bc + - sqrt(1-b^2) * sqrt (1-c^2) if b=0.6 and=0.8, then a can be between 0 and 0.96

the determinant of the 3x3 correlation matrix is 1 + 2 (ab) (ac) (bc) - (ab^2 + ac^2 + bc^2) which must be positive (or zero), where "ab" is the correlation between A and B, etc. Note, bc can be negative. Based on this 1 + 2 * 0.6 * 0.8 (bc) - (0.6^2 + 0.8^2 + bc^2) >= 0 this simplifies to: bc^2 - 0.96 *bc = 0

... bc^2 - 0.96 *bc <= 0 so 0 <= bc <= 0.96 to keep the determinant of the correlation matrix greater than or equal to zero.

edit: a can be between bc ± sqrt (1-b^2) * sqrt(1-c^2) if b=0.6 and c=0.8, then a can be between 0 and 0.96

Between -0.29 or 0.41

Edit: -0.29 or 0.411

suppose EA=EB=EC=0. Then E(A)E(B)/sqrt(E(A^2)E(B^2))=0.6, E(A)E(C)/sqrt(E(A^2)E(C^2))=0.8. Multiplying these gives Cor(B,C)*E(A)^2/E(A^2)=0.48. Since E(A)^2=0.48. Note that by E(A)^2/E(A^2)=0.8^2*E(C^2)/E(C)^2>=0.64, we have Cor(B,C)<=0.48/0.64=0.75. So between 0.48 and 0.75.