Let π΄ and π΅ be independent random variables with means 1, and variances 1 and 2, respectively.
Let π=π΄βπ΅ and π=π΄+π΅.
Find the coefficients π1 and π2 of the Linear Least Mean Squares (LLMS) estimator πΛπΏπΏππ=π1π+π2 of π based
To find the coefficients c1 and c2 of the Linear Least Mean Squares (LLMS) estimator Y^LLMS = c1X + c2 of Y based on X, we need to minimize the mean squared error between Y and Y^LLMS.
The mean squared error (MSE) is defined as the expected value of the squared difference between Y and Y^LLMS:
MSE = E[(Y - Y^LLMS)^2]
In this case, we have Y = A + B and X = A - B, so we can substitute these expressions into the MSE equation:
MSE = E[(A + B - (c1(A - B) + c2))^2]
Since A and B are independent random variables, we can calculate the expected value of their product as the product of their expected values:
E[AB] = E[A]E[B] = 1 * 1 = 1
Thus, expanding the MSE equation and applying the linearity of the expected value operator, we can simplify it as follows:
MSE = E[(A + B - (c1(A - B) + c2))^2]
= E[(A + B - c1A + c1B - c2)^2]
= E[(1 - c1)A + (1 + c1)B - c2)^2]
Expanding the square and using the linearity of the expected value operator, we get:
MSE = E[(1 - c1)^2A^2 + (1 + c1)^2B^2 + c2^2 - 2c2(1 - c1)A - 2c2(1 + c1)B + 2c2(1 - c1)(1 + c1)AB]
Now, we want to minimize the MSE with respect to c1 and c2. To do that, we take the partial derivatives of the MSE with respect to c1 and c2, and set them to zero.
Partial derivative with respect to c1:
βMSE/βc1 = -2E[2(1 - c1)A - 2(1 + c1)B + 2(1 - c1)(1 + c1)AB] = 0
Partial derivative with respect to c2:
βMSE/βc2 = -2E[-2A - 2B + 2(1 - c1)(1 + c1)AB] = 0
Simplifying these equations, we get:
E[A - B + (1 - c1)(1 + c1)AB] = 0 --> Equation 1
E[-A - B + (1 - c1)(1 + c1)AB] = 0 --> Equation 2
Now, we solve these two equations to find c1 and c2. However, since these equations involve the means of A and B, we need to know their joint probability distribution or covariance matrix to find the coefficients c1 and c2 explicitly.
If you have additional information about the joint distribution or covariance matrix of A and B, please provide it, and we can continue from there.