0
$\begingroup$

I will cite a paper I am reading: (Bugni, Federico A., Ivan A. Canay, and Xiaoxia Shi. "Specification tests for partially identified models defined by moment inequalities." Journal of Econometrics 185.1 (2015): 259-282.)

For a parameter vector $(\theta, F)$, where $\theta \in \Theta$ is a finite dimensional parameter of interest and $F$ denotes the distribution of the observed data, the model states that $$E_F[m_j(W_i,\theta)]\geq0,\;for\;j=1,\ldots,p,$$ $$E_F[m_j(W_i,\theta)]=0,\;for\;j=p+1,\ldots,k,$$ where $\{W_i\}_{i=1}^n$ is an $i.i.d$ sequence of random variables with distribution $F$ and $m:\mathbb{R}^d \times \Theta \rightarrow \mathbb{R}^k$ is a known measurable function.
The model is said to be correctly specified (or statistically adquate) when the moment (in)equalities hold for at least one parameter value.

Here, I am not sure why that the moment (in)equalities hold for at least one parameter value means the model is correctly specified.

I think, the term specification means any assumption or condition imposed by the model. Thus, the moment (in)equalities are also a kind of a specification. And, if any parameter satisfies the (in)equalities given the distribution $F$, we cannot say that the (in)equalities are true. That is the reason why the term "correctly specified" means "when the moment (in)equalities hold for at least one parameter value.

Is it correct?

$\endgroup$

1 Answer 1

1
$\begingroup$

If there is no parameter value that satisfies the moment conditions, then the model is not consistent with the data. This means the when we collect infinitely many data, we will not find a good parameter value that (statistically) satisfies the moment condition. The specification is then considered wrong in a sense the model does not fit data.

I think in this paper's context the specification is said to be correct if there is at least some parameter value satisfies the moment conditions, i.e. it is possible to find some parameter values that satisfy the conditions when we have infinitely many data. It does not require there is a unique parameter value that satisfies the condition, since their focus in the set identification as opposed to the point identification.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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