Compute the standard deviation of W if W 04X06YvarX10varY5 a

Compute the standard deviation of W if W =0.4X+0.6Y,var(X)=10,var(Y)=5 and cov(X,Y)=

5.

What is the distribution of W if W =0.4X+0.6Y,X =1,Y =3,var (X)=10,var(Y)=5 and

cov ( X, Y ) = 5 where X and Y are normal random variables?

Solution

We first calculate var(W)

= Var ( 0.4X + 0.6Y)

= 0.4² Var(X) + 0.6² Var(Y) + 2 (0.4) ( 0.6) Cov (X,Y)

= 0.16(10) + 0.36(5) + 0.48(-5)

= 1.6 + 1.8 - 2.4

= 1.0

Var (W) = 1

Thus, Stdev = sqrt (Var) = sqrt ( 1 )= 1

b)

W would be a normal distribution.

mean (W) = E ( W ) = E ( 0.4X + 0.6Y )

= 0.4 E(X) + 0.6 E(Y)

= 0.4 (1) + 0.6(3)

= 2.2

Thus,

W ~ N( 2.2, 1)

Hope this helps.