2:29 p.m.
hello everybody, i would like to know whetere there is a way to eliminate single
observations and outliers so that you can fix heteroskedasticity problems
Thanks!
2:47 p.m.
Am 21.04.2020 um 20:29 schrieb robdans2(a)gmail.com:
hello everybody, i would like to know whetere there is a way to
eliminate single observations and outliers so that you can fix heteroskedasticity
problems
Thanks!
Leaving aside the question of whether that would be a proper "fix", you
could add an impulse dummy to your equation for that observation. You
could either slightly abuse the menu entry Add/observation range dummy
(or your translated equivalent) for that, or you could type something like:
series impulsed = (t == 50) # cross-section
or
series impulsed = (t == 2000:1) # time series
cheers
sven
3:54 p.m.
Thanks for the quick answer, what do you think would be a better fix? Changing the
variable (for example making it a log)?
6:32 a.m.
Am 21.04.2020 um 21:54 schrieb robdans2(a)gmail.com:
Thanks for the quick answer, what do you think would be a better fix?
Changing the variable (for example making it a log)?
All I'm saying is that we're talking about an outlier that is going to
be removed. I know that this affects the test outcome of a diagnostic
test for heteroskedasticity, but it is still a different topic, most of
the time.
cheers
sven
9:22 a.m.
On Wed, 22 Apr 2020 at 11:32, Sven Schreiber <svetosch(a)gmx.net> wrote:
Am 21.04.2020 um 21:54 schrieb robdans2(a)gmail.com:
> Thanks for the quick answer, what do you think would be a better fix?
Changing the variable (for example making it a log)?
All I'm saying is that we're talking about an outlier that is going to
be removed. I know that this affects the test outcome of a diagnostic
test for heteroskedasticity, but it is still a different topic, most of
the time.
cheers
sven
Failure of your heteroskedasticity test can be regarded as indicating
misspecification. Have you done any other misspecification tests (e.g.
autocorrelation, functional form, recursive residuals, etc.) and did any of
these indicate misspecification? Is your outlier the result of some policy
change, innovation, change in method of compilation of your time series, or
similar. If so, you can use a point, or step dummy or another intervention
variable to improve your model. Even if you have heteroscedasticity your
coefficient estimates are still consistent and you can use a robust
estimate of their standard deviations. What you do depends on your
economic model, your data, and the aim of your analysis.
When Sven says 'whether that would be a proper "fix" ' I would think
that
he was referring to considerations such as the above.
John C Frain
3 Aranleigh Park
Rathfarnham
Dublin 14
Ireland
www.tcd.ie/Economics/staff/frainj/home.html
mailto:frainj@tcd.ie
mailto:frainj@gmail.com
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4:35 p.m.
On Wed, 22 Apr 2020, John C Frain wrote:
On Wed, 22 Apr 2020 at 11:32, Sven Schreiber <svetosch(a)gmx.net>
wrote:
> Am 21.04.2020 um 21:54 schrieb robdans2(a)gmail.com:
>> Thanks for the quick answer, what do you think would be a better fix?
> Changing the variable (for example making it a log)?
>
> All I'm saying is that we're talking about an outlier that is going to
> be removed. I know that this affects the test outcome of a diagnostic
> test for heteroskedasticity, but it is still a different topic, most of
> the time.
Failure of your heteroskedasticity test can be regarded as indicating
misspecification. [...]
True indeed. I have one more observation to throw in.
In his introductory economics textbook, Wooldridge has a house-price
example which he treats as a poster-case for heteroskedasticity, its
detection and treatment. I was messing around with the dataset in
question (scatter plots and so on) and noticed that it contained one
serious outlier: a house with a huge lotsize but very low price. In
any regression of price on a list of regressors containing lotsize,
the standard tests for heteroskedasticity would light up like
Christmas trees. But remove that one observation and everything was
fine and dandy. This was actually a poster-case for outlier
detection.
So this is a possibility worth considering when heteroskedasticity
is flagged by "mechanical" means. Of course, in gretl you have
direct means of detecting outliers (see the "leverage" command).
Allin
5:32 p.m.
On Wed, 22 Apr 2020, Allin Cottrell wrote:
On Wed, 22 Apr 2020, John C Frain wrote:
> On Wed, 22 Apr 2020 at 11:32, Sven Schreiber <svetosch(a)gmx.net> wrote:
>
>> Am 21.04.2020 um 21:54 schrieb robdans2(a)gmail.com:
>>> Thanks for the quick answer, what do you think would be a better fix?
>> Changing the variable (for example making it a log)?
>>
>> All I'm saying is that we're talking about an outlier that is going to
>> be removed. I know that this affects the test outcome of a diagnostic
>> test for heteroskedasticity, but it is still a different topic, most of
>> the time.
>
> Failure of your heteroskedasticity test can be regarded as indicating
> misspecification. [...]
True indeed. I have one more observation to throw in.
In his introductory economics textbook, Wooldridge has a house-price example
which he treats as a poster-case for heteroskedasticity, its detection and
treatment. I was messing around with the dataset in question (scatter plots
and so on) and noticed that it contained one serious outlier: a house with a
huge lotsize but very low price. In any regression of price on a list of
regressors containing lotsize, the standard tests for heteroskedasticity
would light up like Christmas trees. But remove that one observation and
everything was fine and dandy. This was actually a poster-case for outlier
detection.
While we're at it, let me chime in with a pet peeve of mine. There's no
such a thing as heteroskedasticity "in the data"; heteroskedasticity is,
possibly, "in the model". Example: consider a variable y such that
y_i = b * x_i + e_i * z_i
where b is a constant parameter, e_i is white noise with variance s^2 and
x_i and z_i are observable and independent of each other. Of course the
regression of y on x (aka the conditional expectation of y given x) is (i)
linear and (ii) homoskedastic, since v(y|x) = v(e*z | x) = s^2 * v(z) (a
constant). But if you consider z as an explanatory variable, you move
to a model of y | x,z; the regression stays the same (E(y | x,z) = b*x),
but the model becomes heteroskedastic, since V(y | x,y) = s^2 * z_i.
-------------------------------------------------------
Riccardo (Jack) Lucchetti
Dipartimento di Scienze Economiche e Sociali (DiSES)
Università Politecnica delle Marche
(formerly known as Università di Ancona)
r.lucchetti(a)univpm.it
http://www2.econ.univpm.it/servizi/hpp/lucchetti
-------------------------------------------------------
1556
days inactive
1557
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6 comments
5 participants
participants (5)
-
Allin Cottrell -
John C Frain -
Riccardo (Jack) Lucchetti -
robdans2＠gmail.com -
Sven Schreiber