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RE: st: RE: Re: Calculation of integrals


From   "Mentzakis, Emmanouil" <[email protected]>
To   <[email protected]>
Subject   RE: st: RE: Re: Calculation of integrals
Date   Wed, 12 Dec 2007 13:41:42 -0000

Dear Scott, 

Thank you for the suggestion. Restricting the sample in Limdep works
nicely and the results are the same. 

Thanks again
Manos

-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of Scott
Merryman
Sent: 11 December 2007 19:51
To: [email protected]
Subject: Re: st: RE: Re: Calculation of integrals

I believe with Limdep, in this case, you need to restrict the sample
before Fintegrate command:

Sample ; 1$


Also, try a more basic example:

clear
set obs 100
range x 0 100
integ x x

Do the following Limdep commands (without restricting the sample) also
results in an integral  5000?

 Fintegrate ;fcn = var
    ;labels = var
    ;start = 0
    ;pts =100
    ;limit = 0,100
    ;vary(var)
    $

Scott


On Dec 11, 2007 12:24 PM, Mentzakis, Emmanouil <[email protected]>
wrote:
> Dear Rodrigo,
>
> Thanks for your reply. Differences in the results are not restricted 
> to a specific integral.
>
> E.g. using the command below
>
> clear
> range x 1 100 50
> g double lnx=ln(x)
> g double y=5*lnx
> integ y x, g(result)
>
> If I understand correctly, I compute the integral with lower limit 1 
> and upper limit 100, over 50 points and the function integrated is
5*ln(x).
> Is this correct?
>
> Similarly in Limdep, I thought that the same calculation would be done

> by the command
>
> Fintegrate ;fcn = 5*log(var)
>    ;labels = var
>    ;start = 10
>    ;pts =50
>    ;limit = 1,100
>    ;vary(var)
>    $
>
> Where, again, the limits are from 1 to 100, for 50 points and for the 
> function 5*log(x). What is specified as "start" does not influence the

> calculation. It, actually, is a value given so that it can be assessed

> before the computation is done whether the function evaluation is 
> possible at all.
>
> I'd really appreciate if you could comment on the equality of the two 
> commands, because probably I am estimating two different things 
> without realising it.
>
> Thanks
> Manos
>
>
>
>
> -----Original Message-----
> From: [email protected]
> [mailto:[email protected]] On Behalf Of Rodrigo A.
> Alfaro
> Sent: 11 December 2007 00:47
> To: [email protected]
> Subject: st: Re: Calculation of integrals
>
> Manos,
>
> It is difficult to give an answer with this info. What is exactly the 
> integral that you want to calculate? What is the initial parameter in 
> Limdep? It seems that you are trying to get an integral of the normal 
> cdf, then something like 1000 seems too much as result (just a guess)
>
> Rodrigo.
>
>
> ----- Original Message -----
> From: "Mentzakis, Emmanouil" <[email protected]>
> To: <[email protected]>
> Sent: Monday, December 10, 2007 4:47 PM
> Subject: st: Calculation of integrals
>
>
> > Dear all,
> >
> > I am calculating an integral in Stata and then compare the 
> > calculation with the same calculation in Limdep.
> >
> > However, the difference between the two calculations is very big
> (Stata
> > = 3.233 ; Limdep = 980.525).
> >
> > Below I have the code for Stata and the code for Limdep.
> >
> > Could anyone explain if I am making a mistake or what is the reason
> for
> > this result?
> >
> > Thank you
> > Manos
> >
> >
> > *****************************
> > *********** Stata ***********
> > *****************************
> >
> > clear
> > set obs 300
> > capture drop xb x
> > g xb=invnormal(uniform())
> > range x .001 1000 300
> > g double lnx=ln(x)
> > g double norm=normal(xb + .5*lnx)/300 integ norm x, g(integral)
> >
> > number of points = 300
> >
> > integral         = 3.233067
> >
> >
> >
> > ********************************
> > ************ Limdep ************
> > ********************************
> > Coping the data from Stata, in Limdep we don't have to create a
> variable
> > for the range. It automatically uses the variable that is put in the

> > "label" and its limit is given by the "limit" and the number of 
> > points by "pts". "Start" is the values of the parameters, but 
> > changing it doesn't make any difference in the actual calculation.
> >
> > Fintegrate ;fcn = PHI(xb + .5*log(var))/300
> >    ;labels = var
> >    ;start = 500
> >    ;pts =300
> >    ;limit = 0.001,1000
> >    ;vary(var)
> >    $
> >            +---------------------------------------------------+
> >            | Function integration:                             |
> >            | Grid is  300 points in [      .001 to   1000.000] |
> >            | Value of the integral is       980.52567          |
> >            +---------------------------------------------------+
> >
> >
> >
> >
> > *
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