Complex cross-correlation: conjugate first or second argument?

Question

If I understand correctly the definition of cross-correlation, the complex conjugate is not applied to the same argument depending on whether we're working with deterministic signals or with stochastic processes.

For deterministic signals, here's the the Wikipedia page, citing Rabiner and Schafer:

$${\displaystyle (f\star g)[n]\ \triangleq \sum _{m=-\infty }^{\infty }{\overline {f[m]}}g[m+n]}$$

For random vectors, here's another section of the same Wikipedia page, citing Gubner 2006:

$${\displaystyle \operatorname {R} _{\mathbf {Z} \mathbf {W} }\triangleq \ \operatorname {E} [\mathbf {Z} \mathbf {W} ^{\rm {H}}]}$$

These two definitions are inconsistent with each other. I feel like I would prefer the latter because it is consistent with sesquilinear forms, which are linear in their first argument and semilinear in their second argument.

Here are some DSP.SE answers in which the "conjugate first argument" convention is being used:

• https://dsp.stackexchange.com/a/55087/74603
• Correlation : Cross correlation why we need to multiply the samples
• Auto-correlation of the sum of two generic signals
• Efficient way to calculate $n$ first elements of cross-correlation using FFT (in Fourier domain)

Why has the DSP community adopted another convention than the statistical and mathematical communities? Is it because we prefer to apply both conjugation and time reversal to the same argument before convolving via FFT?

[EDIT: my question is different from this question which asked about the sign of temporal lag when cross-correlating real signals. The OP suggested to replace $(f\star g)[n]$ by $(f\star g)[-n]$. I am not suggesting that at all. What i am suggesting is to replace $(f\star g)[n]$ by its complex conjugate.]

Answer 1

Since there seems to be quite a bit of discussion and previous answers, I'll add a different perspective of that from the PSD.

The PSD has two definitions: \begin{align} \phi(\omega) &= \sum_{n=-\infty}^{\infty}r(n)e^{-j\omega n} \\ &= \lim_{N\to\infty}E\left\{\frac{1}{N}\left\lvert\sum_{n=0}^{N-1}y(n)e^{-j\omega n}\right\rvert^{2}\right\} \end{align} which are equivalent. These can be shown equivalent by [1] \begin{align} \lim_{N\to\infty}E\left\{\frac{1}{N}\left\lvert\sum_{n=0}^{N-1}y(n)e^{-j\omega n}\right\rvert^{2}\right\} &= \lim_{N\to\infty}\frac{1}{N}\sum_{n=0}^{N-1}\sum_{m=0}^{N-1}E\left[y(t)y^{*}(s)\right]e^{-j\omega(t-s)} \end{align} If we define the ACS as \begin{equation} r_{yy} = E\left[y(t)y^{*}(t-s)\right] = E\left[y^{*}(t)y(t+s)\right] \end{equation} then by allowing $\tau = t-s$, after some mild assumptions [1] (not related to the topic at hand) we get \begin{equation} \sum_{\tau=-\infty}^{\infty}r(\tau)e^{-j\omega\tau} \end{equation} On the contrary, if we define the ACS as \begin{equation} r_{yy} = E\left[y^{*}(t)y(t-s)\right] = E\left[y(t)y^{*}(t+s)\right] \end{equation} we instead get \begin{equation} \lim_{N\to\infty}\frac{1}{N}\sum_{n=0}^{N-1}\sum_{m=0}^{N-1}E\left[y^{*}(t)y(s)\right]e^{j\omega(t-s)} = \sum_{\tau=-\infty}^{\infty}r(\tau)e^{j\omega\tau} \end{equation} This corresponds to the PSD at frequency $-\omega$ as opposed to the frequency $\omega$. For real signals, the difference in definition is not a problem, however, it is important for complex signals where the spectrum might not be mirrored. Thus, the first definition of $r_{yy}$ is typically desirable. The proof was adapted from [1], and you can see chapter 1 for more details.

[1] - Stoica, P., & Moses, R. L. (2005). Spectral analysis of signals (Vol. 452, pp. 25-26). Upper Saddle River, NJ: Pearson Prentice Hall.