Showing weighted average is consistent estimate

Question

Here's the problem statement: Let $X_1$, . . . , $X_n$ be independent random variables with common mean $\mu$ and variances $σ_i^2$ . To estimate $μ$, we use the weighted average $T_n$ = $\sum_{i=1}^n w_iX_i$

with weights $w_i$ = $\frac{\sigma_i^{−2}}{\sum_{i=1}^n \sigma_j^{-2}}$

Show that the estimate $T_n$ is consistent (in probability) if $\sum_{i=1}^n \sigma_j^{-2} \Rightarrow \infty$. End Problem.

So I know that a consistent estimator is one that asymptotically approaches the parameter as the sample size goes to infinity. So by the problem statement, if the denominator of each of the weightings goes to infinity, then each of the $w_iX_i$ would just go to 0, yes? I'm a little confused on what this problem is trying to show. Thanks for any help!

Answer 1

Yes, $w_{i}X_{i}$ go to 0, but you are interested in the behaviour of $T_{n}$ . Convergence in probability means .

\begin{equation} \lim_{n \rightarrow \infty }P(|T_{n} - \mu | \ge \epsilon) = 0 , \forall \epsilon > 0 \end{equation}

Try using some concentration inequality involving the variance .

Answer 2

Here are some steps:

1. Without loss of generality, $\mu=0$.
2. Compute $\operatorname{Var}\left(T_n\right)$. Using independence this reduces to $\sum_{i=1}^n\operatorname{Var}\left(\omega_iX_i\right)$. Since $$\operatorname{Var}\left(\omega_iX_i\right)=\omega_i^2\sigma_i^2,$$ it follows that $$ \operatorname{Var}\left(T_n\right)=\sum_{i=1}^n\left(\frac{\sigma_i^{-2}}{\sum_{j=1}^n\sigma_j^{-2}}\right)^2\sigma_i^2=\sum_{i=1}^n\sigma_i^{-2}\frac 1{\left(\sum_{j=1}^n\sigma_j^{-2}\right)^2}=\frac{1}{\sum_{j=1}^n\sigma_j^{-2}}. $$
3. Since $\mathbb E\left[T_n\right]=\mu$, we get that $\mathbb E\left[\left(T_n-\mu\right)^2\right]\to 0$ from which the convergence in probability follows.