How Do We Optimize Weights in a Linear Estimator for Minimum Error?

[purplebird]

I know I already posted this but I made a mistake in the original post which I realized now and I am reposting the correct problem as i am not able to edit it.

Homework Statement

Given
X(i) = u + e(i) i = 1,2,...N
such that e(i)s are statistically independent and u is a parameter
mean of e(i) = 0
and variance = $$\sigma(i)$$^2

Find W(i) such that the linear estimator

$$\mu$$ = $$\sum$$W(i)X(i) for i = 1 to N

has

mean value of $$\mu$$= u

and E[(u- $$\mu$$)^2 is a minimum

The Attempt at a Solution

For a linear estimator:

W(i) = R$$^{}-1$$b

where b(i)= E($$\mu$$(i) X(i)) and R(i) = E(X(i)X(j))

I do not know how to proceed beyond this. Thanks for your help

 

[mmwave]

Thank you for your post and for posting the corrected problem. I will do my best to assist you with finding the solution.

Based on the given information, we can approach the problem in the following steps:

Step 1: Define the problem

The problem is to find the weights W(i) such that the linear estimator \mu = \sumW(i)X(i) has a mean value of u and minimizes the error term E[(u- \mu)^2].

Step 2: Express the linear estimator in terms of the given variables

We can rewrite the linear estimator as:

\mu = \sum W(i) (u + e(i)) = u \sum W(i) + \sum W(i) e(i)

Step 3: Simplify the expression

Since the mean of e(i) is 0, the second term in the above expression becomes 0. Therefore, we can rewrite the linear estimator as:

\mu = u \sum W(i)

Step 4: Solve for W(i)

To ensure that the mean value of \mu is u, we can set the sum of weights \sum W(i) to be equal to 1. Therefore, we have:

\sum W(i) = 1

We can also express the error term as:

E[(u- \mu)^2] = E[(u - u \sum W(i))^2] = u^2 E[(1 - \sum W(i))^2]

To minimize this error term, we can use the method of Lagrange multipliers and minimize the following expression:

L = u^2 E[(1 - \sum W(i))^2] - \lambda(\sum W(i) - 1)

Taking the derivative of L with respect to W(i) and setting it equal to 0, we get:

-2u^2 E[1 - \sum W(i)] + \lambda = 0

Solving for \lambda and substituting it back into the equation, we get:

\sum W(i) = \frac{1}{N}

Therefore, the weights W(i) are:

W(i) = \frac{1}{N}

Step 5: Check the solution

We can check the solution by substituting the weights back into the linear estimator and verifying that it has a mean value of u and minimizes the error term.

\mu = u \sum W(i) = u \frac