4
$\begingroup$

Suppose there is a latent variable, $y^*_i$ defined by, $$y_i^* = x_i'\beta + u_i$$

Consider the probit assumption that $u_i \sim N(0,1)$ (although the question is analogous for a logistically distributed error and logit).

We observe $y_i=1$ if $y_i^* > 0$ and $y_i =0$ otherwise. Given these assumptions, the probability of $y_i$ conditional on $x_i$ and $\beta$ is: $f(y_i|x_i,\beta) = \Phi(x_i'\beta)^{y_i}(1-\Phi(x_i'\beta))^{1-y_i}$

The MLE estimator chooses $b$ to maximize likelihood: $max_b \prod_{i=1}^n f(y|x,b)$

QUESTION: if $Cov(u_i, u_j)\ne 0$ for some $i$ and $j$, then the marginal distribution of $y_i$ is unchanged, $f(y_i |x_i, \beta)$. However, the likelihood function would change, as observations are no longer independent. The probability of $y_i$ and $y_j$ is not just the product of the two separate probabilities.

I have seen people estimate probit with clustered standard errors, essentially admitting to this problem, however I have not seen a proof or discussion that probit is still consistent in such a case.

Is probit consistent under serial correlation if the marginal distribution is correctly specified?

$\endgroup$

1 Answer 1

4
$\begingroup$

The basic idea is this: The "quasi" log-likelihood (based on the wrong independence assumption) $$\ell(\beta) = \sum_{i=1}^n \log L_i(\beta),$$ where $\log L_i(\beta) = y_i \log \Phi(x_i'\beta) + (1-y_i) \log [1-\Phi(x_i'\beta)]$, is obviously not the true log-likelihood for $(y_1,\ldots,y_n)$, but its expectation is still maximized at the true parameter, so the "quasi" MLE is consistent.

The rests are mathematics. Formally, the proof can go this way:

(i) By the law of large numbers, $n^{-1} \sum_{i=1}^n \log L_i(\beta)$ converges in probability to the limit '$\lim n^{-1} \sum_{i=1}^n E[\log L_i(\beta)]$' uniformly in a compact neighborhood of the true parameter. (If $(x_i,u_i)$ is identically distributed across $i$, then the RHS may be written as $E[\log L_i(\beta)]$.) Note that the convergence requires the autocorrelation to be only weak, which is usually only a technical issue. (For example, $u_i = \xi + e_i$ violates it.)

(ii) The maximizer of $E[\log L_i(\beta)]$ is unique and equals the true parameter, and thus the true parameter also uniquely maximizes $\lim n^{-1} \sum_{i=1}^n E[\log L_i(\beta)]$, the probability limit of the normalized quasi log-likelihood function in (i).

(iii) (i) and (ii) imply that the maximizer of $\ell(\beta)$, the quasi-MLE, converges in probability to the true parameter.

References

I've found this article by Estrella and Rodrigues (1998) from Google: https://www.newyorkfed.org/research/staff_reports/sr39.html (see pages 4-5 and references therein), which refers to:

Gourieroux, C., Monfort, A. and Trognon, A. 1984. Estimation and Test in Probit Models with Serial Correlation. In Florens, J.P., Mouchart, M., Raoult, J.P. and Simar, L. (Eds.).

Poirier, Dale J. and Ruud, Paul A. 1988. Probit with Dependent Observations. Review of Economic Studies 55, 593-614.

Wooldridge, Jeffrey. 1994. Estimation and Inference for Dependent Processes. Handbook of Econometrics. Elsevier, Amsterdam.

$\endgroup$
2
  • $\begingroup$ So the true parameter does in fact still maximize log likelihood? Which of the papers you cite proves this? And is it also true for the logit case? $\endgroup$ May 28 at 5:49
  • 1
    $\begingroup$ The marginal distribution is correctly specified, so $E[\log f(y_i|x_i,\beta)]$ in your notations is maximized by the true parameter. The prob limit of the quasi log-likelihood is the sum (or average) of the correct marginal log-likelihoods, which should be maximized at the true parameter too. I would be surprised if not. The rests are maths involving uniform LLN (alpha-mixing stuff) and continuous mapping. See Poirier and Ruud (1988, Theorem 1). Should be the same for logit. $\endgroup$
    – chan1142
    May 28 at 6:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.