\[ I_{N}\left ( \omega \right ) ={\displaystyle \sum \limits _{m=-\left ( N-1\right ) }^{N-1}} c_{xx}\left ( m\right ) e^{-j\omega m}\]
\begin{align} \left \vert X\left ( e^{j\omega }\right ) \right \vert ^{2} & =X\left ( e^{j\omega }\right ) X^{\ast }\left ( e^{j\omega }\right ) \nonumber \\ & =\left ({\displaystyle \sum \limits _{m=0}^{N-1}} x\left ( m\right ) e^{-j\omega m}\right ) \left ({\displaystyle \sum \limits _{n=0}^{N-1}} x\left ( n\right ) e^{-j\omega n}\right ) ^{\ast }\nonumber \\ & =\left ({\displaystyle \sum \limits _{m=0}^{N-1}} x\left ( m\right ) e^{-j\omega m}\right ) \left ({\displaystyle \sum \limits _{n=0}^{N-1}} x^{\ast }\left ( n\right ) e^{j\omega n}\right ) \nonumber \\ & ={\displaystyle \sum \limits _{m=0}^{N-1}}{\displaystyle \sum \limits _{n=0}^{N-1}} x\left ( m\right ) x^{\ast }\left ( n\right ) e^{-j\omega m}e^{j\omega n}\nonumber \end{align}
But
\[ e^{-j\omega m}e^{j\omega n}=e^{-j\omega \left ( m-n\right ) }\]
and
\[ x\left ( m\right ) x^{\ast }\left ( n\right ) =x\left ( m\right ) x^{\ast }\left ( m+\left ( n-m\right ) \right ) \]
So
\[ \left \vert X\left ( e^{j\omega }\right ) \right \vert ^{2}={\displaystyle \sum \limits _{m=0}^{N-1}}{\displaystyle \sum \limits _{n=0}^{N-1}} x\left ( m\right ) x^{\ast }\left ( m+\left ( n-m\right ) \right ) e^{-j\omega \left ( m-n\right ) }\]
Let \(n-m=\tau \) then above can be rewritten as
\[ \left \vert X\left ( e^{j\omega }\right ) \right \vert ^{2}={\displaystyle \sum \limits _{m=0}^{N-1}}{\displaystyle \sum \limits _{n=0}^{N-1}} x\left ( m\right ) x^{\ast }\left ( m+\tau \right ) e^{j\omega \tau }\]
When \(n=0\),\(m=-\tau \) and when \(n=N-1\),\(m=N-\tau -1\), hence the above becomes
\begin{align*} \left \vert X\left ( e^{j\omega }\right ) \right \vert ^{2} & ={\displaystyle \sum \limits _{m=0}^{N-1}}{\displaystyle \sum \limits _{m=-\tau }^{N-\tau -1}} x\left ( m\right ) x^{\ast }\left ( m+\tau \right ) e^{j\omega \tau }\\ & ={\displaystyle \sum \limits _{m=0}^{N-1}} \left ({\displaystyle \sum \limits _{m=-\tau }^{-1}} x\left ( m\right ) x^{\ast }\left ( m+\tau \right ) e^{j\omega \tau }+{\displaystyle \sum \limits _{m=0}^{N-\left \vert \tau \right \vert -1}} x\left ( m\right ) x^{\ast }\left ( m+\tau \right ) e^{j\omega \tau }\right ) \\ & ={\displaystyle \sum \limits _{m=0}^{N-1}} \left ({\displaystyle \sum \limits _{m=-1}^{-\tau }} x\left ( m\right ) x^{\ast }\left ( m+\tau \right ) e^{j\omega \tau }+N\ c_{xx}\left ( m\right ) e^{j\omega \tau }\right ) \end{align*}
I made another attempt at the end,
We see that \(S_{xx}\left ( \omega \right ) \) is the Fourier transform of \(c_{xx}\left ( m\right ) w\left ( m\right ) \). i.e.
\[ S_{xx}\left ( \omega \right ) =\digamma \left [ c_{xx}\left ( m\right ) w\left ( m\right ) \right ] \]
Where \(\digamma \) is the Fourier transform operator. Using modulation property
\[ S_{xx}\left ( \omega \right ) =\frac{1}{2\pi }\left ( \digamma \left [ c_{xx}\left ( m\right ) \right ] \otimes \digamma \left [ w\left ( m\right ) \right ] \right ) \]
But \(I_{N}\left ( \omega \right ) =\) \(\digamma \left [ c_{xx}\left ( m\right ) \right ] \) and let \(W\left ( \omega \right ) =\digamma \left [ w\left ( m\right ) \right ] \), then the above becomes
\begin{align*} S_{xx}\left ( \omega \right ) & =\frac{1}{2\pi }\left ( I_{N}\left ( \omega \right ) \otimes W\left ( \omega \right ) \right ) \\ & =\frac{1}{2\pi }{\displaystyle \int _{-\pi }^{\pi }} I_{N}\left ( \theta \right ) W\left ( \omega -\theta \right ) d\theta \end{align*}
Hence, taking expectation of LHS, and since only \(I_{N}\left ( \theta \right ) \) is random, then the above becomes (after moving expectation inside the integral in the RHS)
\[ E\left [ S_{xx}\left ( \omega \right ) \right ] =\frac{1}{2\pi }{\displaystyle \int _{-\pi }^{\pi }} E\left [ I_{N}\left ( \theta \right ) \right ] W\left ( \omega -\theta \right ) d\theta \]