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st: how to do endogeniety test with two suspect endogenous variables


From   gjhxmu@sina.com
To   statalist<statalist@hsphsun2.harvard.edu>
Subject   st: how to do endogeniety test with two suspect endogenous variables
Date   Thu, 21 May 2009 22:47:26 +0800

<style id="sinamailpaperfilter">.sinamailpaper-0{cursor:text;}.sinamailpaper-0 td,.sinamailpaper-0 textarea,.sinamailpaper-0 input,.sinamailpaper-0 br,.sinamailpaper-0 div,.sinamailpaper-0 span{font-size:14px;font-family:"宋体",Verdana,Arial,Helvetica,sans-serif;line-height:1.5;}.sinamailpaper-0 p{*margin:0.2em auto;}.sinamailpaper-0 img{border:0;}.sinamailpaper-0 pre{white-space:normal;}.sinamailpaper-0 form{margin:0;}</style>Dear&nbsp;all,<BR>I&nbsp;encountered&nbsp;a&nbsp;problem&nbsp;with&nbsp;endogeniety.Concretely,&nbsp;I&nbsp;have&nbsp;a&nbsp;equation&nbsp;below:<BR>y=al+a2x1+a3x2+a4z+e<BR>where&nbsp;z&nbsp;is&nbsp;exogenous&nbsp;definitely.However,&nbsp;I&nbsp;suspect&nbsp;whether&nbsp;x1&nbsp;and&nbsp;x2&nbsp;are&nbsp;endogenous.<BR>and&nbsp;there&nbsp;are&nbsp;two&nbsp;instrument&nbsp;variables,&nbsp;z1&nbsp;for&nbsp;x1&nbsp;and&nbsp;z2&nbsp;for&nbsp;x2.<BR><BR>In&nbsp;wooldridge&nbsp;book,&nbsp;there&nbsp;is&nbsp;an&nbsp;introduction&nbsp;about&nbsp;how&nbsp;to&nb!
 sp;test&nbsp;whether&nbsp;one&nbsp;variable&nbsp;is&nbsp;endogenous.i.e,hausman&nbsp;test.<BR>First&nbsp;get&nbsp;the&nbsp;residual&nbsp;and&nbsp;then&nbsp;add&nbsp;the&nbsp;residual&nbsp;to&nbsp;the&nbsp;equation.If&nbsp;the&nbsp;p&nbsp;value&nbsp;of&nbsp;residual&nbsp;is&nbsp;significant,&nbsp;the&nbsp;suspect&nbsp;variable&nbsp;is&nbsp;endogenous.<BR><BR>However,&nbsp;how&nbsp;to&nbsp;test&nbsp;with&nbsp;two&nbsp;suspect&nbsp;endogenous&nbsp;variables?&nbsp;What&nbsp;confuses&nbsp;me&nbsp;is&nbsp;how&nbsp;to&nbsp;treat&nbsp;the&nbsp;other&nbsp;suspect&nbsp;endogenous&nbsp;variable&nbsp;when&nbsp;testing&nbsp;one&nbsp;endogenous&nbsp;varible.Treat&nbsp;it&nbsp;exogenous&nbsp;or&nbsp;endogenous?&nbsp;BecausTake&nbsp;the&nbsp;above&nbsp;equation&nbsp;for&nbsp;example,&nbsp;maybe&nbsp;I&nbsp;can&nbsp;compute&nbsp;joint&nbsp;F&nbsp;statistic&nbsp;of&nbsp;the&nbsp;two&nbsp;residuals.&nbsp;If&nbsp;the&nbsp;p&nbsp;value&nbsp;of&nbsp;residuals&nbsp;is&nbsp;not&nbsp;significant,&n!
 bsp;I&nbsp;can&nbsp;conclude&nbsp;both&nbsp;of&nbsp;x1&nbsp;and&nbsp;x

2
&nbsp;are&nbsp;exogenous.&nbsp;However,&nbsp;if&nbsp;the&nbsp;p&nbsp;value&nbsp;is&nbsp;significant,&nbsp;it&nbsp;shows&nbsp;that&nbsp;at&nbsp;least&nbsp;one&nbsp;of&nbsp;the&nbsp;two&nbsp;variables&nbsp;are&nbsp;endogenous.&nbsp;Then&nbsp;next&nbsp;what&nbsp;should&nbsp;I&nbsp;do?&nbsp;I&nbsp;want&nbsp;to&nbsp;judge&nbsp;the&nbsp;endogeniety&nbsp;for&nbsp;each&nbsp;of&nbsp;the&nbsp;two.&nbsp;Can&nbsp;I&nbsp;just&nbsp;take&nbsp;the&nbsp;residual&nbsp;of&nbsp;each&nbsp;of&nbsp;x1&nbsp;and&nbsp;x2&nbsp;for&nbsp;judgement?&nbsp;(I&nbsp;note&nbsp;the&nbsp;results&nbsp;are&nbsp;not&nbsp;equal&nbsp;to&nbsp;ivendog&nbsp;x1&nbsp;or&nbsp;ivendog&nbsp;x2&nbsp;in&nbsp;stata.)<BR><BR>By&nbsp;the&nbsp;way,&nbsp;in&nbsp;stata&nbsp;-ivendog-&nbsp;can&nbsp;give&nbsp;the&nbsp;result&nbsp;of&nbsp;endogenous&nbsp;test&nbsp;after&nbsp;-ivreg2-&nbsp;and&nbsp;-vreg-.&nbsp;<BR>I&nbsp;type&nbsp;ivreg2&nbsp;y&nbsp;z&nbsp;(x1&nbsp;x2=z1&nbsp;z2&nbsp;z);<BR>then&nbsp;type&nbsp;ivendog;<BR>the&nbsp;res!
 ult&nbsp;of&nbsp;the&nbsp;joint&nbsp;F&nbsp;statistic&nbsp;is&nbsp;equal&nbsp;to&nbsp;the&nbsp;"residual&nbsp;method"&nbsp;above&nbsp;in&nbsp;which&nbsp;I&nbsp;compute&nbsp;manually.<BR>then&nbsp;type&nbsp;ivendog&nbsp;x1&nbsp;or&nbsp;ivendog&nbsp;x2;<BR>I&nbsp;get&nbsp;some&nbsp;results&nbsp;of&nbsp;p&nbsp;values&nbsp;for&nbsp;x1&nbsp;or&nbsp;x2,&nbsp;but&nbsp;I&nbsp;do&nbsp;not&nbsp;know&nbsp;how&nbsp;stata&nbsp;get&nbsp;them,&nbsp;I&nbsp;can&nbsp;not&nbsp;get&nbsp;the&nbsp;statistics&nbsp;manually.&nbsp;<BR><BR>could&nbsp;anyone&nbsp;tell&nbsp;me&nbsp;how&nbsp;to&nbsp;do&nbsp;endogeniety&nbsp;test&nbsp;with&nbsp;two&nbsp;suspect&nbsp;endogenous&nbsp;variables&nbsp;manually&nbsp;or&nbsp;by&nbsp;stata?<BR><BR>Hope&nbsp;sincerely&nbsp;replies.&nbsp;Thank&nbsp;you&nbsp;in&nbsp;advance.<BR><BR>Rose&nbsp;&nbsp; 

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