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Re: st: theory reg vs. qreg
From
Kasal Roman <[email protected]>
To
[email protected]
Subject
Re: st: theory reg vs. qreg
Date
Mon, 29 Apr 2013 13:30:07 +0200
but do you agree with:
if "reg" gives me the general results:
A) Y1=b1
B) Y2=a2*x+b2
on the same data
then must be:
Y2=mean(x)*x+b2=b1=Y1 ?
Is the assumption clear now?
thx
On Mon, Apr 29, 2013 at 1:21 PM, Nick Cox <[email protected]> wrote:
> I see no evidence here for that extraordinary claim.
>
> Sorry, I am still lost on what you want.
> Nick
> [email protected]
>
>
> On 29 April 2013 12:17, Kasal Roman <[email protected]> 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 <[email protected]> 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
>>> [email protected]
>>>
>>>
>>> On 29 April 2013 11:23, Kasal Roman <[email protected]> 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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