Solution to Review Question by Qiang Gao, updated at Sep 17, 2017 Chapter 2 Large-Sample Theory Section 1 Review of Limit Theorems for Sequences of Random Variables ... Review Question 2.1.4 Suppose $$ \sqrt{n} ( \hat\theta_n - \theta ) \to_d N(0, \sigma^2) $$. Does it follow that $$ \hat\theta_n \to_p \theta $$? Hint: $$ \hat\theta_n - \theta = \frac{1}{\sqrt{n}} \cdot \sqrt{n} ( \hat\theta_n - \theta ) \text{, } \operatorname*{plim}_{n \to \infty} \frac{1}{\sqrt{n}} = 0. $$ Solution Using the multiply-and-divide strategy, $$ \hat\theta_n - \theta = \frac{1}{\sqrt{n}} \cdot \sqrt{n} ( \hat\theta_n - \theta ). \tag{1} $$ Because $$ 1 / \sqrt{n} \to 0 $$, as is shown in Review Question 2.1.1, $$ \frac{1}{\sqrt{n}} \to_p 0. \tag{2} $$ Because $$ \sqrt{n} ( \hat\theta_n - \theta ) \to_d N(0, \sigma^2) $$, combining (2) into (1) and Lemma 2.4(b), $$ \begin{gather} \hat\theta_n - \theta \to_p 0, \ \hat\theta_n \to_p \theta. \end{gather} $$ Copyright ©2017 by Qiang Gao