asbio (version 1.8-3)

chi.plot: Chi plots for diagnosing multivariate independence.

Description

Chi-plots (Fisher and Switzer 1983, 2001) provide a method to diagnose multivariate non-independence among Y variables.

Usage

chi.plot(Y1, Y2, ...)

Value

Returns a chi-plot.

Arguments

Y1

A Y variable of interest. Must be quantitative vector.

Y2

A second Y variable of interest. Must also be a quantitative vector.

...

Additional arguments from plot.

Author

Ken Aho and Tom Taverner (Tom provided modified original code to eliminate looping)

Details

The method relies on calculating all possible pairwise differences within y\(_1\) and within y\(_2\). Let pairwise differences associated with the first observation in y\(_1\) that are greater than zero be transformed to ones and all other differences be zeros. Take the sum of the transformed values, and let this sum divided by (1 - n) be be the first element in the 1 x n vector z. Find the rest of the elements (2,..,n) in z using the same process.

Perform the same transformation for the pairwise differences associated with the first observation in y\(_2\). Let pairwise differences associated with the first observation in y\(_2\) that are greater than zero be transformed to ones and all other differences be zeros. Take the sum of the transformed values, and let this sum divided by (1 - n) be be the first element in the 1 x n vector g. Find the rest of the elements (2,..,n) in g using the same process.

Let pairwise differences associated with the first observation in y\(_1\) and the first observation in \(\bold{y}_2\) that are both greater than zero be transformed to ones and all other differences be zeros. Take the sum of the transformed values, and let this sum divided by (1 - n) be be the first element in the 1 x n vector h. Find the rest of the elements (2,..,n) in h using the same process. We let:

$$S = sign((\bold{z} - 0.5)(\bold{g} - 0.5))$$ $$\chi =(\bold{h} - \bold{z} \times \bold{g})/\sqrt{\bold{z} \times (1 - \bold{z}) \times \bold{g} \times (1 - \bold{g})}$$ $$\lambda = 4 \times S \times max[(\bold{z} - 0.5)^2,(\bold{g} - 0.5)^2]$$

We plot the resulting paired \(\chi\) and \(\lambda\) values for values of \(\lambda\) less than \(4(1/(n - 1) - 0.5)^2\). Values outside of \(\frac{1.78}{\sqrt{n}}\) can be considered non-independent.

References

Everitt, B. (2006) R and S-plus Companion to Multivariate Analysis. Springer.

Fisher, N. I, and Switzer, P. (1985) Chi-plots for assessing dependence. Biometrika, 72: 253-265.

Fisher, N. I., and Switzer, P. (2001) Graphical assessment of dependence: is a picture worth 100 tests? The American Statistician, 55: 233-239.

See Also

bv.boxplot

Examples

Run this code
Y1<-rnorm(100, 15, 2)
Y2<-rnorm(100, 18, 3.2)
chi.plot(Y1, Y2)

Run the code above in your browser using DataLab