# Weights of a linear estimator

1. Apr 15, 2008

### purplebird

Given
Y(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

2. Jul 25, 2008

There may be other ways to do this, but one of the most direct is outlined below.
$$\hat\mu = \sum w(i) x_i$$
then in order for $$E(\hat \mu) = \mu$$ to be true you must have

$$\sum w(i) = 1$$

Next, note that $$E(\hat \mu - \mu)^2$$ is simply the $$\textbf{variance}$$ of your estimate (since your estimate has expectation $$\mu$$).

Since the $$x_i$$ are independent, the variance of $$\hat \mu$$ is

$$\mathbf{Var}{\hat \mu} = \sum w(i)^2 \mathbf{Var}(x_i) = \sigma^2 \sum w(i)^2$$

You want to choose the $$w(i)$$ so that the most recent expression is minimized, subject to the constraint that $$\sum w(i) = 1$$

From here on use the method of undetermined multipliers.