When deriving the power spectral density of stochastic processes, why does taking an expectation allow the $T\rightarrow\infty$ limit to be taken? I am following the arguments presented in the paper AN-255 Power Spectra Estimation, from Texas Instruments, to learn how to derive the power spectral density for a stationary stochastic process, and do not understand one of their steps. I am looking at Page 4 only of the the document for the purposes of this question.

The derivation is done by considering only a single realisation (sample function), denoted by $x(t)$, of the stochastic process. Then they do the usual thing of only considering a truncated version of this signal $x_T(t)$, which allows the Fourier transform of it to exist, denoted by $X_T(f)$ (because one of the requirements for FT to exist is absolute integrability, which isn't satisfied for random signals).

I am confused with how they justify taking the limit $T\rightarrow\infty$ in the right hand column of page 4, shown here:

$\hskip2in$

$\hskip2in$

I understand that you cannot take the limit $T\rightarrow\infty$ directly on the truncated sample function's Fourier transform $X_T(f)$, because it is not defined in that limit (indeed, that was the reason for doing the truncation in the first place). 

I also understand that $X_T(f)$ results from a single realisation only of the stochastic process, and if you were to repeat the experiment again you would obtain a different $X_T(f)$. As such, this function is actually a random variable itself (at each frequency $f$), and therefore it makes sense to do an ensemble average over many realisations, and consequently motivates taking an expectation.

What I don't understand is why the expectation solves the problem, and allows you to take the limit $T\rightarrow\infty$ after you have taken the expectation:

$\hskip2in$

What is it about the expectation of $X_T(f)$ that means the limit $T\rightarrow\infty$ can then be taken? I'm having a hard time seeing it... Could someone spell this out plainly for me?
It is not the expectation operator that makes sure that the limit exists. The expectation just results in an ensemble average, which we need to obtain a deterministic function $S(f)$ for the power spectrum.

Assume we're given a deterministic power signal $x(t)$, i.e., a signal with finite non-zero power, and, consequently, infinite energy. Its Fourier transform generally doesn't exist. We can define a truncated Fourier transform

$$X_T(\omega)=\int_{-T}^Tx(t)e^{-j\omega t}dt\tag{1}$$

By assumption, $\lim_{T\to\infty}X_T(\omega)$ doesn't exist. However, the limit

$$\lim_{T\to\infty}\frac{\left|X_T(\omega)\right|^2}{2T}\tag{2}$$

exists because of the finite power assumption, and this is also the way the power spectrum is defined for such deterministic power signals.

If $x(t)$ is modeled as a random signal, we only need to modify $(2)$ by taking the expectation of the numerator to obtain the power spectrum of $x(t)$.