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With $x_1\sim N(\mu,\sigma^2)$ and a population model...

$Y=\alpha_0+\alpha_1X_1+\alpha_2X_1^2+\epsilon$

...if I run OLS omitting the square term...

$y_i=\beta_0+\beta_1x_{1,i}+u_i$

...the $x_1$ slope has a bias of $2\alpha_2E[x_1]$. I think the bias should be:

$\frac{\partial E[u|x_1]}{\partial x_1}=\frac{\partial E[\alpha_2x_1^2|x_1]}{\partial x_1}=\frac{\partial \alpha_2E[x_1^2|x_1]}{\partial x_1}=\frac{\partial \alpha_2x_1^2}{\partial x_1}=2\alpha_2 x_1$

which doesn't make a lot of sense, but where exactly does the expectation in $2\alpha_2E[x_1]$ come from?

Similarly, with $x_2\sim N(\mu,\sigma^2)$, $\rho_{x_1 x_2} \neq0$, and a population model...

$Y=\alpha_0+\alpha_1X_1+\alpha_2X_2+\alpha_3X_1X_2+\epsilon$

...if I run OLS omitting the interaction, the bias for the slope of $x_1$ is $\alpha_3 E[x_2]$ and the bias for the slope of $x_2$ is $\alpha_3 E[x_1]$, but I would also like to know how the math works:

$\frac{\partial E[u|x_1,x_2]}{\partial x_1}=\frac{\partial E[\alpha_3x_1x_2|x_1,x_2]}{\partial x_1}=\frac{\partial \alpha_3E[x_1x_2|x_1,x_2]}{\partial x_1}=?$

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  • $\begingroup$ When $\hat \theta$ is used to estimate $\theta$, the bias is defined as $E(\hat\theta) - \theta$. Two questions: 1)which one is parameter $\theta$, and which one is its estimate? 2) Any relation between definition of bias and partial derivative? $\endgroup$ – user158565 Nov 21 '18 at 18:53
  • $\begingroup$ I see.. If you post this as an answer, I will accept it: $bias(\hat{\beta_1})=E[\hat{\beta_1}]-\alpha_1=E[\alpha_1 +2\alpha_2x_1]-\alpha_1=2\alpha_2E[x_1]$ $\endgroup$ – Alvaro Fuentes Nov 21 '18 at 19:27
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When $\hat θ$ is used to estimate $θ$, the bias is defined as E(\hat θ)−θ$. Two questions: 1)which one is parameter θ, and which one is its estimate? 2) Any relation between definition of bias and partial derivative?

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