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I have a question regarding linear regression. Suppose we have the following regression model: $$ y_{it}=\alpha+x_{it}'\beta+u_{it} $$ where say $i$ represents individual $i$ at time period $t.$ The conditional mean function is in general defined as: $$ E[Y|X=x]=\int yf(y|x)dy $$ In the linear regression, assuming conditional exogeneity, this simplifies to: $$ E[y_{it}|x_{it}]=\alpha+x_{it}'\beta $$ My question is how is the defintion of the mean function related to the above expression? Do we assume that for ${each }$ value of $x$ , we have infinite possible values of $y$ and we can estimate the average? Thanks!

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You have posted the conditional expectation for continuous random variables. The conditional expectation for discrete random variables is

$\mathbb E(Y|X=x)=\sum\limits_{i=1}^n y_i\cdot f(y_i|x_i)$

In linear regression this conditional expectation is the estimated regression line:

$\mathbb E(Y|X=x)=\alpha+\beta x_i=\hat y_i$

$\alpha$ and $\beta$ are the estimated parameter of the regression line. Every value $x_i$ have a related $y_i$. They are given by the sample.

And the difference between the observed $y$-values and the $y$-values of the regression line is $y_i-\hat y_i=u_i$. The parameters $\alpha$ and $\beta$ are estimated by minimizing $\sum\limits_{i=1}^n u_i^2$.

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Each $y_{it}$ is understood a random variable; so there is no problem in writing $\Bbb{E}(y_{it} \mid x_{it})$.

It seems an abuse of notation to write $y_{it}$ for both a random variable and a particular realization; but this is not uncommon in applied fields.

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  • $\begingroup$ I understand that the yit is a random variable, but its conditional mean function is assumed to be constant.Then for two datapoints with the same value of the x but different values of y, can we infer anything about the conditional distribution? $\endgroup$
    – ChinG
    Nov 9, 2015 at 15:14
  • $\begingroup$ @ChinG We can learn about the conditional distribution of $y_{it}$. $\endgroup$
    – Yes
    Nov 9, 2015 at 15:18
  • $\begingroup$ You may want to try CrossValidation S.E., which is for statistics questions. :) @ChinG $\endgroup$
    – Yes
    Nov 9, 2015 at 15:19

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