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Here is a interview question I head from others, but I think the information may be not complete and correct. Could anyone help me to modify it?

Question: Suppose $X\sim N(0,1), \epsilon\sim N(0,1)$ and $Y = X+\epsilon,$ calculate the linear regression (OLR) coefficient of $Y$ respect to $X.$ How about $Y = X^2 + \epsilon?$

In the EIV (errors in variable) I know the assumption is

$$E Y=\eta,\ E X = \xi;\ \eta = \alpha + \beta \xi,$$ $$Y = \alpha + \beta \xi + \epsilon,\ \epsilon\sim n(0,\sigma^2_{\epsilon});\quad X = \xi + \delta,\ \delta\sim n(0,\sigma^2_{\delta})$$ then use MLE to solve the coefficient. In the question, directly write $Y$ as a function of $X,$ then how to define Linear?

kjetil b halvorsen
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user6703592
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  • There are various definitions of "linear" in this context: see https://stats.stackexchange.com/questions/148638/. But the default assumption is it means that you seek a model for the conditional expectation $E[Y\mid X]$ (that's what "regression" means) as a linear function of $X.$ – whuber Nov 12 '20 at 16:30
  • but if we use $Y|X,$ then the randomness of $X$ is meanless. – user6703592 Nov 12 '20 at 16:34
  • For regression you are correct that the randomness of $X$ is irrelevant, because *by definition* regression concerns the conditional response of $Y,$ not the joint distribution of $(X,Y).$ – whuber Nov 12 '20 at 17:09
  • @whuber that is why I mentioned it may be related to the `EIV (errors in variable)` or $X\sim N$ is just a typo... – user6703592 Nov 12 '20 at 17:13
  • I don't think $X\sim N(0,1)$ is a typographical error: *it simply is not relevant to questions of regressing $Y$ against $X.$* – whuber Nov 12 '20 at 17:46

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