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So I'm trying to show that regressing y on x in this case (fixed effects model):
$y_{it} = \beta x_{it} + \theta_{i} + \epsilon_{it}$

is the same as this regression:
$y_{it} - \bar{y_{l}} = \beta(x_{it} - \bar{x_{l}}) + \epsilon_{it}$

EDIT: enter image description here

financial_physician
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    what does that $l$ stand for? – user158565 Jan 19 '19 at 21:59
  • I think it's a capital "i" to account for the means that fixed effects pull out. I'll edit the problem – financial_physician Jan 19 '19 at 22:30
  • At the moment you have written both models with the same error term, so clearly they are *not* the same. Even if you fix this, without specifying an error distribution for the two models, you have not even fully specified the models, so it is not possible to compare them and say if they are different. I suggest you start by changing the error notation on one of the models and then specify your error distributions, so that you have fully specified your two models. – Ben Jan 20 '19 at 07:32
  • Here I'll post the problem exactly how it's written in my homework. Hopefully that will help to clarify. If we can't figure out the answer I'll talk to the TA on Tuesday and get back with you guys about it so we have the answer and can discuss it! – financial_physician Jan 20 '19 at 08:13
  • This is demonstrated (in passing) at https://stats.stackexchange.com/questions/196807/explicit-solution-for-linear-regression-with-two-predictors/197788#197788 using a technique that (obviously) generalizes to multiple regression. – whuber Jan 20 '19 at 20:45

2 Answers2

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You can rewrite your first equation as $\bar{y}_i = \beta \bar{x}_i + \theta_i + \bar{\epsilon}_i $. Moreover, notice that $\bar{\epsilon}_i = 0$, because you have $\theta_i$ in the model.

So that the only thing you have to do is to subtract this equation from your first equation so you can reach your second equation. I hope it helped.

Henrique
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    This makes a lot of sense to me. I'm not sure if taking the difference between the original and the expected value of the original conditioned on t would suggest the betas are identical. – financial_physician Jan 20 '19 at 08:22
  • What do the bars on the variables mean? – whuber Jan 20 '19 at 16:04
  • The bars are the sample mean. – Henrique Jan 20 '19 at 16:23
  • Considering the question, there is another way to prove it, using the Frisch Waugh Lovell theorem (this is what the question provided you in parenthesis). 1st you regress y and x on all $\theta_i$. The residual terms from these two regressions are the demeaned variables $y_i - \bar{y}$ and $x_i - \bar{x}$ (I let you to check it by yourself, but it is a very standard result in OLS regression). 2nd you take these residuals and regress $y_i - \bar{y}$ on $x_i - \bar{x}$ . The Frisch Waugh Lovell theorem states that the $ – Henrique Jan 20 '19 at 16:24
  • If that's what the bars are intended to denote, then could you clarify what exactly the other variables refer to? In particular, would $\beta,$ $\theta,$ and the $\epsilon_i$ be the *model parameters* and *errors* or would they be the *least squares estimates* and *residuals*? – whuber Jan 20 '19 at 20:41
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An alternative answer would use the Frisch Waugh Lovell theorem (this is what the question provided you in parenthesis).

1st you regress y and x on all $\theta_i$. The residual terms from these two regressions are the demeaned variables $y_i - \bar{y}$ and $x_i - \bar{x}$ (It is a standard result in OLS regression). 2nd you regress these residuals each other. I.e. regress $y_i - \bar{y}$ on $x_i - \bar{x}$. The Frisch Waugh Lovell theorem states that the $\hat{\beta}$ from this regression is numerically identical to the original regression.

Henrique
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