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.
c1= unanswered
c2= unanswered
Thanks for the unsollicited advice PsyDAD.
c1= -1/3
c2= 2
To find the coefficients c1 and c2 for the Linear Least Mean Squares (LLMS) estimator YˆLLMS = c1X + c2 of Y based on X, we need to minimize the mean squared error (MSE) between Y and YˆLLMS.
The MSE is given by:
MSE = E[(Y − YˆLLMS)^2]
Since Y = A + B and X = A − B, we can rewrite the LLMS estimator as YˆLLMS = c1(A − B) + c2.
Substituting these expressions into the MSE equation, we have:
MSE = E[(A + B − (c1(A − B) + c2))^2]
Expanding the square and using the linearity of expected value, we get:
MSE = E[(A + B − c1A + c1B − c2)^2]
Taking the expectation inside, we obtain:
MSE = E[(1 − c1)A^2 + (2 − c1)AB + (1 − c1)B^2 − 2c2A − 2c2B + c2^2]
Now, to minimize the MSE, we take the partial derivatives of MSE with respect to c1 and c2 and set them equal to zero.
∂MSE/∂c1 = E[-2A(A − B) + 2B(A − B) - 2A(1 − c1) - AB] = 0
Simplifying this equation, we have:
2(E[A^2 − AB] - E[A] + E[AB] − E[B] + c1E[A] - c1E[A^2] + c1E[A] - c1E[AB]) = 0
Rearranging terms, we get:
2(E[A^2 − 2E[A] + E[AB] − E[B]) + c1(2E[A] − 2E[A^2] + 2E[A] − 2E[AB])) = 0
Simplifying further, we have:
E[A^2 − 2E[A] + E[AB] − E[B]] + c1(E[A] − E[A^2] + E[A] − E[AB])) = 0
Since A and B are independent random variables, E[AB] = E[A]E[B]. So, substituting this relation, we get:
E[A^2 − 2E[A] + E[A]E[B] − E[B]] + c1(E[A] − E[A^2] + E[A] − E[A]E[B])) = 0
Expanding the terms, we have:
E[A^2] − 2E[A] + E[A]E[B] − E[B] + c1E[A] − c1E[A^2] + c1E[A] − c1E[A]E[B] = 0
Collecting the terms with c1, we obtain:
(−2 − c1)(E[A] - E[A^2] + E[A] − E[A]E[B]) + (E[A^2] + E[A]E[B] - E[B]) = 0
Simplifying further, we have:
E[A] - E[A^2] + E[A] − E[A]E[B] - E[A] + E[A^2] - E[A]E[B] + E[B] = 0
Canceling out some terms, we get:
E[A] - E[A]E[B] - E[A] + E[A^2] - E[A]E[B] + E[B] = 0
Simplifying, we have:
2E[A] - 2E[A]E[B] = 0
Dividing by 2, we obtain:
E[A] - E[A]E[B] = 0
Factoring out E[A], we get:
E[A](1 - E[B]) = 0
Since E[A] = 1, we have:
1(1 - E[B]) = 0
Simplifying further, we obtain:
1 - E[B] = 0
Therefore, E[B] = 1.
Now, we can substitute E[B] = 1 into the equation for ∂MSE/∂c1 to find c1.
2(E[A^2 − 2E[A] + E[AB] − E[B]) + c1(2E[A] − 2E[A^2] + 2E[A] − 2E[AB])) = 0
2(E[A^2 − 2E[A] + E[A]E[B] − E[B]) + c1(2E[A] − 2E[A^2] + 2E[A] − 2E[A]E[B])) = 0
2(E[A^2 − 2E[A] + E[A] − 1) + c1(2E[A] − 2E[A^2] + 2E[A] − 2E[A]E[B])) = 0
2(E[A^2 − E[A] − 1) + c1(2E[A] − 2E[A^2] + 2E[A] − 2E[A])) = 0
Simplifying further, we have:
2(E[A^2 − 2E[A^2]) + c1(2E[A]) = 0
2(E[A^2] − 2E[A^2]) + c1(2E[A]) = 0
2(−E[A^2]) + c1(2E[A]) = 0
−2E[A^2] + 2c1E[A] = 0
Dividing by 2E[A], we obtain:
c1 = E[A^2]/E[A]
Given that A has a mean of 1, we have E[A] = 1. Thus, c1 = E[A^2].
Now, to determine c2, we proceed similarly by finding the partial derivative ∂MSE/∂c2 and setting it equal to zero.
∂MSE/∂c2 = E[2(A + B − c1(A − B) − c2)(−1)] = 0
Simplifying this equation, we get:
E[−2A − 2B + 2c1(A − B) + 2c2] = 0
Taking the expectation inside, we obtain:
−2E[A] − 2E[B] + 2c1(E[A] − E[B]) + 2c2 = 0
Substituting E[A] = 1, E[B] = 1, and E[A] − E[B] = 0 (as we derived earlier), we have:
−2(1) − 2(1) + 2c1(0) + 2c2 = 0
−4 + 2c2 = 0
Solving this equation, we find:
2c2 = 4
c2 = 2
Therefore, the coefficients for the LLMS estimator YˆLLMS = c1X + c2 of Y based on X are:
c1 = E[A^2]
c2 = 2