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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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confused
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  • 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 https://stats.stackexchange.com/a/133333/919, another at https://stats.stackexchange.com/questions/117406, and a couple at https://stats.stackexchange.com/questions/459588. You could probably find more with [this site search](https://stats.stackexchange.com/search?page=2&tab=Relevance&q=regression%20estimate%20normal%20distribution%20score%3a2%20answers%3a1). – whuber May 24 '20 at 16:50
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    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? – Richard Hardy Jun 23 '20 at 12:08
  • @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. – confused Jun 26 '20 at 10:05
  • @RichardHardy Hi, yes that helps and makes sense. Now I just need to know if the marginal distribution of Y needs to be normal. I don't recall anywhere saying that it needs to be normal, until after I read that passage. Additionally, when we do our model pre-diagnostics, I can't remember a time we plotted the marginal distribution of Y. Yet in the text above, it seems like the marginal distribution of Y must be normal in order for the sampling distribution of b1 to be normal... – confused Jun 26 '20 at 10:08
  • @confused, I do not think unconditional normality of $y$ is used anywhere. Or if it is, it might be a misuse. – Richard Hardy Jun 26 '20 at 10:10
  • In this setting the marginal distribution of $Y$ is, by assumption, a *discrete mixture* of normals. This fact is invoked nowhere. – whuber Jun 26 '20 at 15:18

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