topic Re: Box Cox in Statistical Procedures

Box Cox

jacksonan123 — Fri, 26 Aug 2016 10:39:17 GMT

I conducted a BOX-COX analysis using the following code:

ods rtf file='/folders/myfolders/regr4BCoxNresuls/CLnrGFR1.rtf' style=journal;

ods TRACE ON;

ods output FitStatistics=FitStat ParameterEstimates=param;

proc transreg data=boot2 plots=all; /*plots(only)=FITPLOT(stats=none); */

model boxcox(CLnr/convenient lambda=-3 to 3 by 0.125)=identity(GFR1)/cl; /*SPEC;*/

output out=pred ;

proc print data=fitstat;run;

proc print data=param;run;

proc print data=pred;run;

run;

quit;

ods rtf close;

The best lamda was =0 which means that the log transformation for CLnr was best. The data is posted below (Nstudyr4.csv)set which if I use log transformation for CLnr would give negative values. On line it was suggested that I can do log transformation if I add a constant to each value. My question is how do I interpret the value for the intercept if I do a regression when I take the antilog with the added constant?

I have attached the data set and the BOX-Cox output.

Subject	CL	CLr	CLnr	GFR1	GFR2
1	2.02	1.73	0.29	125.5	124.1
2	2.18	1.63	0.55	114.3	113.9
3	1.95	1.28	0.67	102.2	102
4	1.83	1.16	0.67	101.6	100.7
5	1.64	1.47	0.17	96.6	95.6
6	1.94	1.72	0.22	90.7	89.2
7	1.98	1.47	0.51	87.8	86.2
8	1.67	1.07	0.6	84.5	83.6
9	1.25	1.19	0.06	84.2	83.4
10	0.9	0.74	0.16	77.7	76.5
11	1.53	1.06	0.47	75.6	75.2
12	1.52	0.82	0.7	68	67.5
13	1.58	1.23	0.35	66.5	65.3
14	0.93	0.75	0.18	64.4	64.2
15	0.93	0.95	.	60	59.6
17	0.61	0.48	1.2	52.3	52.5
18	0.76	0.46	0.15	52	52.3
19	0.57	0.44	0.32	51.7	52.3
20	0.66	0.4	0.17	49.3	49.7
21	0.51	0.15	0.51	37.5	37.7
22	0.41	0.32	0.19	35.5	35.7
23	0.65	0.57	.	34.3	34.4
24	0.9	0.58	0.07	31	31.4
25	0.5	0.29	0.61	28.1	29.3
26	0.31	0.22	0.28	26.6	26.8
27	0.3	0.21	0.1	24.8	24.7
28	0.48	0.38	.	18	18.6
29	0.42	0.19	0.29	16.3	16.6
30	0.31	0.15	0.27	16	16.6
31	0.2	0.12	0.19	12.7	12.7
32	0.18	0.2	.	12.7	12.7
33	0.25	0.18	.	12.7	12.7
34	0.35	0.13	0.12	10.3	12.7
35	0.2	0.12	0.23	9.8	12.7
36	0.15	0.08	0.12	8	11.5
37	0.15	0.21	.	7.7	10.2
38	0.11	0.092	0.058	6.5	10.2
39	0.046	0.092	0.018	4.7	8.9
40	0.076	0.061	.	4.1	6.4
41	0.14	0.034	0.042	4.1	3.8
42	0.16	0.018	0.122	4.1	3.8
43	0.18	0.015	0.145	1.8	1.5
44	0.15	0.018	0.162	1.5	1.5

Re: Box Cox

SteveDenham — Tue, 30 Aug 2016 18:15:31 GMT

So what if ln(CLnr) is negative? There is no requirement that a dependent variable be positive. If you are truly worried, rescale the dependent variable--say by multiplying by 1000. Then the ln values will also be positive. Since I suppose that CLnr is clearance of some metabolite, it is like shifting the measurement from millimolar to micromolar.

Steve Denham

Re: Box Cox

Rick_SAS — Tue, 30 Aug 2016 18:29:07 GMT

Because your response data are positive, the log transformed response model is

log(Y) = a + b*x

Y = exp(a + b*x) where a is the estimate for the intercept and b is the estimate for the explanatory coefficient.

If you define A=exp(a), you get Y = A*exp(b*x)

So the intercept term multiplies the model: a unit change in the intercept estimate results in a mulitplicative change (a factor of e) in the predicted response.

You can do the same computation for other models. If you add c to the response before you fit the model, then

Y = -c + A*exp(b*x)

Re: Box Cox

jacksonan123 — Tue, 30 Aug 2016 19:30:35 GMT

Thanks for the response.

Re: Box Cox

jacksonan123 — Tue, 30 Aug 2016 19:31:23 GMT

Thanks for the response and it addresses my issue.