From: Multiple phenotype association tests based on sliced inverse regression
Case 1 | \(k=40,q=50\) | \(\varvec{\beta }_{3}=c(1.10,-1.10,1.10,-1.10,1.10,0.00,0.00,\dots ,0.00)\) |
\(\varvec{\beta }_{4}=c(0.00,0.02,0.00,0.00,\dots ,0.00)\) | ||
Case 2 | \(k=40,q=100\) | \(\varvec{\beta }_{3}=c(1.10,-1.10,1.10,-1.10,1.10,0.00,0.00,\dots ,0.00)\) |
\(\varvec{\beta }_{4}=c(0.00,0.02,0.00,0.00,\dots ,0.00)\) | ||
Case 3 | \(k=100,q=50\) | \(\varvec{\beta }_{3}=c(1.10,-1.10,1.10,-1.10,1.10,0.00,0.00,\dots ,0.00)\) |
\(\varvec{\beta }_{4}=c(0.00,0.02,0.00,0.00,\dots ,0.00)\) | ||
Case 4 | \(k=100,q=100\) | \(\varvec{\beta }_{3}=c(1.10,-1.10,1.10,-1.10,1.10,0.00,0.00,\dots ,0.00)\) |
\(\varvec{\beta }_{4}=c(0.00,0.02,0.00,0.00,\dots ,0.00)\) |