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My OLS regression looks like this:

$$y=\beta_0+\beta_1x_1+\beta_2x_2+\beta_3(x_3)^2+\epsilon.$$

I do not include $x_3$ linear because it vifs with squared value and without squared value I have wrong specification according to RESET.

How do I interpret the $\beta_3$ in my equation (in general)? I know it should be something like: $$\Delta y=(2\beta_3x_3)\Delta x$$ But what should I use as an $x_3$? Should it be an average $x_3$ value?

whuber
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Emil
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    I would suggest you include both linear and quadratic terms otherwise we risk biasing our model. Please see: https://stats.stackexchange.com/questions/28730 for more details. – usεr11852 Jun 29 '19 at 09:05
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    You may use $y = \beta_0 + \beta_1 x_1 + \beta_2 x_2 + \beta_3 x_3 + \beta_{32} (x_3 - \overline{x_3})^2$ where $\overline{x_3}$ is the mean of $x_3$ or an estimate of the extreme of the parabolic relation between $x_3$ and $y$. In this latter case, it can be chosen such that $\beta_3 \approx 0$. – Ertxiem - reinstate Monica Jun 29 '19 at 11:54

2 Answers2

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The interpretation of the $\beta$'s is the same for each $\beta_i$; when $y=\beta_0+\beta_1x_1+\beta_2x_2+\beta_3(x_3^2)+ε$, we have that $y$ increases by $\beta_1$ when $x_1$ increases by one and all the other covariates stay the same. For $\beta_3$ this is similar: $y$ increases by $\beta_3$ as $x_3^2$ increases by one.

If you want to compute the value for $y$ given the values you have for $x_1,x_2,x_3$, you can simply use the OLS regression formula you have created but square $x_3$ when multiplying by $\beta_3$. Hope this answers your question.

mariekejee
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$\beta_3$has no meaningful interpretation itself, its more so what changes with respect to the interpretation of the marginal effect. In your example, if we obtain a positive value for $\beta_3$then the marginal effect of $x_3$ on $y$ is increasing with $x_3$. That is, the effect is larger for larger values of $x_3$. But as mentioned in the comments, i would include a linear term whenever using a quadratic term except for in very special circumstances.

Brennan
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