Logit Regression: Calculating the value of a continuous independent var corresponding to a 0.5 probability value of an event occuring

[Jack Treliving]

Hi there,

I have run a logit regression and I am now trying to determine the value of a continuous variable that corresponds to a 0.5 probability of an event occurring. In my code below, I am trying to find the value of cont1 that corresponds to 0.5 probability of event == 1. I realise that this will depend on cont2, so it would be useful if I could include a way of setting cont2 to a specific, but adjustable value, then being able to calculate the value of cont1 at prob 0.5 from that.

logit event cont1 cont2
margins, at(cont1=(0(0.1)6))

I would be very grateful for some advice!

Many thanks,
Jack

[Maarten Buis]

The predicted probability is 0.5 when the linear predictor is 0 So

\[\beta_0 + \beta_1 x_1 + \beta_2 x_2 = 0\]

And we are looking for a value of \(x_2\) where this is true for a given value of \(x_1\)

\[\beta_2 x_2 = -\beta_0 - \beta_1 x_1 \Rightarrow x_2 = - \frac{\beta_0 + \beta_1 x_1}{\beta_2}\]

In an example:

. clear all

. sysuse nlsw88
(NLSW, 1988 extract)

.
. logit union grade tenure

Iteration 0:  Log likelihood = -1040.9971  
Iteration 1:  Log likelihood = -1019.7881  
Iteration 2:  Log likelihood = -1019.5953  
Iteration 3:  Log likelihood = -1019.5953  

Logistic regression                                     Number of obs =  1,866
                                                        LR chi2(2)    =  42.80
                                                        Prob > chi2   = 0.0000
Log likelihood = -1019.5953                             Pseudo R2     = 0.0206

------------------------------------------------------------------------------
       union | Coefficient  Std. err.      z    P>|z|     [95% conf. interval]
-------------+----------------------------------------------------------------
       grade |   .0720392    .021608     3.33   0.001     .0296882    .1143902
      tenure |   .0485961   .0092647     5.25   0.000     .0304376    .0667545
       _cons |   -2.41764   .2967544    -8.15   0.000    -2.999268   -1.836012
------------------------------------------------------------------------------

. local at = 5

. di -(_b[_cons] + _b[tenure]*`at')/_b[grade]
30.187173

Since the maximum education in the data is 18, this 50% mark happens clearly outside the range of data and is thus quite an extrapolation.

If you want a confidence interval around this number, then that is a bit complicated as there is a division involved. I would use the bootstrap for that.

program toboot, rclass
    syntax [if] , at(real)
    marksample touse
    logit union grade tenure if `touse'
    return scalar p50 = -(_b[_cons] + _b[tenure]*`at')/_b[grade]
end

bootstrap p50=r(p50), reps(10000) bca : toboot, at(5)

estat bootstrap, bca  



Bootstrap results                               Number of obs     =      1,866
                                                Replications      =      10000

      Command: toboot, at(5)
          p50: r(p50)

------------------------------------------------------------------------------
             |    Observed               Bootstrap
             | coefficient       Bias    std. err.  [95% conf. interval]
-------------+----------------------------------------------------------------
         p50 |   30.187173   3.000599   22.514546    23.60066   57.12771 (BCa)
------------------------------------------------------------------------------
Key: BCa: Bias-corrected and accelerated