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I was trying to find the OLS estimator for the model:

$Y$ = $\beta_0$ + $\beta_1X_{1t}$ + $\beta_2X_{2t}$ +.......+ $\beta_5X_{5t}$ + $e$

t = 1,2,3 ......, 50 time ordered observations

X is a full ordered rank matrix with 5 columns

and error term is distributed as:

$e_t = \lambda + \gamma*e_{t-1} + \rho*x_{3t} + u_t$

$u_t = N(0, \sigma^2$)

This is how I was doing it:

$Var(e_t) = Var(\lambda + \gamma*e_{t-1} + \rho*x_{3t} + u_t)=$

$Var(\gamma*e_{t-1} + \rho*x_{3t} + u_t) =$

$\gamma^2*Var(e_{t-1}) + \rho^2*Var(x_{3t}) + \sigma^2 + 2*\gamma*cov(e_{t-1}, u_t) + 2*\rho*\gamma*cov(e_{t-1},x_{3t}) + 2*\rho*cov(x_{3t}, u_t)$

$Cov(e_{t-1}, u_t) = 0$

$Cov(x_{3t}, u_t) = 0$

How do I proceed further?

Since once we have the sample, can we simply say that $X_{3t}$ will be known and hence a constant

so

$Var(x_{3t}) = 0$

$cov(e_{t-1},x_{3t}) = 0$

So, the variance equation is reduced to :

$Var(e_t) =\gamma^2*Var(e_{t-1}) + \sigma^2$ and we just proceed similar to AR(1)

Is this correct or I am awfully wrong here? Please help me out!

Also, how do we correct for mean since the conditional mean independence assumption is violated here. I am not sure how to correct for it

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  • $\begingroup$ Anyone can help me out here? $\endgroup$
    – Raghav Goyal
    Mar 13, 2019 at 15:52
  • $\begingroup$ I am confused: What is the model equation? I get you have an equation for the error...but where is the equation including dependent variable? Also why use $x_3$ where is $x_1,x_2$ are they anywhere? $\endgroup$
    – Jesper for President
    Mar 13, 2019 at 19:20
  • $\begingroup$ @JesperHybel made the changes. Sorry for the confusion $\endgroup$
    – Raghav Goyal
    Mar 13, 2019 at 20:11

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