Problem
Let $\{X_t\}$ be a real-valued, wide-sense stationary, continuous stochastic process and consider the autocorrelation of samples of $\{X_t\}$ taken in time windows of width $T$:
$$R_{X; T}(\tau) = E\left[Y_tY_{t + \tau}\right].$$
Here $Y_t = X_t\chi_{T}$, where $\chi$ is the indicator function on the interval $[-T/2,T/2]$. Can I express $R_{X; T}(\tau)$ in terms of the autocorrelation of $\{X_t\}$, $R_{X}(\tau) = \lim_{T\to\infty}R_{X; T}(\tau)$? What if $\{X_t\}$ is strictly stationary?
Background:
I am trying to calculate the effects of windowing a time signal on its observed autocorrelation. In particular, I assume the power spectral density (PSD) $S_X(\omega) = \mathcal{F}\left[R_X(\tau)\right](\omega)$ of the process known (where $\mathcal{F}$ is the Fourier transform) and I am trying to predict the form of $S_{X; T}(\omega) = \mathcal{F}\left[R_{X; T}(\tau)\right]$.
For a deterministic signal $x(t)$ we can define the function $y(t) = x(t)\chi_T(t)$ whose autocorrelation is given by
$$R_Y(\tau) = R_{X; T}(\tau) = ((y(t)\ast y(-t))(-\tau) = \int_{-\infty}^\infty dty(t)y(t+\tau);$$ by repeated application of the convolution theorem we then obtain
$$S_{X; T}(\omega) = \mathcal{F}\left[R_{X; T}(\tau)\right](\omega) = \left\lvert \mathcal{F}[x] \ast \mathcal{F}[\chi_T]\right\rvert^2(\omega).$$
That is, I need to obtain the transform of $\mathcal{F}\left[x\right](\omega)$ of the signal and then convolute it with the transform of the window $\mathcal{F}\left[\chi_T\right](\omega) = T \text{sinc}\left(T\omega/2\right)$ to obtain the expected PSD of the windowed signal. However, recovering $\mathcal{F}\left[x\right](\omega)$ from $S_X(\omega)$ is not possible without additional phase information even for a deterministic signal; what's worse, the Fourier transform of a stochastic process is not necessarily well-defined so these formulas are useless for my problem.
I read in a book (Glatter & Kratky, "Small Angle X-Ray Scattering", p. 439) that in the particular case of the indicator function the autocorrelation factorizes which implies:
$$R_{X; T}(\tau) = R_{X}(\tau)R_{\chi_T}(\tau) \iff S_{X; T}(\omega) = \left(S_X \ast \left\lvert \mathcal{F}\left[\chi_T\right]\right\rvert^2\right)(\omega).$$
Clearly this would be great, as I can simply use the PSD of the signal measured at long times and convolute it with $\left(T\text{sinc}\left(T\omega/2\right)\right)^2$ to obtain the PSD of the windowed signal. Is this factorization correct?
Thank you in advance to anybody who will take time to reply! And apologies for the notational abuse, I am not an expert in signal processing.