On wide sense stationary processes over finite non-abelian groups

Abstract: Let X be a real-valued wide sense stationary process over a finite nonabelian group G. We provide results on optimal orthogonal decomposition of X into real-valued mutually orthogonal components and using this decomposition we develop a test for correlation of X over the group G. Applications of these results to the analysis of variance of the carry-over effects in the cross-over designs in clinical studies are given. Our focus will be on groups S 3 , S 4 , and A 4 .

Let G denote a finite non-abelian group and let X denote a zero mean real-valued wide sense stationary process (WSS) over the group G. In particular, we have E(X(t)) = 0 for all t ∈ G. The autocorrelation function R X of X is defined as follows, set = t −1 s, where t, s ∈ G Suppose X and Y are two zero mean WSS processes over G. We define the cross-correlation function of X and Y as follows: denote a family of zero mean WSS processes over G. Such a family is said to be mutually orthogonal if R X i X j ( ) = 0 for all ∈ G and i ≠ j. In such case, we refer to the WSS process over G as white. For more information of random processes in general, we refer the reader to Bartoszynski and Niewiadomska-Bugaj (2008).
A natural place where WSS processes over non-abelian groups arise are the cross-over designs in clinical trials. We will refer to this later on in this paper.
Non-abelian Fourier Transform. Let ℂ n denote the n-dimensional vector space over the complex numbers. The standard basis for ℂ n is identified with the ordered group elements of G, where |G| = n.
A finite dimensional representation of a finite group G over ℂ is a group homomorphism where GL(d j , ℂ) denotes the general linear group, the set of all d j × d j invertible matrices. We refer to d j as the degree of the group representation.
Two group representations are said to be equivalent if there exists an d j × d j invertible matrix T such that for all g ∈ G.
An irreducible group representation of G is a group representation of G, for which there is no nontrivial subspace W of ℂ j for which for all g ∈ G.
Let ℂ[G] be the algebra of complex-valued functions on G with respect to G-convolution. Let = (c 0 , c 1 , … , c n−1 ) ∈ ℂ n and identify the function with its symbol A G-convolution of and is defined by the following action, ∈ G Let Ĝ be the set of all (equivalence classes) irreducible representations of the group G. WLOG we can assume these representations are unitary (please see Hazewinkel (2001) for further details). Let ∈Ĝ be of degree d j and let ∈ ℂ n . Then, the Fourier transform of at is the d j × d j matrix The Fourier inversion formula, s ∈ G, is given by Observe the the switch s → s −1 in the above functions. We refer the reader to Stankovic, Radomir, Moraga, and Astola (2005) for further reading on this subject.
Let and be two elements in ℂ n . We have a natural identification The action of on via G-convolution is delivered by the matrix multiplication by the G-circulant matrix C G ( ) Definition 1 For given vectors , ∈ l 2 (G) the G cross-correlation function is defined by We have, see Zizler (2014), The character of a group representation is the complex-valued function defined by We call a character irreducible if the underlying group representation is irreducible. We define an inner product on the space of class functions, functions on G that are constant on its conjugacy classes A character is a class function. There are as many irreducible characters as there are conjugacy classes of G(please see Dummit (1999), p. 870 for details). Let r denote the number of conjugacy classes of G and we have r irreducible characters { 1 , 2 , … , r } for the group G. We have, with respect to the usual inner product, where i, j is the Kronecker delta. The set of all irreducible characters form a basis for the space of class functions on G, see Dummit (1999) for more details.
Let ∈ ℂ n and define for s ∈ G Note = ∑ r j=1 j . We are able to decompose the function into a sum of r functions which is the number of conjugacy classes of G.
We define an (orthogonal) projection P j on ℂ n by the following action, ∈ ℂ n The action of the linear operator P j in the Fourier domain is given by the (matrix) multiplication by the vector where the d j × d j identity matrix j is in the jth position. The inverse Fourier transform of this vector is the function (evaluated at g ∈ G) Therefore, for all ∈ ℂ n , we have Observe the image and the kernel of P j are given by and Note dim Im(P j ) = d 2 j . Moreover, the functions form an orthonormal basis for Im(P j ). Also note Im(P j ) ⟂ Im(P i ) for i ≠ j. We also have The important observation here is the fact that the space Im(P j ) is invariant under the group circulant matrix C = C G ( ) and is also independent. We have an orthonormal basis for each Im(P j ) , and thus for ℂ n , but these vectors no longer need to be eigenvectors for the group circulant matrix C = C G ( ). Still, this orthonormal basis is independent. For more information we refer the reader to Zizler (2013). For more details on group representations, we refer the reader to Dummit (1999). We refer the reader to Stankovic et al. (2005) or An and Tolimieri (2003) for more material on the non-abelian Fourier transform and its applications.

Main results
We need the orthogonal components {X j } to be real valued for real-life applications. An element of a group G is said to be real if it is conjugate to its inverse. Recall that two elements a and b of a group G are said to be conjugate if there exists c ∈ G such that c −1 ac = b. A conjugacy class is said to be real if it has a real element. Note that if a conjugacy class has a real element then all the elements of this conjugacy class are real. It is a known result, see James and Liebeck (1993), for example, that the number of real irreducible characters of a group G is equal to the number of real conjugacy classes of G. Therefore, a group G has all irreducible characters real if and only if all the elements of that group are real. Therefore, the symmetric group S n has all the irreducible characters real as all of its elements are real. However, this is no longer true for the alternating groups.
Recall r denotes the number of conjugacy classes for G and {̃j} r j=1 denotes the set of all the (inverted) irreducible characters of G. Let d j denote the degree of the irreducible representation j .
Define a vector valued zero mean WSS process over G as where G = {g 0 , g 1 , … , g n−1 } and each X(g i ) is a zero mean WSS process. Consider a decomposition of X = ∑ p j=1 X j where {X j } is a mutually orthogonal set of zero mean real-valued WSS processes. We say that the value p is optimal if it is impossible to decompose X into more zero mean real-valued mutually orthogonal WSS processes. Here, of course, the value p should be independent of X. Thus this decomposition is requested for all X and is only group G dependent. Note that a specific process X could potentially be decomposed into more mutually orthogonal components.
Theorem 1 Let X denote a zero mean real-valued WSS process over the group G. Let p = r+s 2 where s is the number of real conjugacy classes of G. Then, we can write where {X j } r j=1 is a mutually orthogonal set of zero mean real-valued WSS processes and the value p is optimal. Here, we have and Moreover, we have for all t ∈ G.
Proof Define the Fourier transforms of the two zero mean WSS processes X and Y evaluated at the irreducible representation j Consider the cross-correlation function R XY and let  denote the Fourier transform operator. Define We have we obtain Define Note, here the components X j can be complex valued. Set Z = X − ∑ r i=1 X i and let R Z denote the autocorrelation function for Z. Observe for any j ∈ {1, 2, … , r} and consider, for any j ∈ {1, 2, … , r} Now we have for all n = {1, 2, … , r} as long as i ≠ j. Thus, we have, for i ≠ j, R X i X j ( ) = 0 for all ∈ G. In this manner, we have a decomposition into potentially complex components. We have X = ∑ r j=1 X j as zero mean WSS processes over G.
To analyze the real case, we make a key observation. If is an irreducible character of G then so is it conjugate. Note that (g) =̃(g) = (g −1 ) = (g). Consider the complex mutually orthogonal decomposition as above If j is a real-valued irreducible character we set Y j = X j in the sum. If j 1 is complex valued, we pair it up with the irreducible representation j which is another irreducible character in the list of irreducible characters of G, say j 2 . In this case we set Therefore, the value p equals to s, the total number of real irreducible characters of G, which equals to the number of real conjugacy classes, plus half of the remaining complex-valued irreducible characters of G. Thus p = r+s 2 .
We now address the optimality of the decomposition. Observe that we are requesting, for all X and for all n = {1, 2, … , r} We can associate X j with the following element in M(Ĝ) Suppose we can decompose an arbitrary X further. This would mean there exist square matrices E and F so that The interpretation of var(X 1 ) = E(X 1 (t)X 1 (t)) is straightforward since 0 denotes the trivial group representation. This quantity refers to the amount of variance of X(t) captured by the variance of the mean of X over G. The above decomposition X = ∑ r j=1 X j into complex components will be referred to as the optimal complex orthogonal decomposition, similarly, the above decomposition X = ∑ p j=1 X j into real components will be referred to as the optimal real orthogonal decomposition.

Theorem 2 We have E(X(t)X(s)) = 0 for all t, s ∈ G such that t ≠ s if and only if for any t ∈ G, we have
where {X j } r j=1 is the optimal complex orthogonal decomposition.
Proof Consider the decomposition X = ∑ r j=1 X j into possibly complex valued components and suppose E(X(t)X(s)) = 0 for all t ≠ s. Then, we have Conversely, suppose var(X j (t)) = d 2 j |G| varX(t) for all t ∈ G. Define n = |G| unknowns {x } ∈G , where x = E(X(t)X(s)) = R X ( ) with = t −1 s. Now consider the following linear equations, and note that the number of these equations is equal to rn. However, we have only as many linearly independent equations as there are linearly independent columns in the matrices E + F = I and EQFQ = 0, for all square Q.
This number is equals to ∑ r j=1 d 2 j = n, the number of unknowns. The result follows with the unique solution E(X(t)X(s)) = 0 for t ≠ s. ✷ As an application of the above result we develop a hypothesis testing for correlation of the WSS processes X(t) and X(s) based on testing equality of variances of the following possibly complexvalued random variables Testing the above random variables might be better that testing correlations of X(t) and X(s) due to averaging process that could bring the random variables close to normality.

The group S 3
We will consider the symmetric group S 3 in our example. The group G = S 3 consists of elements The group S 3 has three conjugacy classes We have three irreducible representations, two of which are one-dimensional, 1 is the identity map, 2 is the map that assigns the value of 1 if the permutation is even and the value of −1 if the permutation is odd. Finally, we have 3 , the two-dimensional irreducible representation of S 3 , defined by the following assignment The irreducible characters of S 3 are given by where 1 and 2 are also multiplicative characters. Moreover, we have The G-convolution by a function = (c 0 , c 1 , c 2 , c 3 , c 4 , c 5 ) T ∈ l 2 (G) can be induced by a G-circulant matrix C G ( ) given by . g 0 = (1); g 1 = (12); g 2 = (13) g 3 = (23); g 4 = (123); g 5 = (132).

Carry-over effects in the cross-over designs
The above orthogonal decomposition of the WSS processes appear naturally in the cross-over designs in clinical trials, in particular, the William's 6 × 3 design with 3 treatments. During a cross-over trial every patient receives more than one treatment in a certain pre-specified sequence. Thus, each 1 1 1 1  1 1 1 1 1 1  1 1 1 1 1 1  1 1 1 1 1 1  1 1 1 1 1 1  1 1 1 1 1 subject then acts as his or her own control. Each treatment is administered for a pre-selected time period. In these experiments a washout period is established between the last administration of one treatment and the first administration of the next treatment. The intention is for the effect of the preceding treatment to wear off during the trial. However, there will be some carry-over effects in all the specified treatment sequences, clearly starting with the second treatment. For more information on cross-over designs in clinical trials see Senn (2002) or Chow and Liu (2013), for example.
Consider a certain sequence of treatments. By a carry-over effect within these treatments we understand the total sum of all effects arising from all the treatments within this sequence. In our example, we follow the William's 6 × 3 design with 3 treatments A, B, and C, in particular, the underlying group is the symmetric group of three elements S 3 . In particular, we have six treatment sequences ABC, ACB, BAC, BCA, CAB, CBA. For example, suppose the order of treatment administration is BCA, with B first. We decide to collect the sum of all carry-over effects of the treatments in this sequence (starting with the second one), X BCA . We observe the sequence BCA as a permutation of the sequence ABC by the permutation g 4 = (123), an element of the group S 3 . Thus, we can write X BCA = X(g 4 ) = X(4). Similarly, a permutation sequence ACB would result in X ACB = X(g 3 ) = X(3).
We treat X as a zero mean WSS process, if necessary we can subtract the mean to ensure zero mean WSS process. In particular, we assume the variance of the cross-over effect is the same for all treatment sequences. Moreover, we assume E(X(g i )X(g j )) only depends on the sequence g −1 i g j , in another words, depends only on the permutation that gets us from the sequence g i to the sequence g j . However, note that the random processes X(g i ) and X(g j ) are in general dependent and, in general, might have different distributions for i ≠ j.
We can now decompose the variance of the zero mean WSS process X over S 3 as follows. For all t ∈ G we have In this context, the real zero mean WSS processes over S 3 , {X(t)} t∈S 3 , are un-correlated if and only if have the same variance.

The group S 4
The symmetric group of four elements has five conjugacy classes represented by with sizes 1, 6, 8, 6, and 3, respectively. We have two irreducible characters of degree 1, namely the principal one and the one whose value at a permutation is its sign. There is one-degree two irreducible character 3 and two-degree 3 irreducible representations 4 and 5 . The value of 4 at a permutation is the number of fixed points in that permutation minus one. Finally, we have 5 = 4 2 . For reference, see Dummit (1999). Find below a character table for S 4 .
Suppose we have a cross-over design experiment with 4 treatments where we have all permutations of treatments allowed. We study the variance of the carry-over effects from the sequence of var(X(t)) = var(X 1 (t)) + var(X 2 (t)) + var(X 3 (t)).
(1); (12); (123); (1234); (12)(34) conj. classes (1) (12) (123) (1234) (12)(34) treatments. Once, again, we define the carry-over effect for the treatment sequence as the sum of all the carry-over effects in that specific sequence. We can now decompose the variance of the zero mean WSS process X over S 4 as follows. We define and obtain In this context, the zero mean WSS processes {X(t)} t∈S 4 are un-correlated if and only is have the same variance.

The group A 4
The case of the alternating group A 4 is a little different as r ≠ p. Here, we would still have 4 treatments, but only even permutations of the treatments are allowed in the cross-over design. We have three irreducible representations of degree 1 and one of degree 3, see Dummit (1999) for reference. We have 4 conjugacy classes represented by of sizes 1, 3, 4, and 4, respectively. The irreducible characters are not necessarily real, find below the character table.
where = e 2 i 3 . We can now decompose the variance of the zero mean WSS process X over A 4 as follows. We do not have 4 summands in the decomposition as we do not have 4 real irreducible characters. We start with complex decomposition with where X 1 = 1 24 C( 1 )X = 1 6 ⟨ X, 1 ⟩ 1 .