302018-03-12 17:22:22PeterWatson292013-03-08 10:17:25localhostconverted to 1.6 markup282011-02-04 11:28:11PeterWatson272011-02-04 11:27:06PeterWatson262011-02-04 11:25:06PeterWatson252011-02-04 11:24:42PeterWatson242011-02-04 11:24:11PeterWatson232011-02-04 11:22:41PeterWatson222011-02-04 11:13:01PeterWatson212011-02-04 10:40:53PeterWatson202011-02-04 10:40:07PeterWatson192011-02-04 10:38:51PeterWatson182011-02-04 10:30:12PeterWatson172011-02-04 10:25:19PeterWatson162011-02-04 10:23:53PeterWatson152011-02-04 09:59:10PeterWatson142011-02-04 09:57:05PeterWatson132011-02-04 09:55:28PeterWatson122011-01-27 12:37:02PeterWatson112011-01-27 12:36:17PeterWatson102011-01-27 12:35:24PeterWatson92011-01-27 12:34:11PeterWatson82011-01-27 12:33:24PeterWatson72011-01-27 12:33:03PeterWatson62011-01-27 12:31:48PeterWatson52011-01-27 12:27:28PeterWatson42011-01-27 12:26:58PeterWatson32011-01-27 12:25:18PeterWatson22011-01-27 12:23:49PeterWatson12011-01-27 12:22:34PeterWatson

Suppose we have a response Y and two continuous predictors such as age of onset (a) and duration of hearing deficit (b-a) with b representing the individual's current age. Then there is an equivalence between the coefficients in this regression and the ones associated with the same response,y, being predicted using a and b as predictors. In particular if $$B_text{i}$$ represents the regression coefficient for variable i then in a regression using a and b-a as predictors: Predicted y = $$B_text{a}$$a + $$B_text{b-a}$$(b-a) = $$B_text{a}$$a + $$B_text{b-a}$$b - $$B_text{b-a}$$a = $$(B_text{a}$$ - $$B_text{b-a}$$)a + $$B_text{b-a}$$b So it follows that if $$B_text{i|i,j}$$ represents the regression coefficient of variable i in a regression with i and j as predictors being used to predict a response, y, that $$B_text{a|a,b-a}$$ - $$B_text{b-a|a,b-a}$$ = $$B_text{a|a,b}$$ and $$B_text{b-a|a,b-a}$$ = $$B_text{b|a,b}$$ In other words subtracting the regression coefficients for a and b-a in a regression using a and b-a as predictor is equivalent to the regression coefficient for a in a regression with a and b as predictors and the regression coefficient for b-a with a and b-a as predictors is the same as the regression coefficient for b in a regression with a as the other predictor. It also follows that the standard errors of the regression coefficients for a and b respectively can be derived using the standard errors of the regression coefficients for a and b-a. se($$B_text{a|a,b}$$) = se($$B_text{a|a,b-a}$$ - $$B_text{b-a|a,b-a}$$) = $$\sqrt{V(B_text{a|a,b-a}) \mbox{ + } V(B_text{b-a|a,b-a}) \mbox{ - } 2\mbox{Cov}(B_text{a|a,b-a},B_text{b-a|a,b-a})}$$ and se($$B_text{b|a,b}$$) = $$B_text{b-a|a,b-a}$$ Example For one study involving a response y and variables a and b-a we have regression coefficients (s.es) of 1.170 (0.446) for a and 1.023 (0.399) for b-a. It follows in a regression involving a and b on the same response the regression (s.e.) of b equals that of b-a in the a, b-a regression, namely 1.023 (0.399). The regression coefficient for a equals 1.170 - 1.023 = 0.148. Given a covariance of 0.026 between the a and b-a regression coefficients The se(a) in the regression involving a and b is computed using the s.es and covariance from the regression coefficients in the regression with a and b-a as predictors. se(a) = $$\sqrt{0.4462 + 0.3992 - 2(0.026)}$$ = $$\sqrt{0.306}$$ = 0.553. The zero-order correlations have the same t-values as the regression estimates used to obtain them and their zero-order correlations correspond to the signed square root of the change in R-squareds. Example showing the extraction of zero-order correlations from the above regressions We can obtain the zero-order correlations of a and b with y from the regressions involving a and the a+b sum and b with the a+b sum by evaluating R-squareds and regression coefficient t-values associated with the 'a+b' sum regression term. These results confirm that the zero-order correlation of a (b) with y can be obtained indirectly from the b (a) scores and the sum of a and b. Examples below are for a randomly generated data set. The zero-order correlation of b with y is the signed square root of the change in R-squared adding 'a+b' to a regression already containing 'a' predicting y = $$\sqrt{0.066-0.050}$$ = -0.12 where the R-squared of 0.066 corresponds to a regression on y using 'a' and 'a+b' as predictors of y and 0.050 is the R-squared of a regression with only 'a' used to predict y. The t-value for 'a+b' =0.34, p=0.75 which equals the p-value for the zero-order correlation of b with y of -0.12. The zero-order correlation of a with y is the signed square root of the change in R-squared adding 'a+b' to a regression already containing 'b' predicting y = $$\sqrt{0.066-0.014}$$ = 0.22 where the R-squared of 0.066 corresponds to a regression on y using 'a' and 'a+b' as predictors of y and 0.014 is the R-squared of a regression with 'a' as the only predictor of y. The t-value for 'a+b' =-0.62, p=0.55 which equals the p-value for the zero-order correlation of a with y of 0.22. Relationships between a,b and a+b It also follows Predicted y = $$B_text{a}$$a + $$B_text{a+b}$$(a+b) = $$(B_text{a}$$ + $$B_text{a+b}$$)a + $$B_text{a+b}$$b and Predicted y = $$B_text{b}$$b + $$B_text{a+b}$$(a+b) = $$(B_text{b}$$ + $$B_text{a+b}$$)b + $$B_text{a+b}$$a So it follows that knowing the relationship between the response with both a+b and a is enough to give the relationship between the response and b and the relationship between the response and both a+b and b is enough to give the relationship with a. It is also true that unless a and b are highly correlated so a $$\approx$$ $$\pm$$b, $$B_text{a}$$a + $$B_text{b}$$b $$\ne$$ $$B_text{a+b}$$(a+b) because $$B_text{a}$$ and $$B_text{b}$$ will not in general be equal. This means you cannot obtain a zero-order correlation between y and the sum of a and b, indirectly, using the separate a and b scores. One can also interpret this as knowing the a+b sum does not tell you the numbers (a and b) that were added together to give it if these number were weighted unequally (so that $$B_text{a}$$ is not equal to $$B_text{b}$$). If a and b are highly correlated then the relationships between y and a and y and b are nearly equal and the relationship between y and a+b will then be equal to the relationship between y and either a or b. Example If b =-3a then for a Pearson correlation, r, r(a,y)=-r(b,y). r(a+b,y) = r(b,y)= - r(a,y) since the b values are higher in absolute value than those of a so the summation will have the same sign of relationship with y as the b values.