Equality of Covariances Matrices Test in R (varcomp)

Equality of Covariances Matrices Test in R (varcomp)

Posted by on Oct 9, 2011 in R, Statistics Tags: , , | 9 comments

This is a piece of code I implemented in 2004, which was supposed to be part of an R-package in multivariate testing (to be named, rather creatively, mvttests).

Time has flown, I haven’t still got around to implementing the said package, but people keep asking me for the varcomp function, so here it is, for posterity:

Download varcomp.R
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40

varcomp <- function(covmat,n) {
   if (is.list(covmat)) {
    if (length(covmat) < 2)
        stop("covmat must be a list with at least 2 elements")
    ps <- as.vector(sapply(covmat,dim))
    if (sum(ps[1] == ps) != length(ps))
        stop("all covariance matrices must have the same dimension")
    p <- ps[1]
        q <- length(covmat)
        if (length(n) == 1)
        Ng <- rep(n,q)
    else if (length(n) == q)
        Ng <- n
    else
        stop("n must be equal length(covmat) or 1")
 
    DNAME <- deparse(substitute(covmat))
   }
 
   else
    stop("covmat must be a list")
 
   ng <- Ng - 1
   Ag <- lapply(1:length(covmat),function(i,mat,n) { n[i] * mat[[i]] },mat=covmat,n=ng)
   A <- matrix(colSums(matrix(unlist(Ag),ncol=p^2,byrow=T)),ncol=p)
   detAg <- sapply(Ag,det)
   detA <- det(A)
   V1 <- prod(detAg^(ng/2))/(detA^(sum(ng)/2))
   kg <- ng/sum(ng)
   l1 <- prod((1/kg)^kg)^(p*sum(ng)/2) * V1
   rho <- 1 - (sum(1/ng) - 1/sum(ng))*(2*p^2+3*p-1)/(6*(p+1)*(q-1))
   w2 <- p*(p+1) * ((p-1)*(p+2) * (sum(1/ng^2) - 1/(sum(ng)^2)) - 6*(q-1)*(1-rho)^2) / (48*rho^2)
   f <- 0.5 * (q-1)*p*(p+1)
   STATISTIC <- -2*rho*log(l1)
   PVAL <- 1 - (pchisq(STATISTIC,f) + w2*(pchisq(STATISTIC,f+4) - pchisq(STATISTIC,f)))
   names(STATISTIC) <- "corrected lambda*"
   names(f) <- "df"
   RVAL <- structure(list(statistic = STATISTIC, parameter = f,p.value = PVAL, data.name = DNAME, method = "Equality of Covariances Matrices Test"),class="htest")
   return(RVAL)
}


tabs-top

9 Responses to “Equality of Covariances Matrices Test in R (varcomp)”

  1. Sara says:
    March 21, 2012 at 8:00 am

    Hi Fernando,

    I was wondering what test for equality of covariance matrices your varcomp function is based on? I cant really figure it out myself.

    Sara

    Reply
  2. Sara says:
    March 21, 2012 at 10:03 am

    And also what is the parameter n?

    Reply
  3. Evan says:
    April 24, 2012 at 3:27 pm

    Hi Fernando,

    I am trying to implement the above function but keep receiving this error message:
    Error in ps[1] == ps : comparison of these types is not implemented

    The code I am trying to use is this:
    mn.test <- t(cbind(Shelter$Toptosub, Shelter$Dbot,
    Shelter$Rtopratio, Shelter$Rbotratio))
    covmat <- as.list(cov(mn.test))
    varcomp(covmat, 133) # this is my overall n. I have also tried
    # it with length(covmat) and 1.

    Thanks for any help and also for sharing this function!

    Reply
    • fernandohrosa says:
      August 17, 2012 at 10:32 pm

      Hi Evan,

      Sorry for taking so long to reply. My WordPress install went berserk and I stopped receiving e-mail notices upon new comments.

      As to your question: you have only one covariance matrix to compare, that’s why it’s not working. The function is supposed to work on two or more (potentially several) covariance matrices. cov(mn.test) is a single covariance matrix, you’ve got nothing to compare it against!

      Fernando,

      Reply
  4. Watheq Al-Mudhafar says:
    March 2, 2013 at 9:35 pm

    This function has been implemented correctly. Do you have R code example to work on it after this function in order to understand the steps of covariance testing?

    Regards

    Reply
    • fernandohrosa says:
      March 5, 2013 at 7:43 pm

      You just call the function with two arguments, the first is a list of covariance matrices, and n is a vector of the number of observations used to calculate each of the covariance matrices. If n is of length 1, it is assumed that all matrices were calculated from data sets with the same length.

      Reply
  5. ivo kwee says:
    March 20, 2013 at 4:45 am

    Hi. Thanks. I changed the code to use the logarithm of the determinant in the computations because for larger dimensions and n it wouldnt compute anymore. Below is the change. Seems to work. Ivo.

    ————– snippet ————–
    ng <- Ng – 1
    Ag <- lapply(1:length(covmat),function(i,mat,n) { n[i] * mat[[i]] },mat=covmat,n=ng)
    A <- matrix(colSums(matrix(unlist(Ag),ncol=p^2,byrow=T)),ncol=p)
    log.detAg <- sapply(Ag,function(x) determinant(x, logarithm=TRUE)$modulus)
    log.detA <- determinant(A, logarithm=TRUE)$modulus
    log.V1 <- as.numeric( sum((ng/2)*log.detAg) – (sum(ng)/2)*log.detA )
    kg <- ng/sum(ng)
    log.l1 <- (p*sum(ng)/2) * sum( kg*log(1/kg)) + log.V1
    rho <- 1 – (sum(1/ng) – 1/sum(ng))*(2*p^2+3*p-1)/(6*(p+1)*(q-1))
    w2 <- p*(p+1) * ((p-1)*(p+2) * (sum(1/ng^2) – 1/(sum(ng)^2)) – 6*(q-1)*(1-rho)^2) / (48*rho^2)
    f <- 0.5 * (q-1)*p*(p+1)
    STATISTIC <- -2*rho*log.l1

    Reply

Leave a Reply