When working with probit models in stata the first line of the output is (for a sample of 583 with 3 variables):

Iteration 0: log likelihood = -400.01203

If I understand this correctly the iteration 0 is the log likelihood when the parameter for my 3 variables = 0.

The log likelihood function I'm working from is:

\begin{equation} \ln L(\boldsymbol{\alpha}, \beta|\boldsymbol{y,z,t}) = \sum\limits_{j=1}^T I_j ln[\Phi(\frac{\boldsymbol{\alpha}\boldsymbol{z_j}}{\sigma} - \frac{\beta t_j}\sigma)] + (1 - i_j) ln[1 - \Phi(\frac{\boldsymbol{\alpha}\boldsymbol{z_j}}{\sigma} - \frac{\beta t_j}\sigma)] \end{equation}

where $T =$ total # of observations and $\boldsymbol{\alpha}$ and $\beta$ are my parameters

I assumed that by setting my parameters from $\boldsymbol{\alpha}$ and $\beta$ to 0 would be the same as setting my $\Phi$ to 0 and hence the value per observation to 0.5. this would then be multiplied by my number of observations, $\ T$:

ln L["# of observations" * ln(0.5)]

I.e. 583 * ln(0.5) = -404.1048

However, as we see, I was wrong. I assume this has to do with a constant being estimated for each parameter that changes the probability slightly (roughly to 50.35%). Where is this constant coming from? what kind of estimation is being done with my parameters?


1 Answer 1


By running the example below, you can see that the constant is based on the mean of the dependent variable transformed to the probit metric.

// get some example data
sysuse auto, clear

// estimate a probit model, use the trace
// option to see the estimates at each iteration
probit foreign price mpg, trace

// compute the mean on the estimation sample
sum foreign if e(sample)

// transform that mean to the probit metric
di invnormal(r(mean))

// compute the log likelihood
gen double ll = cond(foreign,         ///
                     ln(r(mean)),     /// ll for foreing cars
                     ln((1-r(mean)))) /// ll for domestic cars
                     if e(sample)

// compute the sum                   
sum ll, meanonly
di r(sum)
  • $\begingroup$ thanks a lot, that helped to understand part of the puzzle. however, what I don't fully understand is how to go from the iteration 0 constant of -.5321897 to the log-likelihood value for iteration 0 of -45.03321. I thought that would be # of observations * ln(cdfnormal(constant))? $\endgroup$
    – TTNor
    Jan 24, 2014 at 16:29
  • $\begingroup$ I edited the answer $\endgroup$ Jan 24, 2014 at 16:37

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