Katrien, Even though one can calculate the fractional covariance as Philipp described, one has to consider whether there is much overall covariance between matrices. The following is an example of how the fractional covariances based on squared singular values can be deceiving, in R.
> set.seed(22) > n <- 20 > p <- 3 > X <- matrix(rnorm(n * p), n, p) > Y <- matrix(rnorm(n * p), n, p) > > X <- scale(X, scale = FALSE) > Y <- scale(Y, scale = FALSE) > > sxy <- svd(crossprod(X, Y)/(n-1)) > > sxy$d^2 # squared singular values [1] 0.2689045372 0.0543512900 0.0001299134 > sum(sxy$d^2) # their sum [1] 0.3233857 > sxy$d^2/sum(sxy$d^2) # their fraction [1] 0.8315287394 0.1680695319 0.0004017288 > > sx <- svd(crossprod(X)/(n-1)) > sy <- svd(crossprod(Y)/(n-1)) > > sx$d # eigenvalues for X [1] 1.6143390 1.1115365 0.2711724 > sum(sx$d) # their sum [1] 2.997048 > > sy$d # eigenvalues for Y [1] 1.5372356 0.9224574 0.4432287 > sum(sy$d) # their sum [1] 2.902922 So, by generating two matrices of random data, and any covariation between them is incidental, we find that 83.15% of the overall covariance is explained in the first PLS vectors. However, the summed squared singular values (overall covariance) is 0.32, which is quite small compared to the 2.99 or 2.90 summed eigenvalues (total variances) of either matrix. It is possible to have much of the very little covariance explained by the first axis. Focusing on the 83.15%, alone, could be misleading about the amount of covariance between X and Y. Cheers! Mike > On May 14, 2020, at 2:59 AM, [email protected] wrote: > > Dear Katrien, > > > > The sum of the squared singular values in PLS equals the sum of all the > squared pairwise covariances between the two blocks of variables. Therefore, > what you can compute is the fraction of "squared" total covariance that each > PLS dimension accounts for. You can do this by dividing each squared singular > value by the summed squared singular values. > > Also, if one insists on a significance test, it should be based on the > distribution of these singular values. > > > > Best, > > > > Philipp > > > > Am Donnerstag, 14. Mai 2020 03:31:12 UTC+2 schrieb katrien.janin: > Hello everybody, > > I am hoping you can help me figure out the following: after having used the > integration.test function (geomorph), I would like to know the % each PLS > captures of the total covariance but I am kinda stumped on what might be a > good way how to go about it. > > I was initially thinking along the lines of generating the covariance matrix > of the two block (e.g. cov.il.is <http://cov.il.is/> <-cov(il.is.int > <http://il.is.int/>$A1.matrix,il.is.int <http://il.is.int/>$A2.matrix) and > computing the eigenvalue for each PLS , and then regress these. But of course > the cov matrix has a very different structure, so that will not work. I am > clearly thinking in the wrong direction, but for now I can’t seem to see the > tree in the forest. Any ideas? > > Best wishes, > Katrien > > -- > You received this message because you are subscribed to the Google Groups > "Morphmet" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected] > <mailto:[email protected]>. > To view this discussion on the web visit > https://groups.google.com/d/msgid/morphmet2/76c0b670-b6ed-448a-b80d-4d662263aa83%40googlegroups.com > > <https://groups.google.com/d/msgid/morphmet2/76c0b670-b6ed-448a-b80d-4d662263aa83%40googlegroups.com?utm_medium=email&utm_source=footer>. -- You received this message because you are subscribed to the Google Groups "Morphmet" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/morphmet2/0F0BC6C1-D02A-46FE-847F-C8F6BE069DD4%40gmail.com.
