2.1.5.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 2.1.5 (Combine Delta method with Lindeberg-Levy)

Let $$ { z_i } $$ be a sequence of i.i.d. (independently and identically distributed) random variables with $$ \operatorname{E} ( z_i ) = \mu \neq 0 $$ and $$ \operatorname{Var} ( z_i ) = \sigma^2 $$, and let $$ \bar{z}_n $$ be the sample mean. Show that

$$ \sqrt{n} \left( \frac{1}{\bar{z}_n} - \frac{1}{\mu} \right) \to_d N \left( 0, \frac{ \sigma^2 }{ \mu^4 } \right). $$

Hint: In Lemma 2.5, set $$ \boldsymbol{\beta} = \mu $$, $$ \mathbf{a} ( \boldsymbol{\beta} ) = 1 / \mu $$, $$ \mathbf{x}_n = \bar{z}_n $$.

Solution

By Linderberg-Levy CLT,

$$ \sqrt{n} ( \bar{z}_n - \mu ) \to_d N(0, \sigma^2). $$

Set $$ \boldsymbol{\beta} = \mu $$, $$ \mathbf{a} ( \boldsymbol{\beta} ) = 1 / \mu $$, $$ \mathbf{x}_n = \bar{z}_n $$,

$$ \mathbf{A} ( \boldsymbol{ \beta } ) =

• \frac{ 1 }{ \mu^2 }, $$

using Lemma 2.5,

$$ \begin{align} \sqrt{n} \left( \frac{ 1 }{ \bar{z}_n } - \frac{ 1 }{ \mu } \right) & = \sqrt{n} [ \mathbf{a} ( \mathbf{x}_n ) - \mathbf{a} ( \boldsymbol{ \beta } ) ] \\ & \to_d N \left( 0, \left( - \frac{1}{ \mu^2 } \right) \sigma^2 \left( - \frac{1}{ \mu^2 } \right) \right) \\ & = N \left( 0, \frac{ \sigma^2 } { \mu^4 } \right) \end{align} $$

Appendix

Lemma 2.5 (the “delta method”): Suppose $$ \mathbf{x}_n $$ is a sequence of $$K$$-dimensional random vectors such that $$ \mathbf{x}_n \to_p \boldsymbol{\beta} $$ and

$$ \sqrt{n} ( \mathbf{x}_n - \boldsymbol{\beta} ) \to_d \mathbf{z}, $$

and suppose $$ \mathbf{a} (\cdot): \mathbb{R}^K \to \mathbb{R}^r $$ has continuous first derivatives with $$ \mathbf{A} ( \boldsymbol{\beta} ) $$ denoting the $$ r \times K $$ matrix of first derivatives evaluated at $$ \boldsymbol{\beta} $$:

$$ \underset{ ( r \times K ) }{ \mathbf{A} ( \boldsymbol{\beta} ) } \equiv \frac{ \partial \mathbf{a} ( \boldsymbol{ \beta } ) } { \partial \boldsymbol{ \beta }' }. $$

Then

$$ \sqrt{n} [ \mathbf{a} ( \mathbf{x}_n ) - \mathbf{a} ( \boldsymbol{ \beta }) ] \to_d \mathbf{A} ( \boldsymbol{ \beta } ) \mathbf{z}. $$

In particular:

$$ \sqrt{n} ( \mathbf{x}_n - \boldsymbol{ \beta } ) \to_d N( \mathbf{0}, \boldsymbol{\Sigma} ) \implies \sqrt{n} [ \mathbf{a} ( \mathbf{x}_n ) - \mathbf{a} ( \boldsymbol{ \beta } ) ] \to_d N( \mathbf{0}, \mathbf{A} ( \boldsymbol{ \beta } ) \boldsymbol{ \Sigma } \mathbf{A} ( \boldsymbol{ \beta } )' ). $$

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Copyright ©2017 by Qiang Gao