*Bounty: 50*

*Bounty: 50*

I have two non-parametric rank correlations matrices `emp`

and `sim`

(for example, based on Spearman’s $rho$ rank correlation coefficient):

```
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), ncol=4)
sim <- matrix(c(
1.0000000, 0.7616454, 0.6545774, 0.3081403,
0.7616454, 1.0000000, 0.5360392, 0.3146167,
0.6545774, 0.5360392, 1.0000000, 0.3739758,
0.3081403, 0.3146167, 0.3739758, 1.0000000), ncol=4)
```

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 $H_0$: 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 $M(1-c)simchi^2(df)$.

After that I have defined the 5-% quantile of Chi-squared statistic $M(1-c)simchi^2(df)$. 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 $M(1-c)simchi^2(df)$ (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 $0.05$.

**Edit 4.**

Dominik Wied (2014): A Nonparametric Test for a Constant Correlation

Matrix, Econometric Reviews, DOI: 10.1080/07474938.2014.998152

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.

**Edit 5.**

**The first founded paper that has no the assumption about normal distribution.**

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