Closed famuvie closed 7 years ago
Nobs <- 1e4 ## residual covariance matrix S_res <- matrix(c( 9, 3, -3, 3, 9, 9, -3, 9, 14 ), nrow = 3, ncol = 3) ## simulated residual-only dataset testdat <- data.frame(breedR.sample.ranef(3, S_res, Nobs, vname = 'e')) ## fitted model res <- remlf90( cbind(e_1, e_2, e_3) ~ 1, data = testdat, method = "ai" ) #> Using default initial variances given by default_initial_variance() #> See ?breedR.getOption. summary(res) #> Formula: cbind(e_1, e_2, e_3) ~ 0 + Intercept #> Data: testdat #> AIC BIC logLik #> 83966 84031 -41974 #> #> #> Variance components: #> Estimated variances S.E. #> Residual.e_1 9.021 0.12758 #> Residual.e_1_Residual.e_2 3.098 0.09474 #> Residual.e_2 -2.900 0.11476 #> Residual.e_1_Residual.e_3 8.886 0.12567 #> Residual.e_2_Residual.e_3 8.787 0.14095 #> Residual.e_3 13.666 0.19327 #> #> Fixed effects: #> value s.e. #> Intercept.e_1 -0.042498 0.0300 #> Intercept.e_2 0.020878 0.0298 #> Intercept.e_3 0.045872 0.0370 res$var[["Residual", "Estimated variances"]] #> e_1 e_2 e_3 #> e_1 9.0205 3.0978 -2.9004 #> e_2 3.0978 8.8856 8.7875 #> e_3 -2.9004 8.7875 13.6660
Notice how the value summarized as Residual.e_2 corresponds in fact to the residual covariance between e_2 and e_3. In particular, it is negative!. The values are fine, but the labeling is incorrect.
Residual.e_2
e_2
e_3
Thanks to Vincent Segura for reporting.
Notice how the value summarized as
Residual.e_2
corresponds in fact to the residual covariance betweene_2
ande_3
. In particular, it is negative!. The values are fine, but the labeling is incorrect.