# Ox: Cross-correlation coefficients with lags

Working on the homework for my Monetary Economics class I realized that Ox has different mechanisms to calculate correlation coefficients like Mathlab has.

Our assignment was to replicate the Dynamic Correlations graph in Walshs’ Monetary Theory and Policy for European Data, which shows the cross-correlation of different monetary aggregates (M0,M1,M2) with GDP over lagging periods of -8 to +8 quarters.

What was missing for me to do that assignment in Ox was a function to calculate cross-correlation coefficients for two vectors for a specified lag length:

#include <oxfloat.h> /** * Computes the cross-correlation coefficient for two vector series allowing * to optionally specify the number of positive and negative lags the ccc * should be calculated for. * * @author Benjamin Eberlei (kontakt at beberlei dot de) * @param vVarY Tx1 vector of variable related to * @param vVarX Tx1 vector of variable which is tested in different lags * @param lags Integer indicating the number of negative and positive lags. * @returns (1+lags*2)*1 vector of correlations from -lags to lags **/ ccf(const vVarY, const vVarX, const lags) { // Generate positive and negative lags of given length and fill with NaN, so // that rows with not available numbers can be dropped from calculating the CCF later on. decl mXLag = lag0(vVarX, range(lags, -lags), M_NAN); decl mCorr, sCov; // initialize result vector holding one crosscorrelation per lag decl vCorrLags = zeros(1+lags*2, 1); // sadly there are no matrix operations to ease this computation, loop over all lags for(decl i = 0; i < 1+lags*2; i++) { mCorr = deleter(vVarY ~ mXLag[][i]); // stick y and current lagged x together and delete NaN rows // calculate covariance of both time series sCov = 1/(rows(mCorr)-1) * sumc( (mCorr[][0]-meanc(mCorr[][0]))' * (mCorr[][1]-meanc(mCorr[][1])) ); // calculate correlation coefficient vCorrLags[i][0] = sCov / ( sqrt(varc(mCorr[][0])) * sqrt(varc(mCorr[][1])) ); } return vCorrLags; }

It may prove useful to someone.