Question

If E(X)=100, E(Y)=120, E(Z) = 130, Var(X) = 9, Var(Y) = 16, Var(Z) = 25, Cov(X, Y)= - 10 Cov(X,Z) = 12, and Cov(Y,Z) = 14, then answer the followings: 1) Corr(X,Y) 2) Corr(X,Z) 3) Corr(Y,Z) 4) E(3X + 4Y – 3Z) 5) Var(3X – 3Z) 6) Var(3X + 4Y – 3Z) 7) Cov(3X, 2Y+3Z)

Answer 1

Answer:

(1) -0.833

(2) 0.80

(3) 0.70

(4) 390

(5) 90

(7) 48

Step-by-step explanation:

Given:

E (X) = 100, E (Y) = 120, E (Z) = 130

Var (X) = 9, Var (Y) = 16, Var (Z) = 25

Cov (X, Y) = -10, Cov (X, Z) = 12, Cov (Y, Z) = 14

The formulas used for correlation is:

$Corr (A, B) = \frac{Cov (A, B)}{\sqrt{Var (A)\times Var(B)}}$

(1)

Compute the value of Corr (X, Y)-

$Corr (X, Y) = \frac{Cov (X, Y)}{\sqrt{Var (X)\times Var(Y)}} \\=\frac{-10}{\sqrt{9\times16}} \\=-0.833$

(2)

Compute the value of Corr (X, Z)-

$Corr (X, Z) = \frac{Cov (X, Z)}{\sqrt{Var (X)\times Var(Z)}} \\=\frac{12}{\sqrt{9\times25}} \\=0.80$

(3)

Compute the value of Corr (Y, Z)-

$Corr (Y, Z) = \frac{Cov (Y, Z)}{\sqrt{Var (Y)\times Var(Z)}} \\=\frac{14}{\sqrt{16\times25}} \\=0.70$

(4)

Compute the value of E (3X+4Y-3Z)-

$E(3X+4Y-3Z)=3E(X)+4E(Y)-3E(Z)\\=(3\times100)+(4\times120)-(3\times130)\\=390$

(5)

Compute the value of Var (3X-3Z)-

$Var (3X-3Z)=[(3)^{2}\times Var(X)]+[(-3)^{2}\times Var (Z)]+(2\times3\times-3\times Cov(X, Z)]\\=(9\times9)+(9\times25)-(18\times12)\\=90$

(6)

Compute the value of Var (3X+4Y-3Z)-

$Var (3X+4Y-3Z)=[(3)^{2}\times Var(X)]+[(4)^{2}\times Var(Y)]+[(-3)^{2}\times Var (Z)]+[(2\times3\times4\times Cov(X, Y)]+[(2\times3\times-3\times Cov(X, Z)]+[(2\times4\times-3\times Cov(Y, Z)]\\=(9\times9)+(16\times16)+(9\times25)+(24\times-10)-(18\times12)-(24\times14)\\=-230$

But this is not possible as variance is a square of terms.

(7)

Compute the value of Cov (3X, 2Y+3Z)-

$Cov(3X, 2Y+3Z)=Cov(3X,2Y)+Cov(3X, 3Z)\\=6Cov(X, Y)+9Cov(X,Z)\\=(6\times-10)+(9\times12)\\=48$