If $\hat{\theta_1}$ is consistent, and $\hat{\theta_1}-\hat{\theta_2}$ converge in probability, is $\hat{\theta_2}$ also consistent?

Question

I'm trying to describe the consistency of $\hat{\theta_2}$. It is given that $\hat{\theta_1}$ ~ AN($\hat{\theta_0}$, A1/n). From this, we know that $\hat{\theta_1}$ is a consistent estimator of $\hat{\theta_0}$ (by Chebychev's Inequality).

What can I say about the consistency of $\hat{\theta_2}$ if $\hat{\theta_1}-\hat{\theta_2} \rightarrow 0$ ?

Intuitively, it seems that $\hat{\theta_2}$ would also be consistent, but I am not sure how to show this.

Thank you in advance!

Answer 1

Your intuition is correct. Note that $\hat{\theta_2} = \hat{\theta_1} + (\hat{\theta_2} - \hat{\theta_1})$. Now, you can apply Slutzky's inequality because you know $\hat{\theta_1} \to \hat{\theta_0}$ and $(\hat{\theta_2} - \hat{\theta_1}) \to 0$.