Putting limits on estimated coefficient values

[acemi]

Hello,
I did a logit transformation and want to manually calculate untransformed coefficient value.
I want to verify does @logit(c(2)) = LN(c(2)/(1-c(2))) ?
Thank you.
acemi

[EViews Gareth]

http://www.eviews.com/help/helpintro.ht ... tions.html

[acemi]

I had looked at the explanation of special functions>logit transformation explanation but cannot map it to how it transforms my c(2).

1/(1+exp(-x))) = exp(x)/(1+exp(x))

c(2) has no relation whatsoever with the expression on LHS..

can elaborate please ?

[startz]

I had looked at the explanation of special functions>logit transformation explanation but cannot map it to how it transforms my c(2).

1/(1+exp(-x))) = exp(x)/(1+exp(x))

c(2) has no relation whatsoever with the expression on LHS..

can elaborate please ?

You get 1/(1+exp(-c(2))))

[acemi]

Thank you!

The whole purpose for @logit transformation is to impose/force limits for c(2). For the @logit(c(2)) conversion, the NLS estimation of c(2) is 3,475962 which is clearly not within intended range limits of 0<c(2)<1. So in order to obtain value for @logit(c(2)), substitute this value in 1/(1+exp(-3,475962)) = 0,969996 which is between intended limits. But is it OK to conclude that I have constrained c(2) in this case ?

acemi

[startz]

Thank you!

The whole purpose for @logit transformation is to impose/force limits for c(2). For the @logit(c(2)) conversion, the NLS estimation of c(2) is 3,475962 which is clearly not within intended range limits of 0<c(2)<1. So in order to obtain value for @logit(c(2)), substitute this value in 1/(1+exp(-3,475962)) = 0,969996 which is between intended limits. But is it OK to conclude that I have constrained c(2) in this case ?

acemi

In most cases like this people want the coefficient to between 0 and 1, not specifically the parameter c(2). As you say, using @logit does this. So think of the coefficient as being the value of @logit(c(2)).

[acemi]

Hello,

Is there a way to restrict the parameter c(2) in eviews9 ?Thank you!

The whole purpose for @logit transformation is to impose/force limits for c(2). For the @logit(c(2)) conversion, the NLS estimation of c(2) is 3,475962 which is clearly not within intended range limits of 0<c(2)<1. So in order to obtain value for @logit(c(2)), substitute this value in 1/(1+exp(-3,475962)) = 0,969996 which is between intended limits. But is it OK to conclude that I have constrained c(2) in this case ?

acemi

In most cases like this people want the coefficient to between 0 and 1, not specifically the parameter c(2). As you say, using @logit does this. So think of the coefficient as being the value of @logit(c(2)).

[startz]

The answer is that same as for earlier versions. You cannot restrict c(2) per se, but generally that's not what's desired. Instead, replace every appearance of c(2) with @logit(c(2)).

[rmoralesaramburu]

Hi,

I am doing the reparameterization shown in this forum to coinstraint my coefficients.

CODE: SELECT ALL

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[list]ai > 0 and a1 + a2 + ... + ak = 1[/list]

A reparameterization that satisfies these constraints is

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ai = exp( bi ) / ( exp( b1 ) + exp( b2 ) + ... + exp( bk ) )

I am having an error with my code and I don´t know why. The error is:

Maximum number of iterations exceeded in "OPTIMIZE(MIN)
OPTIVOL(Z1,VDUM,MCOV)" on line 1.

My code is

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subroutine optivol(scalar !scmvol, vector vdum, matrix mmcov)
   scalar sc1=@csum(vdum)
   vector(6) wmvol

      wmvol(1)=exp(vdum(1))/exp(sc1)
      wmvol(2)=exp(vdum(2))/exp(sc1)
      wmvol(3)=exp(vdum(3))/exp(sc1)
      wmvol(4)=exp(vdum(4))/exp(sc1)
      wmvol(5)=exp(vdum(5))/exp(sc1)
      wmvol(6)=exp(vdum(6))/exp(sc1)

   !scmvol=@transpose(wmvol)*mmcov*wmvol
endsub


vector(6) vdum
vdum=1
scalar z1
optimize(min) optivol(z1,vdum,mcov)



Please help!

[tvonbrasch]

Hi,

I want to estimate this equation:

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YK=C(1)*XK1+C(2)*XK2+C(1)*C(2)*XK3

in a way that ensures that the coefficients are positive. However, I get a "near singular matrix" message when using the @exp function to estimate the equation:

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   equation _test2.ls YK=EXP(C(1))*XK1+EXP(C(2))*XK2+EXP(C(1))*EXP(C(2))*XK3


If I instead use the function @abs, there is no singularity issue:

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   equation _test1.ls YK=abs(C(1))*XK1+abs(C(2))*XK2+abs(C(1))*abs(C(2))*XK3
   show _test1

That said, I am wondering if the @abs function is not in general suitable as the derivative is not always defined.

Attached is the workfile with the data.

Can you please take a look and suggest ways to ensure that the coefficients are positive?
t

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close @all
wfopen .\exp2.wf1


'first equation - no singularity
   equation _test1.ls YK=abs(C(1))*XK1+abs(C(2))*XK2+abs(C(1))*abs(C(2))*XK3
   show _test1

'second equation - singularity ...?
   equation _test2.ls YK=EXP(C(1))*XK1+EXP(C(2))*XK2+EXP(C(1))*EXP(C(2))*XK3
   show _test2s


[EViews Glenn]

You are correct on the issue of differentiability of abs. I always use the exp function but you have to make sure starting values are appropriate. What are the values when you have issues?

[edit:]

I spent a little more time looking at this. I downloaded the workfile and ran the example at various starting values. I get singularity in all cases because the estimation is driving the coefficients on XK2 to zero (e.g., C(2) is a very large negative number after a few iterations) which I am pretty certain is giving you zero gradients for C(2) and hence, singularity.

As a sanity check, I did a quick simulation of the model and then estimated the coefficients for the simulated data

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c(1) = .5
c(2) = -1
series z = exp(c(1))*xk1 + exp(c(2))*xk2 + exp(c(1)+c(2))*xk3 + 10*nrnd
nrnd(c)
equation eqtest.ls z = exp(c(1))*xk1 + exp(c(2))*xk2 + exp(c(1)+c(2))*xk3


and the estimation produces sensible results.

At this point, I'd say that it looks to be a data issue, not a specification issue. Along these lines, note that the unrestricted linear regression produces estimated coefficients on XK2 and XK3 that are both negative values.

[tvonbrasch]

Thanks for looking into this, Glenn. I will take a second look at the data.
t