OLS vs WLS in a Heteroskedastic Trending Regression

Question

Suppose we have the following heteroskedastic trending regression model: $$y_i=bi+a_i u_i$$ for some sequence of non-zero constants $a_i$ and $u_i$ an i.i.d. sequence of mean 0 and variance 1 random variables. I want to find the asymptotic distributions of the OLS and unfeasible WLS estimators and compare their efficiency.

To do this, I calculate the normalisation constants for both asymptotic distributions. I use the Lindeberg-Feller CLT and assume the relevant Grenander conditions hold. I calculate that for $\hat{b}_{OLS}$ and $\hat{b}_{WLS}$, $$\left(\sum_{i=1}^n i^2\right)\left(\sum_{i=1}^n a_i^2 i^2\right)^{-1/2}(\hat{b}_{OLS}-b)\xrightarrow{d} N(0,1)$$ $$\left(\sum_{i=1}^n i^2\right)\left(\sum_{i=1}^n \frac{i^2}{a_i^2} \right)^{-1/2}(\hat{b}_{WLS}-b)\xrightarrow{d} N(0,1)$$

However, I think this must be incorrect as I cannot see why the WLS is necessarily more efficient than the OLS estimator. I think there are cases when either normalisation is bigger, implying that either estimator could be efficient in different cases. For example, I think these imply that if $a_i>1$ for all $i$ then the WLS is more efficient, but if $a_i<1$ for all $i$ then the OLS is more efficient.

Am I doing something wrong in my calculations? I expect the WLS to be more efficient but can't see where the error is coming into my work.

Answer 1

I realised I made a mistake in my derivation of the normalisation for the WLS estimator. The right asymptotic should be $$\left(\sum_{i=1}^n \frac{i^2}{a_i^2}\right)^{1/2}(\hat{b}_{WLS}-b)\xrightarrow{d} N(0,1)$$

Then, since our OLS asymptotic is $$\left(\sum_{i=1}^n i^2\right)\left(\sum_{i=1}^n i^2a_i^2\right)^{-1/2}(\hat{b}_{OLS}-b)\xrightarrow{d} N(0,1)$$

WLS is more efficient when $$\left(\sum_{i=1}^n \frac{i^2}{a_i^2}\right)^{1/2}\geq \left(\sum_{i=1}^n i^2\right)\left(\sum_{i=1}^n i^2a_i^2\right)^{-1/2}$$ And this always holds by the Cauchy-Schwarz inequality since it states that $$\left(\sum_{i=1}^n i^2 \right) \leq \left(\sum_{i=1}^n i^2a_i^2\right)^{1/2}\left(\sum_{i=1}^n \frac{i^2}{a_i^2}\right)^{1/2}$$ So this shows that the WLS estimator is more efficient whenever its Grenander condition is satisfied.