Answer:
13,57,31
Step-by-step explanation:
Given that X, Y , Z be three random variables which satisfy the following conditions:
Var(X) = 4, Var(Y ) = 9, Var(Z) = 16. Cov(X, Y ) = −2, Cov(Z, X) = 3,
Var(y,z) =0 since given as independent
To find
[tex]a) Cov (x+2y, y-z)\\ \\= cov (x,y) +cov (2y,y) -cov (x,z) -cov(2y,z)\\= cov (x,y) +2cov (y,y) -cov (x,z) -2cov(y,z)\\=-2+2 var(y) -3-0\\= -2+18-3\\=13[/tex]
b) [tex]Var(3X − Y ).\\= 9Var(x)+var(y) -6 covar (x,y)\\= 36 +9+12\\= 57[/tex]
c) Var(X + Y + Z)[tex]=Var(x) = Var(Y) +Var(z) +2cov (x,y) +2cov (y,z) +2cov (x,z)\\= 4+9+16+(-4) +6\\= 31[/tex]
Note:
Var(x+y) = var(x) + Var(Y) +2cov (x,y)
Var(x+2y) = Var(x) +4Var(y)+4cov (x,y)