9:16 p.m.
Dear forum,
I am struggling with my variable lists once again. I expected the following two ML
estimations to give the same results, but they don't. Any ideas?
#estimation 1
scalar c_ = 0.1
scalar rng = 0.1
scalar err = 0.2
mle ll = -ln(lambda) - sqrtPark/lambda
series lambda = mean(sqrtPark)
series lambda = c_ + rng * sqrtPark(-1) + err * lambda(-1)
params c_ rng err
end mle --robust
#estimation 2
list xDepVars = sqrtPark
list xExpVars = sqrtPark(-1)
c_ = 0.1
rng = 0.1
err = 0.2
series xDepVar = xDepVars[1]
series xExpVar = xExpVars[1]
mle ll2 = -ln(lambda2) - xDepVar/lambda2
series lambda2 = mean(xDepVar)
series lambda2 = c_ + rng * xExpVar + err * lambda2(-1)
params c_ rng err
end mle --robust
Many thanks in advance,
Daniel
1:21 a.m.
On Tue, 6 Jan 2015, Daniel Bencik wrote:
Dear forum,
I am struggling with my variable lists once again. I expected the following
two ML estimations to give the same results, but they don't. Any ideas?
#estimation 1
scalar c_ = 0.1
scalar rng = 0.1
scalar err = 0.2
mle ll = -ln(lambda) - sqrtPark/lambda
series lambda = mean(sqrtPark)
series lambda = c_ + rng * sqrtPark(-1) + err * lambda(-1)
params c_ rng err
end mle --robust
#estimation 2
list xDepVars = sqrtPark
list xExpVars = sqrtPark(-1)
c_ = 0.1
rng = 0.1
err = 0.2
series xDepVar = xDepVars[1]
series xExpVar = xExpVars[1]
mle ll2 = -ln(lambda2) - xDepVar/lambda2
series lambda2 = mean(xDepVar)
series lambda2 = c_ + rng * xExpVar + err * lambda2(-1)
params c_ rng err
end mle --robust
Many thanks in advance,
Thank you for posting. The facility of "indexing into" a list is
fairly new, and although it is now mentioned in the documentation it
is not adequately explained.
Let's start from basics. In the gretl main window you can see that
each series (whose name appears in the second column from the left)
is associated with an index number (in the leftmost column). Now,
internally, a gretl "list" is just an array of integers, each
integer being the index number of a series. So when you "index into"
a list, what you get back is an integer index number.
Suppose we have lists Y and X, each containing at least one series.
Then if we do, for example,
ols Y[1] 0 X[1]
this will work OK. The expressions "Y[1]" and "X[1]" will each yield
a series index number, and such index numbers are valid input for
the "ols" command. However, now suppose we try
series x1 = X[1]
You might think this would give us a series, x1, whose values are
those of the first member of the list named "X", but that's not so.
Let's say the first member of list X is the series with index number
3; in that case the statement above is equivalent to
series x1 = 3
That is, x1 will be a series with constant value 3.0. This explains
the results you got via mle.
So how can you get what you want, namely the actual values of the
series at position i in a list? For that you need a "foreach" loop
across the list, using "$i" to get the name of the series in
question, as in
loop foreach i List
series x = $i
print x
endloop
Allin Cottrell
5:51 p.m.
On Tue, 6 Jan 2015, Daniel Bencik wrote:
I am struggling with my variable lists once again. I expected the
following
two ML estimations to give the same results, but they don't. Any ideas?
#estimation 1
scalar c_ = 0.1
scalar rng = 0.1
scalar err = 0.2
mle ll = -ln(lambda) - sqrtPark/lambda
series lambda = mean(sqrtPark)
series lambda = c_ + rng * sqrtPark(-1) + err * lambda(-1)
params c_ rng err
end mle --robust
#estimation 2
list xDepVars = sqrtPark
list xExpVars = sqrtPark(-1)
c_ = 0.1
rng = 0.1
err = 0.2
series xDepVar = xDepVars[1]
series xExpVar = xExpVars[1]
mle ll2 = -ln(lambda2) - xDepVar/lambda2
series lambda2 = mean(xDepVar)
series lambda2 = c_ + rng * xExpVar + err * lambda2(-1)
params c_ rng err
end mle --robust
Sorry, I realize that the answer I gave yesterday in
http://lists.wfu.edu/pipermail/gretl-users/2015-January/010564.html
will not do!
My explanation was that an expression such as "xDepVars[1]" gives the
integer ID number of the series at position 1 in the named list, not
the series itself. Well, that was true, but it's not consistent with
the documentation I added recently on indexing into a list, and it's
surely not what users will expect.
So today I've modified the result of indexing into a list to match the
doc and the most likely expectation. You now get a series, and your ML
script should behave as you intended. That's in CVS and snapshots.
Note that if you wish to retrieve the ID number of the series at a
certain position in a list this can be done by converting the list
into a matrix. The following little script illustrates this point as
well as the effect of indexing into a list directly.
<hansl>
open data4-1
list L = sqft price
series x = L[1]
print sqft x --byobs
ols L[2] 0 L[1]
matrix m = L
printf "The series at position 1 in L has ID %d\n", m[1]
</hansl>
Allin
6:02 p.m.
-gretl- 1.9.92, Kubuntu 14.10 Plasma 4
On 7 January 2015 at 17:51, Allin Cottrell <cottrell(a)wfu.edu> wrote:
[...]
<hansl>
open data4-1
list L = sqft price
series x = L[1]
print sqft x --byobs
ols L[2] 0 L[1]
matrix m = L
printf "The series at position 1 in L has ID %d\n", m[1]
</hansl>
[...]
Produces an error for me when I run it:
? ols L[2] 0 L[1]
'L[2]' is not the name of a variable
Error executing script: halting
--
Clive Nicholas
"My colleagues in the social sciences talk a great deal about methodology.
I prefer to call it style." -- Freeman J. Dyson
6:12 p.m.
On Wed, 7 Jan 2015, Clive Nicholas wrote:
-gretl- 1.9.92, Kubuntu 14.10 Plasma 4
On 7 January 2015 at 17:51, Allin Cottrell <cottrell(a)wfu.edu> wrote:
[...]
<hansl>
> open data4-1
> list L = sqft price
> series x = L[1]
> print sqft x --byobs
> ols L[2] 0 L[1]
> matrix m = L
> printf "The series at position 1 in L has ID %d\n", m[1]
> </hansl>
>
[...]
Produces an error for me when I run it:
? ols L[2] 0 L[1]
'L[2]' is not the name of a variable
Error executing script: halting
Right, that won't work with gretl 1.9.92. The syntax is new in gretl
CVS.
Allin Cottrell
8:51 p.m.
On 7 January 2015 at 18:12, Allin Cottrell <cottrell(a)wfu.edu> wrote:
Right, that won't work with gretl 1.9.92. The syntax is new in gretl
CVS.
It still doesn't work for me after updating -gretl- from CVS, so I'll look
forward to a time when it does!
--
Clive Nicholas
"My colleagues in the social sciences talk a great deal about methodology.
I prefer to call it style." -- Freeman J. Dyson
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5 comments
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participants (3)
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Daniel Bencik