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I got an assignment to solve, which states the following:

Consider the model with heteroskedasticity: $$ Y_i = X_i'\beta + \epsilon_i $$ where $\epsilon_i$ is explicitly assumed as i.i.d. with $$ \epsilon_i \sim N(0,\sigma_i^2),\sigma_i^2 = \exp(Z_i'\gamma) $$ $Z_i'$ is a $m x 1$ vector including a constant term. $X_i$ is a $kx1$ vector of explanatory variables. No further assumptions are stated.

I am supposed to solve for the parameters of the error term distribution by (unconditional) Maximum Likelihood. In the solution, the classical result of equivalence of the ML estimator and the WLS estimator turns out as a result. This is not the problem, however, I am struggling with the setup. As far as I understood from resources such as Wooldridge's 'Econometrics for Cross-Section and Panel' or Hayashi's 'Econometrics', the i.i.d. assumption of $\epsilon_i$ implies that $Var(\epsilon_i)$ is homoskedastic by construction. In the assignment, the error term is assumed to have identical distribution, however is heteroskedastic. How does this make sense?

My guess would be that something is missing in the exercise. I am also a bit confused that the ML estimator coincides with the WLS estimator since I thought the problem of Heteroskedasticity in OLS can be solved by WLS only if the error term is heteroskedastic conditionally on the explanatory variables $X_i$. In addition, the above mentioned textbooks show the equivalence of OLS and MLE mainly using conditional distributions and conditional MLE. I think I am missing something here but I am not quite sure what.

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    $\begingroup$ I was about to ask how the error can be heteroskedastic yet iid and was quite delighted to see you point out this apparent contradiction a few sentences later! I think there is a typo and that you have a legitimate gripe to discuss with your instructor. $\endgroup$
    – Dave
    Commented Mar 2, 2022 at 10:42
  • $\begingroup$ @Dave thank you for confirming my issues with the exercise. But would the exercise make sense if we leave out the iid assumption and simply say the errors are independent? Or would it only work if we assume that the error is conditional on $X_i'$ normally distributed with conditional mean zero and conditional heteroskedasticity, where the variance is some function of the $X_i'$ ? And then use Conditional Maximum likelihood to estimate the parameters? $\endgroup$
    – Hokkaido21
    Commented Mar 2, 2022 at 11:36
  • $\begingroup$ I’d run it by your instructor, who can comment on what the question was supposed to ask. $\endgroup$
    – Dave
    Commented Mar 2, 2022 at 11:39
  • $\begingroup$ Hi: I think you are right that he didn't mean to include the term "identical". But, as far as solving for the parameters, you don't have to think of it as conditional MLE. Even, when you use MLE instead of OLS ( in the classical formulation ), the likelihood is dependent on the X's yet they don't call it conditional MLE. So, I would just think of it as the instructor making a mistake by using the term "identical" and the solve it by writing out the likelihood and maximizing it. ( and call it the MLE ). $\endgroup$
    – mlofton
    Commented Mar 2, 2022 at 14:44
  • $\begingroup$ @mlofton thanks. However, I am still thinking about one more thing: I assumed, it is possible to estimate the model above with MLE also using $Y_i$ as random variable and its' distribution. That is, use $E[Y_i]=E[X_i'\beta + \epsilon_i] = E[X_i'\beta]$ as $E[\epsilon_i]=0$. Then the mean of the distribution of $Y_i$ is $E[X_i'\beta]$ and the variance can be obtained similarly. However, the issue is, that $X_i$ is also random. and I cannot work with the expectation of it to obtain the OLS estimator. Thus, the Q is, whether we need to treat X as nonrandom to use the distribution of $Y_i$ here? $\endgroup$
    – Hokkaido21
    Commented Mar 3, 2022 at 10:08

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