I have a problem that asks me the following:

" Consider the linear probability model, in which we specify the regression equation to be linear in X,

E(Y |X = x) = Pr(Y = 1|X = x) = x'β

We can accordingly express the regression equation by Y = X'β + e with E(e |X = x) = 0 for all x. Show that the conditional variance of e given X = x depends on x, i.e., e is heteroskedastic. "

Intuitively, I visualise why this is the case - but cannot figure out how to demonstrate this formally. Could someone help? Thanks in advance!

  • $\begingroup$ It would be nice, however, to let us see what you visualize. What does heteroskedasticity mean? $\endgroup$
    – Bertrand
    Nov 27 '20 at 12:40
  • 1
    $\begingroup$ youtube.com/watch?v=pgPhbVEbYqw $\endgroup$
    – EB3112
    Nov 27 '20 at 12:42
  • $\begingroup$ in my head i interpret it as: given that y's can only take two values - 1 and 0 - and the variance is the square of the distance between the actual observations and the regression line, when the the regression line is at y=0 - or to 1 - then, for that precise X=x, there are many values that cancel out $\endgroup$ Nov 27 '20 at 13:16
  • $\begingroup$ @jamesstealth, did you watch the vid? $\endgroup$
    – EB3112
    Nov 27 '20 at 14:46
  • $\begingroup$ @EB3112: consider adding an answer here (based on the video you shared) so that the thread can be considered complete? $\endgroup$
    – Dayne
    Nov 28 '20 at 2:57

In relation to the problem of heteroskedasticity in linear probability models, the following Ben Lambert video is a useful link:



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