Respuesta :
Answer:
a = -6
Cov (2X+Y, 3X-4Z) = 30
Step-by-step explanation:
Key points:
- Cov (aX, bY) = a·b·Cov (X, Y)
- Cov (X, X) = V (X)
- Cov (X, a) = 0
- If X and Y are independent then Cov (X, Y) = 0.
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 value of a = -6. And, Cov (2X+Y, 3X-4Z) = 30.
Calculation of the value of a and cov:
Since
Cov (aX, bY) = a·b·Cov (X, Y)
Cov (X, X) = V (X)
Cov (X, a) = 0
In the case when X and Y are independent so Cov (X, Y) = 0.
Now
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)
here
Suppose that X, Y, and Z are independent, with a common variance of 5, i.e.
V (X) = V (Y) = V (Z) = 5
So,
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.
Learn more about variance here: https://brainly.com/question/24138432
