MATH SOLVE

3 months ago

Q:
# Let 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, and Y and Z are independent. Find: (a) Cov(X + 2Y, Y − Z). (b) Var(3X − Y ). (c) Var(X + Y + Z)

Accepted Solution

A:

Answer:13,57,31Step-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 independentTo 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)