# Weighted least squares estimator - proof of weights

Let $$\{X_i : i = 1, 2, \dots ,n\}$$ be independent random variables with finite second moments. They are not necessarily identically distributed. They do have the same mean,$$\mu=E(X_i)$$ for all i, but possibly different variances, $$\sigma^2_i > 0$$. Assume these variances are known. Let $$p$$ be the estimator that solves the problem $$\min \Sigma[(X_i-m)^2/\sigma^2_i]$$ with respect to $$m$$. Each squared term is weighted by the inverse of the variance. Show that $$p = \Sigma(w_i X_i)$$ for weights $$w_i > 0$$ such that $$\Sigma w_i = 1$$.

I know we have must find the weights. I get the weights as follows: $$w_i = {(\Sigma[1/\sigma^2_i])}^{-1} / \sigma^2_i$$.

Is this right?

Thanks!

• Please see stats.stackexchange.com/questions/243922/…. – whuber Nov 7 '18 at 20:00
• I did see. While it is similar, the other question relates to estimate of variance. I just want to know whether my answer is right – Aishwarya Deore Nov 8 '18 at 14:54
• Also..,how would you prove that these weights are optimal. I know we have to set up the langrangian and differentiate wrt to each wi. But I am stuck at the simplification part – Aishwarya Deore Nov 9 '18 at 15:32
• You can use elementary inequalities to demonstrate optimality, or even Euclidean geometry. I give the geometric argument at stats.stackexchange.com/a/9073/919. – whuber Nov 9 '18 at 15:46