Suppose that I have the following model:

$y_i = \beta_0 + \beta_1 x_{1i} + \beta_2 x_{2i} + u_i$

where $\hat{\beta_k}$, k=0,1,2 , are estimated by the method of least squares, using a sample of size n.

How can I show that if $z_i =a_0 + a_1 x_{1i} + a_2 x_{2i} $, where $a_k$ are constants, then:

$\sum_{i=1}^n z_i(y_i-\hat{\beta_0} - \hat{\beta_1} x_{1i} - \hat{\beta_2} x_{2i})=0$ ?

My attempt was to try to expand and compare to the first order condition: $\sum_{i=1}^n (y_i-\hat{\beta_0} - \hat{\beta_1} x_{1i} - \hat{\beta_2} x_{2i})=0$, but things got messy and lead to nowhere.

  • 1
    $\begingroup$ You probably should have the self study tag. $\endgroup$ Commented Jun 21, 2018 at 21:49

1 Answer 1


Let's recall some features of $\hat{u_{i}}$, which are central to this problem.

By definition, $$\hat{u}_{i} = y_{i} - \hat{\beta}_{0} - \hat{\beta}_{1}x_{1i} - \hat{\beta}_{2}x_{2i}.$$ From the FOC with respect to $\beta_{ki}$, k = 1,2, in the minimization of least-squared errors, one can find that $Cov(x_{ki},\hat{u}_{i}) = 0$. Additionally, minimizing the objective function with respect to $\beta_{0}$ yields $E[\hat{u}_{i}] = 0$. By construction, $z_{i}$ is purely a function of $x_{1i}$ and $x_{2i}$. Hence, it's not stochastic in any channel other than through $x_{ki}$.

If $z_{i}$ is a function of the $x_{ki}$, which are uncorrelated with $\hat{u}_{i}$, then so is $z_{i}$. I'll leave it to you to more rigorously fill in the details I provided.


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