Find the value of a in the relation Cov(2X,−3Y+2)=a⋅Cov(X, Y) . a=
c) Suppose that X , Y , and Z are independent, with a common variance of 5 . Then,
Cov(2X+Y,3X−4Z)=

a = -6

Cov (2X+Y, 3X-4Z) = 30

Step-by-step explanation:

Key points:

Cov(2X, -3Y+2) = a⋅Cov (X,Y)

Cov (2X, -3Y) + Cov (2X, 2) = a⋅Cov (X,Y)

(2)⋅(-3)⋅Cov (X, Y) + 0 = a⋅Cov (X,Y)

-6⋅Cov (X, Y) + 0 = a⋅Cov (X,Y)

⇒ a = -6.

(c)

Suppose that X, Y, and Z are independent, with a common variance of 5, i.e. V (X) = V (Y) = V (Z) = 5

Cov (2X+Y, 3X-4Z) = Cov (2X, 3X) + Cov (2X, -4Z) + Cov (Y, 3X) + Cov (Y, -4Z)

                               = 6⋅Cov (X, X) - 8⋅Cov (X, Z) + 3⋅Cov (Y, X) - 4⋅Cov (Y, Z)

                               = (6 × 5) - 0 + 0 - 0

                               = 30

Thus, the value of Cov (2X+Y, 3X-4Z) is 30.

the answer is (-4,-4)

hope this ! : d

3

step-by-step explanation: