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I am reading DeGroot and we made the assumption that Y | X is normal and each Y | X has the same variance. However, in deriving the sampling distributions of b0 and b1, he says that Y is normal. Do we need to make the additional assumption that Y is normal as well? Or is there some relationship between Y | X and marginal Y that I don't know about?

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  • $\begingroup$ I'm sure this has been asked before--but it's hard to search for. Because $\hat\beta_1$ is a linear combination of iid normal values, it is itself normal, QED. I found one terse demonstration at stats.stackexchange.com/a/133333/919, another at stats.stackexchange.com/questions/117406, and a couple at stats.stackexchange.com/questions/459588. You could probably find more with this site search. $\endgroup$
    – whuber ♦
    May 24, 2020 at 16:50
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    $\begingroup$ Does your title really reflect what you are after? The answer to the title question is a simple "No". Consider $X$ that is binary. Add an independent error, and this will yield $Y$ being a mixture of two normals with different means but equal variances. If $X$ is continuous, $Y$ will be a "continuous mixture" of normals. Should I post this as an answer? $\endgroup$
    – Richard Hardy
    Jun 23, 2020 at 12:08
  • $\begingroup$ @whuber hello, I know sample b1 follows a normal distribution. He is saying Y itself are iid and normal. Doesn't that imply the marginal distribution of Y is normal - which is an assumption I've never heard of for the classical linear model. $\endgroup$
    – confused
    Jun 26, 2020 at 10:05
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    $\begingroup$ @confused, I do not think unconditional normality of $y$ is used anywhere. Or if it is, it might be a misuse. $\endgroup$
    – Richard Hardy
    Jun 26, 2020 at 10:10
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    $\begingroup$ In this setting the marginal distribution of $Y$ is, by assumption, a discrete mixture of normals. This fact is invoked nowhere. $\endgroup$
    – whuber ♦
    Jun 26, 2020 at 15:18

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