Forum: open-discussion

RE: quadratic functional form [ Reply ]
By: Arne Henningsen on 2016-07-06 09:24

Dear João

I suggest that you estimate your models (e.g. quadratic and Translog distance functions) by OLS and then use standard tests for heteroskedasticity (e.g. Breusch-pagan). Furthermore, you could test for the appropriateness of the functional form, e.g. using Ramsey's RESET (test).

I think that efficiencies() works correctly but that for the estimated sigma_v and sigma_u and the obtained residuals in your empirical application, the expected efficiencies are either very close to one or very close to zero.

As the Translog functional form uses a *logarithmic* dependent variable, all deviations from it (i.e. the noise term v and the inefficiency term u) are measured in *relative* terms (log(y) = log(f(x)) + v - u => y = f(x) * e^v * e^(-u) and (v-u) = log(y) - log(f(x)) = log( y / f(x) )), while the quadratic functional form uses the original (non-logarithmic) dependent variable so that all deviations from it are measured absolute terms (y = f(x) + v - u => (v-u) = y - f(x)).

You can use the sfa() function of the "frontier" package to estimate an efficiency effects stochastic frontier model with a non-logarithmic dependent variable (e.g. using a quadratic functional form). However, I am not sure if the efficiencies() method in this package can calculate (in)efficiency estimates from an efficiency effects stochastic frontier model with a non-logarithmic dependent variable. If efficiencies() cannot do this, you can calculate the inefficiencies manually.

Best wishes,
Arne

RE: quadratic functional form [ Reply ]
By: João Silva on 2016-07-05 15:13

[forum:43349]

Dear Arne,

Thanks for your reply and clarifications. I assume the best thing to do first then is to test for heteroskedasticity? What would be your advice to conduct this test? In addition, which other assumptions would you recommend me to check? I am surprised that efficiencies() does not work properly because of this as I though the logDepVar option was not working properly.

I must confess that your explanation about firm size was not very clear to me. How come a quadratic and translog model have such different approaches to deal with the deviation from the output quantity?

In addition, I was wondering if I could use the inefficiency effects model with the quadratic functional form. Is this possible to use it given the fact that the u term needs to estimated through the JMLS approach in case the variables are not logged? I guess that is what happens anyways when I set up logDepVar=F in the efficiencies() function (?).

Thank you,
Joao

RE: quadratic functional form [ Reply ]
By: Arne Henningsen on 2016-07-04 05:57

[forum:43335]

Dear João

Regarding your first question: as you wrote, you can use the sfa() command to estimate a quadratic functional form, whereas the difference between the quadratic and the translog functional form is the way, in which the dependent variable and the explanatory variables are constructed. The stochastic frontier model and thus, the sfa() command assumes that the residuals consist of two parts: a normally distributed noise term (v) and a half-normally or truncated normally distributed inefficiency term (u). If you estimate a Translog functional form, you must assume that log(y) - f(x) follows a combined normal and half/truncated normal distribution. If you estimate a quadratic functional form, you must assume that y - f(x) follows a combined normal and half/truncated normal distribution. In my experience, the latter assumption is violated in almost all empirical specification, e.g. because of very strong heteroskedasticity. For instance, if you estimate (for simplicity of this illustration) a production function with firms that produce between 1 unit and 1000 units of output, a quadratic functional form assumes that the magnitudes of the absolute deviations from the output quantity do not depend on firm size, e.g. the output quantity of a 1-unit producing firm varies between, say, 0.7 and 1.3 units due to noise and can be down to 0.6 units due to inefficiency, while the output quantity of a 1000-unit producing firm varies between, say, 999.7 and 1000.3 units due to noise and can be down to 999.6 units due to inefficiency. In contrast, a Translog functional form assumes that the magnitudes of the *relative* deviations from the output quantity do not depend on firm size, e.g. the output quantity of a 1-unit producing firm varies between, say, 0.7 and 1.3 units due to noise and can be down to 0.6 units due to inefficiency, while the output quantity of a 1000-unit producing firm varies between, say, 700 and 1300 units due to noise and can be down to 600 units due to inefficiency. If the (theoretical) assumptions of a stochastic frontier model with a quadratic functional form hold in your empirical application, you can use the sfa() function to estimate it.

2) Regarding your second question: using efficiencies( ..., logDepVar=FALSE ) is correct when you have estimated a quadratic functional form. I guess that the strange results are caused by model misspecification (see above).

Best regards,
Arne

quadratic functional form [ Reply ]
By: João Silva on 2016-06-12 11:37

[forum:43274]

Dear Arne,

I am writing you because I would like to estimate a quadratic output distance function in R using your frontier package rather than a more conventional translog. The reason for this is that I have a lot of outputs which are difficult to handle when taking the logs for the translog...

My questions are the following:

1) Can I use the sfa() command to estimate a quadratic functional form? If so, I assume the difference between the quadratic and the translog is the way the variables are constructed i.e. the translog has logged variables while the quadratic not. Is this right? Or the sfa() command defining the BC1992 model always assumes that the dependent and independent variables are logged no matter what?

2) How can I estimate the technical efficiency score using the efficiencies() command from a quadratic form? When I define the argument logDepVar=FALSE all farms have a technical efficiency score equal to 1 (apart from one observation which has a negative TE score) which does not sound right to me.

Thank you,
João