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Can you tell me how the $P_x(z)=P_x^*(1/z^*)$ is mathematically correct. I can understand the $P_x(e^{jw})=P_x^*(e^{jw})$ as $P_x$ is real value. But why take the Z domain representation in this way ($P_x(z)=P_x^*(1/z^*)$). Why can't I take it like $P_x(z)=P_x^*(z)$ like the Fourier transform representation?

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The condition $P_x(e^{j\omega})=P^*_x(e^{j\omega})$ corresponds to the time domain symmetry

$$r_x(n)=r^*_x(-n)\tag{1}$$

where $r_x(n)$ is the autocorrelation of $x(n)$, which is the inverse (discrete time) Fourier transform of $P_x(e^{j\omega})$.

Now you just need to figure out the $\mathcal{Z}$-transform of $r_x^*(-n)$ in terms of $P_x(z)$ in order to see the symmetry relationship in the $\mathcal{Z}$-transform domain:

$$\begin{align}\mathcal{Z}\big\{r_x^*(-n)\big\}&=\sum_nr_x^*(-n)z^{-n}\\&=\sum_nr_x^*(n)z^n\\&=\left[\sum_nr_x(n)(z^*)^n\right]^*\\&=\left[\sum_nr_x(n)\left(\frac{1}{z^*}\right)^{-n}\right]^*\\&=P_x^*\left(\frac{1}{z^*}\right)\tag{2}\end{align}$$

Consequently, the symmetry condition $(1)$ implies

$$P_x(z)=P_x^*\left(\frac{1}{z^*}\right)\tag{3}$$

For real-valued sequences, $(3)$ can also be written as

$$P_x(z)=P_x\left(\frac{1}{z}\right)\tag{4}$$

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Simple way of seeing this:

From the definition of the Fourier Transform, when you transform a signal into a linear combination of sines (or cosines), you get the complex conjugates of all the possible values within the spectrum of the input signal. I am not an expert on mathematics but, when you transform the exponential term of your Fourier transform into cosines and sines, you have two sides, a positive side of your spectrum which are all the real values and the negative side of your spectrum which are all the complex values.

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As you see here, they have the same value in magnitude but different signs, that is why you have symmetry with respect to zero (positive and negative).

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Please, correct me if I am wrong! But this is a simple way to explain that symmetry. I am sure someone else could give a bit more detailed explanation

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