Asymptotically unbiased & consistent estimators

[kingwinner]

Theorem: If "θ hat" is an unbiased estimator for θ AND Var(θ hat)->0 as n->∞, then it is a consistent estimator of θ.

The textbook proved this theorem using Chebyshev's Inequality and Squeeze Theorem and I understand the proof.
BUT then there is a remark that we can replace "unbiased" by "asymptotically unbiased" in the above theorem, and the result will still hold, but the textbook provided no proof. This is where I'm having a lot of trouble. I really don't see how we can prove this (i.e. asymptotically unbiased and variance->0 implies consistent). I tried to modify the original proof, but no way I can get it to work under the assumption of asymptotically unbiased.

I'm frustrated and I hope someone can explain how to prove it. Thank you!

 

[micromass]

Hi kingwinner!

What about the following adjustment:

P(|\hat{\theta}_n-\theta_0|\geq \varepsilon)\leq P(|\hat{\theta}_n-E(\hat{\theta}_n)|+|E(\hat{\theta}_n)-\theta_0|\geq \varepsilon)\leq \frac{Var(\hat{\theta}_n)}{(\varepsilon-|E(\hat{\theta}_n)-\theta_0|)^2}\rightarrow 0

 

[kingwinner]

Hi kingwinner!

What about the following adjustment:

P(|\hat{\theta}_n-\theta_0|\geq \varepsilon)\leq P(|\hat{\theta}_n-E(\hat{\theta}_n)|+|E(\hat{\theta}_n)-\theta_0|\geq \varepsilon)\leq \frac{Var(\hat{\theta}_n)}{(\varepsilon-|E(\hat{\theta}_n)-\theta_0|)^2}\rightarrow 0

Thanks for the help, but one of the assumptions of Chebyshev's inequality requires \varepsilon-|E(\hat{\theta}_n)-\theta_0|>0 which is not necessarily true here?

 

[micromass]

Thanks for the help, but one of the assumptions of Chebyshev's inequality requires \varepsilon-|E(\hat{\theta}_n)-\theta_0|>0 which is not necessarily true here?

It's not necessarily true, but it is true for large n. We know that

E(\hat{\theta}_n)\rightarrow \theta_0

So from a certain n₀, we know that

|E(\hat{\theta}_n)-\theta_0|<\varepsilon

So from that certain n₀, we know that

\varepsilon-|E(\hat{\theta}_n)-\theta_0|>0

 

[kingwinner]

Thanks for the help! :) You're a legend...