This question is related to extremum estimators. I am looking to propose an estimator that I think may be a Tobit estimator, but I am unsure. The question is as follows:

Let $z$ be a random variable with a strictly increasing CDF $F( \dot ) $ everywhere on its support. The median of $Z$, $\gamma$ is defined as:

$ \int _{-\infty, \gamma} dF(z) > \frac{1}{2}$ ; $ \int _{\gamma, \infty} dF(z) > \frac{1}{2}$

It is known that $\gamma$ minimizes an expected absolute deviatin criterion: $\gamma = argmin_e E(|Z-e| - |Z|)$.

We consider a model $Y=max\{c, f(X; \theta)+\epsilon\}$ where c is a known constant and the conditional median of $\epsilon$ given $X$ is 0. Suppose we have an i.i.d sample $\{X_i, Y_i\}$, that is generated by this model.

1) Propose an extremum estimator for $\theta$.

Here, I was thinking that a Tobit model would be appropriate since the model is truncated by the constant $c$.

2) Show the estimator you proposed is consistent.

  • $\begingroup$ Yes, that is what I meant, the question is now corrected. $\endgroup$
    – A. D.
    Jul 11, 2018 at 19:24
  • $\begingroup$ Where does the variable $Z$ enter the model? Also is the model $Y = \max \{c, f(X,\theta)\}$ $\endgroup$ Jul 11, 2018 at 20:41
  • $\begingroup$ That is the correct model. I don't understand where Z enters the model, that is part of where my confusion lies. $\endgroup$
    – A. D.
    Jul 11, 2018 at 21:01
  • $\begingroup$ Well, then probably $Z$ is just general notation to introduce the median. In the model it would correspond to the error term. $\endgroup$ Jul 11, 2018 at 21:03

1 Answer 1


First of all, this is not a Tobit-type estimator. A Tobit estimator is a maximum-likelihood estimator, which is based on a full specification of the error distribution. You have only the median specified, so it can't be an MLE.

What can we do? To begin with, note that, for a random variable $Z$, the median of $\max(c,Z)$ is $\max(c, \mathrm{med}(Z))$, which can be proved by using the definition of the median.

Thus, for your problem, you have $$\mathrm{med}(y|X) = \max(c, f(X,\theta))$$ under the assumption that $\mathrm{med}(\epsilon|X)=0$. A natural extremum estimator would be the minimizer of $$\frac{1}{n} \sum_{i=1}^n |y_i - \max(c, f(X_i,\theta)) |.$$ You defined the median as the minimizer of $E(|Z-e|-|Z|)$, but this is the same as the minimizer of $E|Z-e|$ without $-|Z|$, so I dropped the "$-|y_i|$" from the objective function.

This is a generalization of Powell (1984), Least absolute deviations estimator for the censored regression model, Journal of Econometrics, 25, 303-325. See the paper for more details including consistency. https://web.stanford.edu/~doubleh/eco273/powell1984.pdf


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