<p>In relation to the problem of heteroskedasticity in linear probability models, the following Ben Lambert video is a useful link:</p>
<p>youtube.com/watch?v=pgPhbVEbYqw</p>

CC BY-SA 4.0
2020-11-28T13:09:31.597Z
2020-11-28T13:09:31.597Z
EB3112
<p>I have a problem that asks me the following:</p>
<blockquote>
<p>&quot; Consider the linear probability model, in which we specify the
regression equation to be linear in X,</p>
<p>E(Y |X = x) = Pr(Y = 1|X = x) = x'β</p>
<p>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. &quot;</p>
</blockquote>
<p>Intuitively, I visualise why this is the case - but cannot figure out how to demonstrate this formally. Could someone help?
Thanks in advance!</p>

CC BY-SA 4.0
2020-11-27T12:40:09.857Z
It would be nice, however, to let us see what you visualize. What does heteroskedasticity mean?
Bertrand
CC BY-SA 4.0
2020-11-27T12:42:23.117Z
https://www.youtube.com/watch?v=pgPhbVEbYqw
EB3112
CC BY-SA 4.0
2020-11-27T13:16:09.527Z
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
jamesstealth
CC BY-SA 4.0
2020-11-27T14:46:17.817Z
@jamesstealth, did you watch the vid?
EB3112
CC BY-SA 4.0
2020-11-28T02:57:46.423Z
@EB3112: consider adding an answer here (based on the video you shared) so that the thread can be considered complete?
Dayne
<econometrics><statistics><applied-econometrics><probability>
Show that conditional variance of error in linear probability model is heteroskedastic?
jamesstealth
# Show that conditional variance of error in linear probability model is heteroskedastic?

(Asked by: username_0 on 2020-11-27)

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!

**Tags:** econometrics, statistics, applied-econometrics, probability
> **Comments:**
> **username_1**: It would be nice, however, to let us see what you visualize. What does heteroskedasticity mean?  
> **username_2**: https://www.youtube.com/watch?v=pgPhbVEbYqw  
> **username_0**: 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  
> **username_2**: @jamesstealth, did you watch the vid?  
> **username_3**: @EB3112: consider adding an answer here (based on the video you shared) so that the thread can be considered complete?  

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## 1 answers

### Answer by username_2 on 2020-11-28. Score: 1 | Accepted answer

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

youtube.com/watch?v=pgPhbVEbYqw


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