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Having a simple model like: Y = a + bX + u where u represent the error term, if I know that the observation of X and Y are i.i.d, also the error term is i.i.d?

Albert
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    https://stats.stackexchange.com/questions/94872 answers the question about independence, because $u = Y-a-bX$ is a function of the iid variables $(X,Y).$ As far as whether the $u_i$ have identical distributions, perhaps you can answer that part of the question yourself? – whuber Dec 29 '18 at 18:34
  • @whuber ok, thank you, therefore is independent. Although is trivial I still have some doubt about identically distribution. Supposing that both X and Y are N(0,1) then u is N(-a,2). Therefore, are not identically distributed. is it correct? – Albert Dec 29 '18 at 21:01
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    If $X$ and $Y$ are both standard Normal, then perforce $a=0$ and $b=1$ and you must still conclude the $u_i$ are identically distributed. – whuber Dec 29 '18 at 22:57
  • @whuber The variance of the sum of two independent standard normal, should not be the sum of the variance of each r.v.? As a consequence u should not have a different variance? – Albert Dec 30 '18 at 09:04
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    You haven't supplied assumptions that imply $u$ is the sum of independent standard Normal variables. – whuber Dec 30 '18 at 16:07
  • @whuber if a=0, b=1 we have: u = Y - X. Is not this a sum of independent standard Normal variables? – Albert Dec 30 '18 at 17:32
  • @whuber sorry if I return in this question, but in your first answer you say that u, being a trasformation of i.i.d. variables, is still i.i.d.. But if I know that the observation (xi,yi) are i.i.d. this should not imply that the r.v. are independent, indeed there could be some correlation between the same vector of observation i.. so, it still hold your first comment? – Albert Jan 08 '19 at 17:56
  • "i.i.d" refers to the *vector* variables $(x,y)_i.$ It does not state or imply that $x_i$ and $y_i$ are independent. – whuber Jan 08 '19 at 18:11
  • In fact, your link was about independence of r.v.s not about i.i.d. observation. So maybe it does not answer to my question. I’m a little be confused. I need to know it the observation ui are i.i.d. in order to know if I can apply the LLN to ui – Albert Jan 08 '19 at 18:18
  • I cannot understand what distinctions you are making, because iid observations *are* random variables. You might benefit from reviewing definitions of "iid" and "random variable." – whuber Jan 08 '19 at 18:26

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