2.1.3.md

Solution to Review Question

by Qiang Gao, updated at Sep 17, 2017

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Chapter 2 Large-Sample Theory

Section 1 Review of Limit Theorems for Sequences of Random Variables

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Review Question 2.1.3

Prove Lemma 2.4(c)

$$ \mathbf{x}_n \to_d \mathbf{x} \text{, } \mathbf{A}_n \to_p \mathbf{A} \implies \mathbf{A}_n \mathbf{x}_n \to_d \mathbf{A} \mathbf{x} $$

from Lemma 2.4(a)

$$ \mathbf{x}_n \to_d \mathbf{x} \text{, } \mathbf{y}_n \to_p \boldsymbol\alpha \implies \mathbf{x}_n + \mathbf{y}_n \to_d \mathbf{x} + \boldsymbol\alpha $$

and Lemma 2.4(b)

$$ \mathbf{x}_n \to_d \mathbf{x} \text{, } \mathbf{y}_n \to_p \mathbf{0} \implies \mathbf{y}_n' \mathbf{x}_n \to_p \mathbf{0}. $$

Hint: $$ \mathbf{A}_n \mathbf{x}_n = ( \mathbf{A}_n - \mathbf{A} ) \mathbf{x}_n + \mathbf{A} \mathbf{x}_n $$. By (b), $$ ( \mathbf{A}_n - \mathbf{A} ) \mathbf{x}_n \to_p \mathbf{0} $$.

Solution

Using the add-and-subtract strategy,

$$ \mathbf{A}_n \mathbf{x}_n = ( \mathbf{A}_n - \mathbf{A} ) \mathbf{x}_n + \mathbf{A} \mathbf{x}_n. \tag{1} $$

Because $$ \mathbf{A}_n \to_p \mathbf{A} $$, we have $$ \mathbf{A}_n - \mathbf{A} \to_p \mathbf{0} $$. Using $$ \mathbf{x}_n \to_d \mathbf{x} $$ and Lemma 2.4(b),

$$ ( \mathbf{A}_n - \mathbf{A} ) \mathbf{x}_n \to_p \mathbf{0}. \tag{2} $$

Because $$ \mathbf{x}_n \to_d \mathbf{x} $$, by Lemma 2.3(b),

$$ \mathbf{A} \mathbf{x}_n \to_d \mathbf{A} \mathbf{x}. \tag{3} $$

Combining (2) and (3) into (1), using Lemma 2.4(a),

$$ \mathbf{A}_n \mathbf{x}_n \to_d \mathbf{A} \mathbf{x}. $$

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Copyright ©2017 by Qiang Gao