Symmetries of Higher-Order Temporal Probabilistic Parameters in CSP

What are the unique parts of the multidimensional cyclic moments and cyclic cumulants?

In this post, we continue our study of the symmetries of CSP parameters. The second-order parameters–spectral correlation and cyclic correlation–are covered in detail in the companion post, including the symmetries for ‘auto’ and ‘cross’ versions of those parameters.

Here we tackle the generalizations of cyclic correlation: cyclic temporal moments and cumulants. We’ll deal with the generalization of the spectral correlation function, theĀ  cyclic polyspectra, in a subsequent post. It is reasonable to me to focus first on the higher-order temporal parameters, because I consider the temporal parameters to be much more useful in practice than the spectral parameters.

This topic is somewhat harder and more abstract than the second-order topic, but perhaps there are bigger payoffs in algorithm development for exploiting symmetries in higher-order parameters than in second-order parameters because the parameters are multidimensional. So it could be worthwhile to sally forth.

As we did with the second-order parameters, we’ll want to determine the symmetries, if any, relating to the lag (delay) vector $\boldsymbol{u}$ in the temporal parameters and the cycle frequency.

The $n$th-order temporal moment and cumulant functions are functions of the $n$ variables in the set

$\displaystyle X = \left\{ x^{(*)_j} (t + u_j) \right\}_{j=1}^n \hfill (1)$

where $(*)_j$ is the $j$th optionally applied conjugation, and by convention we say that $0 \leq m \leq n$ of these conjugates are used. This notation is often sufficient for our purposes, but not in this post. Let’s define a binary indicator vector

$\displaystyle \boldsymbol{b} = [b_1 \ b_2 \ \ldots b_n] \hfill (2)$

and corresponding conjugation vector $\boldsymbol{h}$, where

$\displaystyle h_j = \left\{ \begin{array}{ll} *, & b_j = 1, \\ 1, & b_j = 0, \end{array} \right. \hfill (3)$

so that our set of variables can be expressed as

$\displaystyle X = \left\{ x^{h_j} (t + u_j) \right\}_{j=1}^n \hfill (4)$

and therefore the specific relationship between the $j$th delay and the conjugation status of $x(t + u_j)$ is made clear through $\boldsymbol{b}$. Key to our understanding of the symmetries of the cyclic moments and cumulants is the notion of permuting the indicator vector. Let $\boldsymbol{q}$ denote a permutation of the index set $\{1\ 2\ \ldots n\}$. For example, for $n = 4$, there are $24$ permutations, and these are easily enumerated or obtained in MATLAB through the function perms.m, as shown here:

We want to consider permutations of the indicator function $\boldsymbol{b}$, so we’ll use the notation

$\displaystyle \boldsymbol{c} = \boldsymbol{b}(\boldsymbol{q}) \hfill (5)$

to indicate that $\boldsymbol{c}$ is a permutation of $\boldsymbol{b}$ specified by the reordering implied by $\boldsymbol{q}$. For example, suppose

$\displaystyle \boldsymbol{z} = [5\ 6\ 7\ 8] \hfill (6)$

and

$\displaystyle \boldsymbol{q} = [1\ 2\ 3\ 4] \hfill (6)$

Then

$\displaystyle \boldsymbol{y} = \boldsymbol{z}(\boldsymbol{q}) = \boldsymbol{z} \hfill (7)$

that is, no reordering is implied by this $\boldsymbol{q}$. For

$\displaystyle \boldsymbol{q} = [2\ 4\ 1\ 3] \hfill (8)$

we have

$\displaystyle \boldsymbol{y} = \boldsymbol{z}(\boldsymbol{q}) = [6\ 8\ 5\ 7] \hfill (9)$

Also, we’ll use a bar to denote an indicator vector that is the complement of another indicator vector

$\displaystyle \bar{\boldsymbol{b}} = [1\ 1\ \ldots 1] - \boldsymbol{b} \hfill (10)$

So when $b_j = 0$, $\bar{b}_j = 1$, and when $b_j = 1$, we have $\bar{b}_j = 0$.

Symmetry in the Delay Variables

If we consider the $n$th-order cyclic temporal moment function,

$\displaystyle R_x^\alpha(\boldsymbol{u}; n, \boldsymbol{b}) = \left\langle \prod_{j=1}^n x^{h_j} (t + u_j) e^{-i2 \pi \alpha t} \right\rangle, \hfill (11)$

it seems intuitive that if we permute the delays $\boldsymbol{u}$ such that each $u_j$ ends up in a $x(\cdot)$ with the same conjugation status that it originally had, then the cyclic moment will be unchanged. For example, consider $n=4$ and $\boldsymbol{b} = [1\ 1\ 0\ 0]$, which implies $m = 2$

$\displaystyle R_x^\alpha(\boldsymbol{u}; 4, \boldsymbol{b}) = \left\langleĀ x^* (t + u_1) x^* (t+u_2) x(t+u_3) x(t+u_4) e^{-i2 \pi \alpha t} \right\rangle, \hfill (12)$

If we use the permutation $\boldsymbol{q} = [2\ 1\ 4\ 3]$, then we obtain

$\displaystyle R_x^\alpha(\boldsymbol{u(\boldsymbol{q})}; 4, \boldsymbol{b}) = \left\langleĀ x^* (t + u_2) x^*(t+u_1) x(t+u_4) x(t+u_3)e^{-i2 \pi \alpha t} \right\rangle, \hfill (13)$

and we can rearrange the factors inside the integral to yield exactly (12). So in this case, the cyclic moment is equal for the two delay vectors $[u_1\ u_2\ u_3\ u_4]$ and $[u_2\ u_1\ u_4\ u_3]$.

Taking a step back, if we consider the set of random variables defined by

$\displaystyle X = \left\{ x^{h_j}(t+u_j) \right\}_{j=1}^n \hfill (14)$

and we permute the delays $u_j$ so that the set is the same before and after the permutation, then the time-varying temporal moment functions corresponding to the two sets are identical. But we can also consider independently permuting the indicator vector as well. Permute the delay vector $\boldsymbol{u}$ by the permutation $\boldsymbol{q}$ and the indicator vector $\boldsymbol{b}$ by the permutation $\boldsymbol{z}$. Define the set of variables that is implied by these two permutations:

$\displaystyle Y = \left\{ x^{h_{z_j}}(t+u_{q_j}) \right\}_{j=1}^n \hfill (15)$

If $X = Y$, then the temporal moment functions for the two sets are equal, and if the temporal moment functions are equal, so are the cyclic moment functions. This same argument holds for the temporal cumulant and temporal cyclic cumulant functions.

We can devise a test for whether or not a particular pair of permutations $\boldsymbol{q}$ and $\boldsymbol{z}$ result in $X=Y$ and therefore in equal cyclic moments and cumulants. If after applying the two permutations, the resulting set of variables in $Y$ can be reordered so that it is identical to an ordering of the variables in $X$, then we have equality, and therefore have found a symmetry.

Define a $2 \times n$ matrix $G_{in}$ by

$\displaystyle \boldsymbol{G}_{in} = \begin{bmatrix} \boldsymbol{b} \\ \boldsymbol{u} \end{bmatrix} = \begin{bmatrix} b_1 & b_2 & \ldots & b_n \\ u_1 & u_2 & \ldots & u_n \end{bmatrix} \hfill (16)$

and a corresponding post-permutation matrix by

$\displaystyle \boldsymbol{G}_{out} = \begin{bmatrix} \boldsymbol{b(\boldsymbol{z})} \\ \boldsymbol{u(\boldsymbol{q})} \end{bmatrix} = \begin{bmatrix} b_{z_1} & b_{z_2} & \ldots & b_{z_n} \\ u_{q_1} & u_{q_2} & \ldots & u_{q_n} \end{bmatrix} \hfill (17)$

If there is a permutation $\boldsymbol{w}$ of the columns of $\boldsymbol{G}_{out}$ such that

$\displaystyle \boldsymbol{G}_{out} (\boldsymbol{w}) = \boldsymbol{G}_{in} \hfill (18)$

then $X = Y$ and the moments and cumulants are equal.

I’ve estimated the fourth-order cyclic moments and cumulants for the rectangular-pulse BPSK signal with bit rate $1/10$, carrier offset $0.05$, and unit power for all indicator vectors $\boldsymbol{b}$ of length $4$ and all integer-valued delay vectors $\boldsymbol{u} = [u_1\ u_2\ u_3\ 0]$ on the three-dimensional hypercube $[-15, 15]^3$. I can use those stored sets of moments and cumulants to verify the symmetries we’ve looked at here, at least for $n = 4$.

In particular, I implemented the $\boldsymbol{G}_{in}$/$\boldsymbol{G}_{out}$ test to find the symmetries, then verified by looking up the corresponding estimates in the stored moment or cumulant files. The results of this verification process do not lend themselves to pretty plots (but do continue reading if you want some pretty plots). I get things like this for the cyclic cumulants:

and similar results for the cyclic moments.

Symmetry in Cycle Frequency

The symmetries developed in the previous section relate to a single cycle frequency value and consider all the different delay vectors that produce identical values of the temporal parameters.

Here we wonder how the cyclic moment for cycle frequency $\alpha$ relates to the cyclic moment for $-\alpha$, and similarly wonder about the cyclic cumulants and negated cycle frequencies.

Looking back at the variable set

$\displaystyle X = \left\{ x^{h_j}(t+u_j) \right\}_{j=1}^n \hfill (19)$

if we permute the indicator vector $\boldsymbol{b}$ such that we obtain the complement

$\displaystyle \bar{\boldsymbol{b}} = \boldsymbol{b}(\boldsymbol{c}) \hfill (20)$

then each variable in $X$ that is conjugated becomes unconjugated, and each variable that is unconjugated becomes conjugated, and the delay vector is unaffected. So, in this sense, we would have

$Y = X^* \hfill (21)$

From elementary considerations, such as the definition of the $n$th-order moment function, we know that the temporal moment and temporal cumulant for such a $Y$ are the conjugated functions for $X$. So we expect a cycle-frequency symmetry of the form

$\displaystyle C_x^\alpha (\boldsymbol{u}; n, \boldsymbol{b}) = \left[C_x^{-\alpha} (\boldsymbol{u}; n, \bar{\boldsymbol{b}})\right]^* \hfill (22)$

Again using the stored sets of cyclic cumulant estimates for the rectangular-pulse BPSK signal, I plotted various cases corresponding to (22) and arranged them in a movie. These all confirm the basic symmetry: negated cycle frequency, complementary indicator function (complementary conjugation configuration), conjugation, and identical delay vector. The individual frames of the movie can also be used to verify correct operation of your cyclic-cumulant estimator. Remember that my cyclic cumulant estimator has been validated by comparing its output to the theoretical cyclic cumulants for those signals for which we know them, such as rectangular-pulse BPSK and several different kinds of QAM and PSK signal types.

Visualizing the Symmetries

What follows are several attempts at visualizing the cyclic moments and cumulants. There are a lot of dimensions to try to include: order $n$, indicator vector $\boldsymbol{b}$, harmonic number $k$, and delay vector $\boldsymbol{u}$.

First let’s look at a couple cyclic moments and cumulants as surfaces, holding the fourth delay at zero, using two other delays as the x- and y-axes variables, and incrementing the final delay to produce each surface. The movie file names indicate whether the movie shows cyclic moments (‘ctmf’) or cyclic cumulants (‘ctcf’).

Here are the cyclic moment and cumulants for order $n=4$, $m=0$, and $k = 0$:

Note how the cyclic moment does not appear to be contained in the range of delays $(\tau_2, \tau_3)$, whereas the cyclic cumulant does. The cyclic moment has the property that in certain directions in the four dimensional delay domain, it does not ever die out (this is intimately related to the notion of impure sine waves).

Here are a couple more in that style:

Here are attempts to simultaneously visualize multiple cyclic-cumulant and cyclic-moment magnitudes and phases:

Finally, here are some movies that show the cyclic cumulant magnitudes and phases for several complementary indicator vectors: