Could someone please check my work?

1.16) If U and Vare independent random variables, both having variance I, find Cov ( x, 4) when X= a U + bV, Y = CU+ dV. Since U and V are independent, Cov ( U, V) = Cor ( V, U) = 0 Since Var (U) = Var(V) = 1 and Cov (U, U) = Var (U) and Cov ( V, v) = Var(V), Then Cov ( U, U) = Cov(V, V) = 1 Cov ( x, 4 ) = Cov ( a U+bv, cutdv) = Cov ( QU, Cutdv) + Cov( bv,cutdV) by linearity property ( 1.9 ) = a Cov ( U, CutdV ) + bCov ( V,cutdv) = a Cov ( cut dv, U ) + b Cov (cutdy, v) = a cov ( c U, U ) + cov(dv, u)| + b/cov ( cu, v) + Cov (dv,v)] by (1.9 = a clov ( U, U) + d cov (v, u)| + bsccov(U, v)+dcov(v,v)] Terms go to zero Since Cov( V, U) = Cov ( U, V ) = 0 = ac lov ( U, U) + bdCov(viv) = Var (U) =1 = Var(V) =1 = ac + bd