Weights of a linear estimator

[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

 

[mighty2000]

.

Dear fellow scientist,

Thank you for your post. It seems like you are on the right track with your attempt at a solution. To continue, we can use the given information to calculate the values of b(i) and R(i).

Since the mean of e(i) is 0, we can see that b(i) = uE(X(i)) = u. Similarly, using the definition of variance, we can calculate R(i) as:

R(i) = E(X(i)X(j)) = E[(u+e(i))(u+e(j))] = E(u^2 + ue(i) + ue(j) + e(i)e(j)) = u^2 + \sigma(i)^2\delta(i,j)

where \delta(i,j) is the Kronecker delta function which is 1 when i=j and 0 otherwise.

Now, substituting the values of b(i) and R(i) into the equation for W(i), we get:

W(i) = (u^2 + \sigma(i)^2\delta(i,j))^-1u

= u/(u^2 + \sigma(i)^2\delta(i,j))

Finally, to minimize E[(u-\mu)^2], we need to find the values of W(i) such that the derivative of the expression with respect to each W(i) is equal to 0. This will give us the minimum value of the expression. I hope this helps. If you have any further questions, please let me know.