6
$\begingroup$

I'm working on a project that involves the use of a Heckman selection model (more specifically a Roy or move-stay model, which is essentially a two-sided Heckman) of the following form:

$$ Y_{i1} = X_i\beta_1 + \varepsilon_{i1} \text{ if } Select_i=1 $$ $$ Y_{i0} = X_i\beta_0 + \varepsilon_{i0} \text{ if } Select_i=0 $$ $$ Select_i^* = X_i\gamma + Z_i\delta + \varepsilon_{i} $$ $$ Select_i = I(Select_i^*>0) $$

The model is identified only by the distributional assumptions on the error terms unless a variable is included in the first stage (the $Select_i^*$ equation) that can be plausibly excluded from the second stage (the $Y_{i1}$ and $Y_{i2}$ equations). Here this variable is $Z_i$.

Is there a rule of thumb for how "strong" the relationship between $Select_i$ and $Z_i$ needs to be for the model to be adequately identified? I'm thinking something like the rule of thumb in the instrumental variables literature that an F-test of the IV's should be 10 or above. I haven't been able to find anything similar for selection models. Is there something?

$\endgroup$
  • $\begingroup$ The notion of identifiability does not go with testing, I believe. In the current setting, the observables are from a mixture$$\Phi(X_i\gamma + Z_i\delta)\mathcal{N}(X_i\beta_1,\sigma^2)+\Phi(-X_i\gamma - Z_i\delta)\mathcal{N}(X_i\beta_2,\sigma^2)$$ so any change of the parameters, individually or jointly, does modify the distribution of the observables. $\endgroup$ – Xi'an Feb 6 '15 at 19:14
  • $\begingroup$ The method I am using is a fairly standard one, as is the argument for its identification (in the econometric sense anyway, see, e.g., the last bullet point en.wikipedia.org/wiki/Heckman_correction#Disadvantages ). I am not expecting a test to give me a formal test of the identification, but wondering if there is anything like the rule-of-thumb test in the IV literature (which checks for "weak instruments") in the selection literature. Sorry for not being clear. $\endgroup$ – NickCHK Feb 6 '15 at 19:43

Your Answer

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

Browse other questions tagged or ask your own question.