I have two non-parametric rank correlations matrices

`emp`

and`sim`

(for example, based on Spearman’s ρ rank correlation coefficient):`library(fungible) emp <- matrix(c( 1.0000000, 0.7771328, 0.6800540, 0.2741636, 0.7771328, 1.0000000, 0.5818167, 0.2933432, 0.6800540, 0.5818167, 1.0000000, 0.3432396, 0.2741636, 0.2933432, 0.3432396, 1.0000000), 4, 4) # generate a sample correlation from population 'emp' with n = 25 sim <- corSample(emp, n = 25) sim$cor.sample [,1] [,2] [,3] [,4] [1,] 1.0000000 0.7221496 0.7066588 0.5093882 [2,] 0.7221496 1.0000000 0.6540674 0.5010190 [3,] 0.7066588 0.6540674 1.0000000 0.5797248 [4,] 0.5093882 0.5010190 0.5797248 1.0000000`

The

`emp`

matrix is the correlation matrix that contains correlations between the emprical values (time series), the`sim`

matrix is the correlation matrix — the simulated values.I have read the Q&A How to compare two or more correlation matrices?, in my case it is known that emprical values are not from normal distribution, and I can’t use the Box’s M test.

I need to test the null hypothesis H0: matrices

`emp`

and`sim`

are drawn from the same distribution.

Question.What is a test do I can use? Is is possible to use the Wishart statistic?

Edit.

Follow to Stephan Kolassa‘s comment I have done a simulation.I have tried to compare two Spearman correlations matrices

`emp`

and`sim`

with the Box’s M test. The test has returned`# Chi-squared statistic = 2.6163, p-value = 0.9891`

Then I have simulated 1000 times the correlations matrix

`sim`

and plot the distribution of Chi-squared statistic M(1−c)∼χ2(df).After that I have defined the 5-% quantile of Chi-squared statistic M(1−c)∼χ2(df). The defined 5-% quantile equals to

`quantile(dfr$stat, probs = 0.05) # 5% # 1.505046`

One can see that the 5-% quantile is less that the obtained Chi-squared statistic:

`1.505046 < 2.6163`

(blue line on the fugure), therefore, my`emp`

‘s statistic M(1−c) does not fall in the left tail of the (M(1−c))i.

Edit 2.

Follow to the second Stephan Kolassa‘s comment I have calculated 95-% quantile of Chi-squared statistic M(1−c)∼χ2(df) (blue line on the fugure). The defined 95-% quantile equals to`quantile(dfr$stat, probs = 0.95) # 95% # 7.362071`

One can see that the

`emp`

‘s statistic M(1−c) does not fall in the right tail of the (M(1−c))i.

Edit 3.I have calculated the exact p-value (green line on the figure) through the empirical cumulative distribution function:`ecdf(dfr$stat)(2.6163) [1] 0.239`

One can see that p-value=0.239 is greater than 0.05.

ReferencesReza Modarres & Robert W. Jernigan (1993) A robust test for comparing correlation matrices, Journal of Statistical Computation and Simulation, 46:3-4, 169-181.

The first founded paper that has no the assumption about normal distribution.There are two different tests were proposed. The quadratic form test is more simple one.Dominik Wied (2014): A Nonparametric Test for a Constant Correlation

Matrix, Econometric Reviews, DOI: 10.1080/07474938.2014.998152

Authors proposed a nonparametric procedure to test for changes in correlation matrices at an unknown point in time.Joël Bun, Jean-Philippe Bouchaud and Mark Potters (2016), Cleaning correlation matrices, Risk.net, April 2016

Li, David X., On Default Correlation: A Copula Function Approach (September 1999). Available at SSRN: https://ssrn.com/abstract=187289 or http://dx.doi.org/10.2139/ssrn.187289

G. E. P. Box, A General Distribution Theory for a Class of Likelihood Criteria. Biometrika. Vol. 36, No. 3/4 (Dec., 1949), pp. 317-346

M. S. Bartlett, Properties of Sufficiency and Statistical Tests. Proc. R. Soc. Lond. A 1937 160, 268-282

Robert I. Jennrich (1970): An Asymptotic χ2 Test for the Equality of Two

Correlation Matrices, Journal of the American Statistical Association, 65:330, 904-912.Kinley Larntz and Michael D. Perlman (1985) A Simple Test for the Equality of Correlation Matrices. Technical report No 63.

Arjun K. Gupta, Bruce E. Johnson, Daya K. Nagar (2013) Testing Equality of Several Correlation Matrices. Revista Colombiana de Estadística

Diciembre 36(2), 237-258Elisa Sheng, Daniela Witten, Xiao-Hua Zhou (2016) Hypothesis testing for differentially correlated features. Biostatistics, 17(4), 677–691

James H. Steiger (2003) Comparing Correlations: Pattern Hypothesis Tests Between and/or Within Independent Samples

It is not the answer.

I have simulated

`n=1000`

times the correlations matrix`sim`

, calculate the statistic M(1−c)i, i=1,2,...,n and ploted the Chi-squared statistic distribution (left) and Cumulative Distribution Function (right).The null hypothesis H0: matrices

`emp`

and`sim`

are drawn from the same distribution.The alternative hypothesis H1: matrices

`emp`

and`sim`

are not drawn from the same distribution.We have a two-tailed test at α=5\%. The critical values are:

`alpha <- 0.05 q025 <- quantile(x, probs = alpha/2);q025 # 2.5% # 1.222084 q975 <- quantile(x, probs = 1 - alpha/2);q975 # 97.5% # 8.170121`

From the calculation one can see:

1.222084 < M(1-c)= 2.6163 < 8.170121,

therefore, H_0 is true.

Counter-example.I have simulated a sample`xx`

from \chi^2(df) distribution and find the sample characteristics:`m <- 2 # number of matrices k <- 4 # size of matrices df <- k*(k+1)*(m-1)/2 # degree of freedom xx <- rchisq(1000, df=df) Mode <- function(x) { ux <- unique(x) ux[which.max(tabulate(match(x, ux)))] } Mode(xx) # [1] 5.845786 mean(xx) # [1] 10.1366808 quantile(xx, probs = alpha/2) # 2.5% # 3.057377 quantile(xx, probs = 1 - alpha/2) # 97.5% # 19.91842`

The sample’s mean

`10.1366808`

falls into the left tail of the statistic`M(c-1)`

distribution, therefore, H_0 is not true.But the sample’s mode

`5.845786`

fails into the middle range.

**Answer**

Since we are working with matrices constructed from the same set of ranks to construct corresponding Spearman correlations matrices, this 2012 simple method presented in this work: A simple procedure for the comparison of covariance matrices, may be of value.

In particular to quote:

Here I propose a new, simple method to make this comparison in two population samples that is based on comparing the variance explained in each sample by the eigenvectors of its own covariance matrix with that explained by the covariance matrix eigenvectors of the other sample. The rationale of this procedure is that the matrix eigenvectors of two similar samples would explain similar amounts of variance in the two samples. I use computer simulation and morphological covariance matrices from the two morphs in a marine snail hybrid zone to show how the proposed procedure can be used to measure the contribution of the matrices orientation and shape to the overall differentiation.

Of particular import is the claimed results and conclusions:

Results

I show how this procedure can detect even modest differences between matrices calculated with moderately sized samples, and how it can be used as the basis for more detailed analyses of the nature of these differences.

Conclusions

The new procedure constitutes a useful resource for the comparison of covariance matrices. It could fill the gap between procedures resulting in a single, overall measure of differentiation, and analytical methods based on multiple model comparison not providing such a measure.

And further comments from the available full text:

In the present work I propose a new, simple and distribution-free procedure for the exploration of differences between covariance matrices that, in addition to providing a single and continuously varying measure of matrix differentiation, makes it possible to analyse this measure in terms of the contributions of differences in matrix orientation and shape. I use both computer simulation and P matrices corresponding to snail morphological measures to compare this procedure with some widely used alternatives. I show that the new procedure has power similar or better than that of the simpler methods, and how it can be used as the basis for more detailed analyses of the nature of the found differences.

If other methods prove less impressive, you may which to further investigate the above for the comparison of rank correlation matrices performing your own simulation testing.

**Attribution***Source : Link , Question Author : Nick , Answer Author : AJKOER*