## Search found 18 matches

- Fri Oct 19, 2018 1:35 pm
- Forum: Estimation
- Topic: Defining derivatives and constrains?
- Replies:
**24** - Views:
**459**

### Re: Defining derivatives and constrains?

The negative R-squared indicates that the program has not converged to a global optimum. You may want to try different starting values. More importantly: Estimating 4 parameters with 8 observations is usually pretty hopeless. Oh, I understand. Thanks for the explanation. Will look for how to do the...

- Fri Oct 19, 2018 12:22 pm
- Forum: Estimation
- Topic: Defining derivatives and constrains?
- Replies:
**24** - Views:
**459**

### Re: Defining derivatives and constrains?

What about the failure notification of non-zero gradient in the results table, wouldnt it question the robustness or validation of the results? This means that EViews may not have converged to the optimum Also for some sections of my data, I get negative R-squared and high probabilities (even 1.00)...

- Fri Oct 19, 2018 6:59 am
- Forum: Estimation
- Topic: Defining derivatives and constrains?
- Replies:
**24** - Views:
**459**

### Re: Defining derivatives and constrains?

c(2) is negative. c(2)^2 is not. What about the failure notification of non-zero gradient in the results table, wouldnt it question the robustness or validation of the results? This means that EViews may not have converged to the optimum Also for some sections of my data, I get negative R-squared a...

- Thu Oct 18, 2018 8:42 pm
- Forum: Estimation
- Topic: Defining derivatives and constrains?
- Replies:
**24** - Views:
**459**

### Re: Defining derivatives and constrains?

c(2) is negative. c(2)^2 is not. Hello Startz, Thanks for your reply. Oh you're right. I should calculate c(2)^2 for the primary constant. What about the failure notification of non-zero gradient in the results table, wouldnt it question the robustness or validation of the results? Also for some se...

- Thu Oct 18, 2018 8:02 pm
- Forum: Estimation
- Topic: Defining derivatives and constrains?
- Replies:
**24** - Views:
**459**

### Re: Defining derivatives and constrains?

Hello, As startz correctly points out, expressing your equation as a list of terms that include coefficients, e.g., w c(1) (@exp(c(2)))*(1/x) (y/x) (z/x), doesn't do what you think it does. Putting the reason aside for a moment, that's why startz recommended that you express the equation as an expl...

- Wed Oct 17, 2018 6:39 pm
- Forum: Estimation
- Topic: Defining derivatives and constrains?
- Replies:
**24** - Views:
**459**

### Re: Defining derivatives and constrains?

People are helping you. You have to help yourself. And help them help you. Hello Gareth, Yes, you're right. But I already mentioned my problem. Okay I will re-explain what I do. I'm having the following equation with variables W, X, Y, and Z, and I need to estimate the constants C1, C2, C3, and C4....

- Wed Oct 17, 2018 4:22 pm
- Forum: Estimation
- Topic: Defining derivatives and constrains?
- Replies:
**24** - Views:
**459**

### Re: Defining derivatives and constrains?

Show people exactly what you've done and maybe someone will chime in having figured out where things have gone astray. I exactly entered the commands that I mentioned above with the data provided in the excel file, and and I got the corresponding results. I also play around by putting space etc. bu...

- Wed Oct 17, 2018 3:56 pm
- Forum: Estimation
- Topic: Defining derivatives and constrains?
- Replies:
**24** - Views:
**459**

### Re: Defining derivatives and constrains?

I think what's happening is that the command you are using puts a coefficient in front of each term. Try writing this out specified as a formula as shown at http://www.eviews.com/help/helpintro.html#page/content%2FRegress1-Specifying_an_Equation_in_EViews.html Thanks very much for your tip. I read ...

- Wed Oct 17, 2018 2:39 pm
- Forum: Estimation
- Topic: Defining derivatives and constrains?
- Replies:
**24** - Views:
**459**

### Re: Defining derivatives and constrains?

You might want to post the exact command you used and your output. Thanks for your reply. The exact equation is W = C1 + C2*(1/X) + (Y/X) + Z/X) with constraint C2>0 . The first command: w c(1) (2*@logit(c(2)))*(1/x) (y/x) (z/x) The first output: http://i65.tinypic.com/262rtit.jpg The second comman...

- Tue Oct 16, 2018 11:48 pm
- Forum: Estimation
- Topic: Defining derivatives and constrains?
- Replies:
**24** - Views:
**459**

### Re: Defining derivatives and constrains?

I'm not sure there are any general tricks, although perhaps someone has a list of useful ones. Dear Startz I'm having the following equation and I need the constant (a) to be no smaller than zero: Z = c + (a/X) + (bY/X) I have already read the related post: viewtopic.php?f=4&t=48&p=99 , whi...

- Tue Oct 16, 2018 12:42 am
- Forum: Estimation
- Topic: Defining constraints for coefficients
- Replies:
**6** - Views:
**287**

### Re: Defining constraints for coefficients

Need urgent help with my problem. Please!!! I'm having the following equation and I need the constant (a) to be no smaller than zero: Z = c + (a/X) + (bY/X) I have already read the related post: viewtopic.php?f=4&t=48&p=99 , which says for one-sided constraints like for the constant c no lar...

- Sat Oct 13, 2018 1:39 am
- Forum: Estimation
- Topic: Defining constraints for coefficients
- Replies:
**6** - Views:
**287**

### Re: Defining constraints for coefficients

EViews Gareth wrote:That post pretty much gives all the details.

Yes, but my background is not at all math and I'm a dummy in math. The post says for a constant no larger than 1 we better use 1-exp(a). Since mine is a constant no smaller than zero, I dont know how to do the conversion using 1-exp(a).

- Fri Oct 12, 2018 7:27 pm
- Forum: Estimation
- Topic: Defining constraints for coefficients
- Replies:
**6** - Views:
**287**

### Re: Defining constraints for coefficients

http://forums.eviews.com/viewtopic.php?f=4&t=48&p=99 Hello Gareth, I read the entire post, but I'm still confused how to make the constraint (a=>0) to my equation. I used both the Exp and Log solutions but either I get errors or negative values which does not make sense. Could you give me s...

- Wed Oct 10, 2018 11:24 pm
- Forum: Estimation
- Topic: Defining constraints for coefficients
- Replies:
**6** - Views:
**287**

### Re: Defining constraints for coefficients

EViews Gareth wrote:http://forums.eviews.com/viewtopic.php?f=4&t=48&p=99

Thanks. I have already read the post, but cannot figure it out and also implement it the 'Quick Estimate' function.

Could you help me out here?

- Wed Oct 10, 2018 10:16 pm
- Forum: Estimation
- Topic: Defining constraints for coefficients
- Replies:
**6** - Views:
**287**

### Defining constraints for coefficients

Hello, I'm trying to estimate the following equation with variables X, Y, and Z, and coefficient a, b, and c, and there is requirement that a>=0: Z = c + (a/X) + (bY/X) I'm using estimation function in the Eview but it gives me negative value for a, which is not acceptable logically for my problem. ...