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shusha [124]
2 years ago
6

Let X, Y, and Z be random variables, and let Cov(⋅,⋅) denote the covariance operator as usual. Suppose that the variance of X is

0.7, Cov(X,Y)=0.4, Cov(X,Z)=1.2, and Cov(Y,Z)=0.8. Find each of the following to two decimal places.
a. Cov(TY,7X)=_______
b. Cov(7X +6,Z)=________
c. Cov(7Y,7X +7Z)=______
d. Cov(7X +6Y,7X+7Z)=_______
Mathematics
1 answer:
bearhunter [10]2 years ago
6 0

By definition of covariance,

Cov[X, Y] = E[(X - E[X]) (Y - E[Y])] = E[XY] - E[X] E[Y]

from which it follows that

Cov[X, X] = E[X^2] - E[X]^2 = V[X]

a. I assume T is supposed to be some scalar factor. I'll use a general scalar k and let you fill in the details for yourself.

Cov[kY, 7X] = E[7k XY] - E[kY] E[7X] = 7k(E[XY] - E[X] E[Y])

= 7k Cov[X, Y]

and plug in 0.4 for Cov[X, Y].

b.

Cov[7X + 6, Z] = E[(7X + 6) Z] - E[7X + 6] E[Z]

= 7 E[XZ] + 6 E[Z] - 7 E[X] E[Z] - 6 E[Z]

= 7 (E[XZ] - E[X] E[Z]) = 7 Cov[X, Z]

= 8.4

c.

Cov[7Y, 7X + 7Z] = E[7Y (7X + 7Z)] - E[7Y] E[7X + 7Z]

= 49 E[XY + YZ] - 49 E[Y] E[X + Z]

= 49 (E[XY] + E[YZ] - E[Y] E[X] - E[Y] E[Z])

= 49 (Cov[X, Y] + Cov[Y, Z])

= 58.8

d.

Cov[7X + 6Y, 7X + 7Z] = E[(7X + 6Y) (7X + 7Z)] - E[7X + 6Y] E[7X + 7Z]

= E[49 X^2 + 49 XZ + 42 XY + 42 YZ] - (49 E[X]^2 + 49 E[X] E[Z] + 42 E[X] E[Y] + 42 E[Y] E[Z])

= 49 (V[X] + Cov[X, Z]) + 42 (Cov[X, Y] + Cov[Y, Z])

= 143.5

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Answer:

58, 37, 9

Step-by-step explanation:

Given:

First term a₁ = 65 and common difference d = - 7

This sequence is arithmetic series and formula for calculating n-th term is:

aₙ = a₁ + (n-1) d

Accordingly

The second term is:

a₂ = 65 + (2-1) (-7) = 65 - 7 = 58

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The fifth term is:

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a₅ = 37

The ninth term is:

a₉ = 65 + (9 - 1) (-7) = 65 + 8 · (-7) = 65 - 56 = 9

a₉ = 9

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