From: Multiple phenotype association tests based on sliced inverse regression
Case 1 | k=5, q=5 | \(\varvec{\beta }_{3}=(1.10,1.10,1.10,1.10,1.10)\) |
\(\varvec{\beta }_{4}=(0.00,0.02,0.00,0.00,0.00)\) | ||
Case 2 | k=5, q=10 | \(\varvec{\beta }_{3}=(1.10,1.10,1.10,1.10,1.10,0.00,0.00,0.00,0.00,0.00)\) |
\(\varvec{\beta }_{4}=(0.00,0.02,0.00,0.00,0.00,0.00,0.00,0.00,0.00,0.00)\) | ||
Case 3 | k=10, q=5 | \(\varvec{\beta }_{3}=(1.10,1.10,1.10,1.10,1.10)\) |
\(\varvec{\beta }_{4}=(0.00,0.02,0.00,0.00,0.00)\) | ||
Case 4 | k=10, q=10 | \(\varvec{\beta }_{3}=(1.10,-1.10,1.10,-1.10,1.10,0.00,0.00,0.00,0.00,0.00)\) |
\(\varvec{\beta }_{4}=(0.00,0.02,0.00,0.00,0.00,0.00,0.00,0.00,0.00,0.00)\) |