without complete mathematical rigor.