SOLUTION: . Independent random variables
X
and
Y
are such that
E X( ) 4 , E Y( ) 5 , Var X( ) 1 ,
Var Y( ) 2
. Find:
A.
E X Y (4 2 ) .
B.
E X Y (5 ) .
C.
Algebra.Com
Question 1207202: . Independent random variables
X
and
Y
are such that
E X( ) 4 , E Y( ) 5 , Var X( ) 1 ,
Var Y( ) 2
. Find:
A.
E X Y (4 2 ) .
B.
E X Y (5 ) .
C.
Var X Y (3 2 ) .
D.
Var Y X (5 3 ) .
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Understanding the Problem:
We are given two independent random variables, X and Y, with their respective means and variances:
E[X] = 4
E[Y] = 5
Var[X] = 1
Var[Y] = 2
Key Property of Independent Random Variables:
For independent random variables X and Y, the following properties hold:
Expected Value of the Sum: E[X + Y] = E[X] + E[Y]
Variance of the Sum: Var[X + Y] = Var[X] + Var[Y]
Expected Value of the Product: E[XY] = E[X] * E[Y]
Variance of the Product: Var(XY) = E[X^2]E[Y^2] - (E[X]E[Y])^2
Solving the Problem:
A. E[4X + 2Y] = 4E[X] + 2E[Y] = 44 + 25 = 26
B. E[5X - Y] = 5E[X] - E[Y] = 5*4 - 5 = 15
C. Var[3X + 2Y] = 9Var[X] + 4Var[Y] = 91 + 42 = 17
D. Var[5Y - 3X] = 25Var[Y] + 9Var[X] = 252 + 91 = 61
Therefore, the answers are:
A. 26
B. 15
C. 17
D. 61
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