c1=-1/3
c2=2
Let A and B be independent random variables with means 1 , and variances 1 and 2 , respectively.
Let X=A−B and Y=A+B .
Find the coefficients c1 and c2 of the Linear Least Mean Squares (LLMS) estimator YˆLLMS=c1X+c2 of Y based on X .
4 answers
i'm getting 0 for cov(x,y), anyone else?
cov(x,y) = cov(A-B, A+B) = cov(A,A) + cov(A,B) + cov(-B,A) + cov(-B,B) = 1 + 0 + 0 -1 = 0 ?
cov(x,y) = cov(A-B, A+B) = cov(A,A) + cov(A,B) + cov(-B,A) + cov(-B,B) = 1 + 0 + 0 -1 = 0 ?
ignore the one above, cov(-B, B) = -var(B) = -2
E(X) = E(A-B) = 0
E(Y) = E(A+B) = 2
Var(X) = Var(A-B) = Var (A) + Var (B) = 3
Var(Y) = Var(A+B) = Var (A) + Var (B) = 3
Cov(X,Y)=E(XY)-E(X)E(Y)
=E(A-B*A+B)- [E(A-B)E(A+B)] = Var(A)-Var(B) = -1
...
Y_llms = 2- x/3
c1 = -1/3
c2 = 2
E(Y) = E(A+B) = 2
Var(X) = Var(A-B) = Var (A) + Var (B) = 3
Var(Y) = Var(A+B) = Var (A) + Var (B) = 3
Cov(X,Y)=E(XY)-E(X)E(Y)
=E(A-B*A+B)- [E(A-B)E(A+B)] = Var(A)-Var(B) = -1
...
Y_llms = 2- x/3
c1 = -1/3
c2 = 2
0 answers