What Is the Optimal Weight in a Linear Estimator to Minimize Mean Squared Error?

[purplebird]

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

 

[KataKoniK]

.



Thank you for posting your question. I am a scientist and I would be happy to help you with your problem. Let's start by defining some terms in your problem. X(i) represents a set of N random variables, each with a mean of u and a variance of \sigma(i)^2. The variable u is a parameter that we are trying to estimate using a linear estimator, which is defined as \mu = \sumW(i)X(i). Our goal is to find the values of W(i) that will result in a minimum error between the estimated value of \mu and the true value of u.

To begin, we can rewrite the linear estimator as \mu = \sumW(i)(u + e(i)), since X(i) = u + e(i). Expanding this equation, we get \mu = u\sumW(i) + \sumW(i)e(i). We want the mean value of \mu to be equal to u, so we can set \sumW(i) = 1. This means that the weights must sum to 1.

Next, we can calculate the expected value of \mu using the linearity property of expected values:

E[\mu] = E[u\sumW(i) + \sumW(i)e(i)] = u\sumW(i) + \sumW(i)E[e(i)]

Since the mean of e(i) is 0, this simplifies to E[\mu] = u\sumW(i).

Now, let's look at the error between the estimated value of \mu and the true value of u. This is given by (u - \mu)^2. We want to minimize this error, so we can take the derivative of this expression with respect to W(i) and set it equal to 0:

\frac{d}{dW(i)}(u - \mu)^2 = -2(u - \mu)\frac{d\mu}{dW(i)} = -2(u - \mu)X(i)

Setting this equal to 0 and solving for W(i), we get W(i) = \frac{X(i)}{\sumX(i)^2}.

Therefore, the weights that will result in a minimum error between the estimated value of \mu and the true value of u are W(i) = \frac{X(i)}{\sumX(i)^2}.

I

 

[piercas]

.

it is important to thoroughly understand the concepts and equations involved in any problem before attempting to solve it. In this case, the problem is asking for a linear estimator that will minimize the mean squared error between the estimated value, \mu, and the true value, u. To solve this, we first need to understand the concept of a linear estimator and how it is calculated.

A linear estimator is a method of predicting a value based on a linear combination of observed data. In this case, the observed data is represented by X(i) and the linear combination is represented by W(i). The goal is to find the values of W(i) that will give us the most accurate prediction of the true value, u.

To do this, we can use the method of least squares. This method involves minimizing the sum of the squared errors between the estimated value and the observed data. In other words, we want to minimize the function E[(u- \mu)^2]. To do this, we need to find the values of W(i) that will make this function as small as possible.

Using the equations provided in the problem, we can rewrite the function as:

E[(u- \mu)^2] = E[(u- \sumW(i)X(i))^2]

= E[(u- \sumW(i)(u + e(i)))^2]

= E[(\sumW(i)e(i))^2]

= \sumW(i)^2E[e(i)^2]

= \sumW(i)^2\sigma(i)^2

Now, to minimize this function, we need to take the derivative with respect to W(i) and set it equal to 0:

\frac{\partial E[(u- \mu)^2]}{\partial W(i)} = 2\sumW(i)\sigma(i)^2 = 0

Solving for W(i), we get:

W(i) = 0

This means that the linear estimator that will minimize the mean squared error is simply a constant value of 0. This may seem counterintuitive, but it is the result of the given information, where the mean of e(i) is 0 and the variance is \sigma(i)^2.

In conclusion, the linear estimator for this problem is W(i) = 0, which means that the estimated value, \mu, will always be equal to the true value, u. This is the best estimator we