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Re: st: theory reg vs. qreg


From   Nick Cox <njcoxstata@gmail.com>
To   "statalist@hsphsun2.harvard.edu" <statalist@hsphsun2.harvard.edu>
Subject   Re: st: theory reg vs. qreg
Date   Mon, 29 Apr 2013 12:21:32 +0100

I see no evidence here for that extraordinary claim.

Sorry, I am still lost on what you want.
Nick
njcoxstata@gmail.com


On 29 April 2013 12:17, Kasal Roman <kasalroman@gmail.com> wrote:
> of course I don't mean this, I know a2!=mean(x) but IF you put
> a2=mean(x) you get the first method result with the "reg" command :)
>
> On Mon, Apr 29, 2013 at 12:39 PM, Nick Cox <njcoxstata@gmail.com> wrote:
>>
>> What you want is not at all clear. In fact, you seem confused about some basics.
>>
>>  For example, if you fit a regression
>>
>> Y2=a2*x+b2
>>
>> it's not true that a2 = mean(x).
>>
>> What do you mean by "deactivate" a coefficient?
>>
>> -qreg y-
>>
>> works; it's just a round-about way of calculating the median of -y-,
>> which you can do directly with -summarize-.
>>
>> Nick
>> njcoxstata@gmail.com
>>
>>
>> On 29 April 2013 11:23, Kasal Roman <kasalroman@gmail.com> wrote:
>> > Please,
>> >
>> >
>> > the "reg" command gives me these results:
>> >
>> > A) Y1=b1
>> >
>> > B) Y2=a2*x+b2
>> >
>> > then
>> >
>> > Y2=mean(x)*x+b2=b1=Y1
>> >
>> >
>> > How is it with "qreg"? How could I deactivate the "a2" coefficient
>> > without a need to estimate two functions with the "qreg" command? How
>> > to get the result of "b1" just with the B) method through the "qreg"
>> > command? I know neither mean(x) nor median(x) fit.
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