mvhtests (version 1.1)

Log-likelihood ratio test for equality of one covariance matrix: Log-likelihood ratio test for equality of one covariance matrix

Description

Log-likelihood ratio test for equality of one covariance matrix.

Usage

equal.cov(x, Sigma, a = 0.05)

Value

A vector with the the test statistic, the p-value, the degrees of freedom and the critical value of the test.

Arguments

x

A matrix containing Euclidean data.

Sigma

The hypothesis covariance matrix.

a

The significance level, set to 0.05 by default.

Author

Michail Tsagris.

R implementation and documentation: Michail Tsagris .

Details

The hypothesis test is that the the sample covariance is equal to some specified covariance matrix: $H_0:\mathrm{\Sigma }\mathrm{\Sigma }=\mathrm{\Sigma }{\mathrm{\Sigma }}_0$, with $\mu \mu$ unknown. The algorithm for this test is taken from Mardia, Bibby and Kent (1979, pg. 126-127). The test is based upon the log-likelihood ratio test. The form of the test is $$-2\log \lambda =n\text{tr}\left\{\mathrm{\Sigma }{\mathrm{\Sigma }}_0^{-1}\mathbf{S}\right\}-n\log \left|\mathrm{\Sigma }{\mathrm{\Sigma }}_0^{-1}\mathbf{S}\right|-np,$$ where $n$ is the sample size, $\mathrm{\Sigma }{\mathrm{\Sigma }}_0$ is the specified covariance matrix under the null hypothesis, $\mathbf{S}$ is the sample covariance matrix and $p$ is the dimensionality of the data (or the number of variables). Let $\alpha$ and $g$ denote the arithmetic mean and the geometric mean respectively of the eigenvalues of $\mathrm{\Sigma }{\mathrm{\Sigma }}_0^{-1}\mathbf{S}$, so that $tr\left\{\mathrm{\Sigma }{\mathrm{\Sigma }}_0^{-1}\mathbf{S}\right\}=p\alpha$ and $\left|\mathrm{\Sigma }{\mathrm{\Sigma }}_0^{-1}\mathbf{S}\right|=g^p$, then the test statistic becomes $$-2\log \lambda =np(\alpha -log(g)-1).$$ The degrees of freedom of the ${\chi }^2$ distribution are $\frac{1}{2}p(p+1)$.

References

Mardia K.V., Kent J.T. and Bibby J.M. (1979). Multivariate Analysis. London: Academic Press.

See Also

likel.cov, Mtest.cov

Examples

x <- as.matrix( iris[, 1:4] )
s <- cov(x) * 1.5
equal.cov(x, s)