X and Y are two random variables with means of 1 and 3 and variances 3 and 7, respectively. W is a random variable which is a weighted average of X and Y, given by
W = aX+(1-a)Y, where a = 0.09. Find:
a) Mean of 11W
b) Variance of 11W, assuming X and Y are independent.
c) Variance of 11W, assuming Cov(X,Y)= -1.7
I have done the following for a) and b) (correct me if I'm wrong):
a) E(W)= (a)1+(1-a)3 = 2.82 hence 11E(W) = 11(2.82) = 31.02
b) Var(W)= (a)3+(1-a)7 = 6.64 hence 11Var(W) = 11(6.64) = 73.04
However, I am unsure how to go about doing c).
The expected value is right. And the variance of $11W$ is
$$Var(11W)=11^2\cdot Var(a\cdot bX+(1-a)\cdot Y)$$
If you factor out a constant, the constant has to be squared.
$$=11^2\cdot \left(Var(a\cdot X)+Var((1-a)\cdot Y)+2\cdot Cov(a\cdot X,(1-a)\cdot Y) \right)$$
Covariance is bilinear. That means, that $Cov(c\cdot X,b\cdot Y)=c\cdot b\cdot Cov(X,Y)$
$$=11^2\cdot \left(a^2\cdot Var(X)+(1-a)^2\cdot Var(Y)+2\cdot a\cdot (1-a)\cdot Cov(X,Y) \right)$$
If $X$ and $Y$ are independent, then $Cov(X,Y)=0$