Question

1. Suppose we have three random variables Y1 , Y2 , and Y3 .

Suppose we have three random variables Y, Y,, and Y,. The standard deviations of Y and Y, are both 3 and the standard deviation of Y is 2. The correlation coefficient between Y and Y, is-0.6. The covariance between Y and Y, is 0.5. Y is independent of Y 1. 1 2 a) (3 pts) Find Var(h + 3%) b) (3 pts) Find Cov(3h + 2⅓'5½-%)

Answer 1

1.

a)

General formula for Variance of sum of two random variables X and Y is :

$Var(aX+bY)=a^2Var(X)+b^2Var(Y)+2abCov(X,Y)$

We need to find,

$Var(Y_1+3Y_2) = Var(Y_1)+3^2Var(Y_2)+2*3*Cov(Y_1,Y_2)$

We know that,

$SD(Y_1) = 3\ \ , SD(Y_2)=3, \ \ SD(Y_3)=2$

So,

$Var(Y_1) = 3^2=9\ \ , Var(Y_2)=3^2=9, \ \ Var(Y_3)=2^2=4$

We are also given,

$Cor(Y_1,Y_2)=-0.6$

Formula for correlation between two Random variables X and Y is :

$Cor(X,Y) = \frac{Cov(X,Y)}{\sqrt{Var(X)*Var(Y)}}$

So,

$Cor(Y_1,Y_2)=\frac{Cov(Y_1,Y_2)}{\sqrt{Var(Y_1)*Var(Y_2)}}=-0.6$

$\frac{Cov(Y_1,Y_2)}{\sqrt{Var(Y_1)*Var(Y_2)}}=-0.6$

$Cov(Y_1,Y_2)=-0.6*(\sqrt{9*9}) = -5.4$

So,

$Var(Y_1+3Y_2) = Var(Y_1)+3^2Var(Y_2)+2*3*Cov(Y_1,Y_2)$

$= 9+3^2*9+2*3*(-5.4)$

$= 57.6$

b)

Now, we need to find,

$Cov(3Y_1+2Y_3,5Y_2-Y_3)$

$= Cov(3Y_1,5Y_2)-Cov(3Y_1,Y_3)+Cov(2Y_3,5Y_2)-Cov(2Y_3,Y_3)$

$= 3*5*Cov(Y_1,Y_2)-3*Cov(Y_1,Y_3)+2*5*Cov(Y_3,Y_2)-2*Var(Y_3)$

Since, Y₂ and Y₃ are independent , so ,

$Cov(Y_2,Y_3)=0$

and we also have,

$Cov(Y_1,Y_2)=-5.4$

$Cov(Y_1,Y_3)=0.5$

Putting all the values we get,

$Cov(3Y_1+2Y_3,5Y_2-Y_3)= 3*5*(-5.4)-3*(0.5)+2*5*0-2*4$$=-90.5$