# Bias of omitting squares and interactions

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:

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

• 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? – user158565 Nov 21 '18 at 18:53
• I see.. If you post this as an answer, I will accept it: $bias(\hat{\beta_1})=E[\hat{\beta_1}]-\alpha_1=E[\alpha_1 +2\alpha_2x_1]-\alpha_1=2\alpha_2E[x_1]$ – Alvaro Fuentes Nov 21 '18 at 19:27

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?