Why correlation property of $\mathcal Z$-transform contains a time reversal operation

Question

I'm reading through Digital Signal Processing, Proakis and Manolakis, third edition. I've reached section 3.2: Properties of $\mathcal Z$-transform.

One property is the convolution:

$$x(n) = x_1(n)\star x_2(n) \longleftrightarrow X(z) = X_1(z)X_2(z)$$

Other property is correlation:

$$r_{x_1,\ x_2}(l) = \sum_{n=-\infty}^{\infty}x_1(n)x_2(n-l) \longleftrightarrow R_{x_1,\ x_2}(z)=X_1(z)X_2\left(z^{-1}\right)$$

I'm confused by the time reversal (folding) which appears in the correlation property at $X_2\left(z^{-1}\right)$. I know that the folding step occurs on convolution but not on cross/auto/correlation. Why is this and what am I missing?

To clarify the "time reversal" expression I'm referring to another property of $\mathcal Z$-transform taken from the same source:

If $$x(n) \longleftrightarrow X(z)$$ Then $$x(-n) \longleftrightarrow X(z^{-1})$$

For the two answers: Matt's one clarifies the matter using a mathematical perspective while Peter's one shows in an intuitive way how the folding occures in polynomial multiplication. So thanks, I wish SE had a way to mark two answers as accepted.

Answer 1

Note that (discrete-time) convolution is defined as

$$x_1[n]\star x_2[n]=\sum_kx_1[k]x_2[n-k]\tag{1}$$

and correlation is defined as

$$r_{x_1,x_2}[n]=\sum_kx_1[k]x_2[k-n]\tag{2}$$

Comparing $(1)$ and $(2)$ we see that correlation can be written as the following convolution:

$$r_{x_1,x_2}[n]=x_1[n]\star x_2[-n]\tag{3}$$

The $\mathcal{Z}$-transform of $x_2[-n]$ is given by

$$\sum_nx_2[-n]z^{-n}=\sum_nx_2[n]z^n=X_2\left(\frac{1}{z}\right)\tag{4}$$

Consequently, from $(3)$ the $\mathcal{Z}$-transform of $r_{x_1,x_2}[n]$ must be

$$R_{x_1,x_2}(z)= X_1(z)X_2\left(\frac{1}{z}\right)\tag{5}$$

Answer 2

Let's look at convolving: $$ x_1 = [1, 2, 3];\\ x_2 = [3, 2, 2]; $$ So: $$ Y(z) = X_1(z) X_2(z) = (1 + 2 z^{-1} + 3z^{-2})(3 + 2z^{-1} + 2 z^{-2})\\ = 3 + 2z^{-1} + 2 z^{-2} + 6z^{-1} + 4z^{-2} + 4 z^{-3} + 9z^{-2} + + 6z^{-3} + 6 z^{-4}\\ = 3 + 8 z^{-1} + 15 z^{-2} + 10 z^{-3} + 6 z^{-4} $$ and then via the reverse-and-multiply:

$$ \begin{array}{ccccccc} n=0 :& 0 & 0 & 0 & 1&2&3& & & = 3\\ & 0 & 2 & 2 & 3\\ n=1 :& 0 & 0 & 0 & 1&2&3& & & = 8\\ & 0 & 0 & 2 & 2 & 3\\ n=2 :& 0 & 0 & 0 & 1&2&3& & & = 15\\ & 0& 0 & 0 & 2 & 2 & 3\\ n=3 :& 0 & 0 & 0 & 1&2&3& & & = 10\\ & 0& 0& 0 & 0 & 2 & 2 & 3\\ n=4 :& 0 & 0 & 0 & 1&2&3& & & = 6\\ & 0& 0& 0& 0 & 0 & 2 & 2 & 3\\ \end{array} $$

So really polynomial multiplication and the "reverse and multiply" convolution procedure produce the same result.