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first question on StackExchange; thank you for having me.

I am trying to really nail the intuition for the Heckman sample selection model. One little thing that is bothering me is the assumption made on the error terms for the case where we estimate using two-steps (i.e., not the ML case).

Our model: $$ {y}_{1}={x}_{1}{\beta}_{1}+{u}_{1} \\ s=1(x{\delta}_{2}+{v}_{2}>0) $$ where $x=({x}_{1},{x}_{2})$.

We assume $({u}_{1},{v}_{2})$ are independent of $x$ and ${v}_{2}\sim N(0,1)$. This implies the selection equation is Probit.

The assumption I am struggling with is: $$ E({U}_{1}|{V}_{2})=\gamma {v}_{2} $$

Why is the conditional mean of the error in the structural equation increasing in the error in the selection equation? Does $\gamma$ need to be greater than 0? I understand that this assumption is important for being able to write:

$$ E({y}_{1}|x,s=1,{v}_{2})={x}_{1}{\beta}_{1}+\gamma {v}_{2} $$

which is intuitive: if we run OLS on the truncated data our slope will be shallower than the ``true" line (for truncation at 0). What I cannot understand is why unobservables in the selection equation have a smaller mean than unobservables in the structural equation? Is it because the selection outcome is binary, so has less variance?

Any insights would be great!

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  • $\begingroup$ Cheers Kjetil!! $\endgroup$ – singlepeaked Nov 11 '14 at 15:12
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In the LIML, two-step version of this estimator, we just make an assumption that $u_1 = \gamma v_2 + \xi$, where $\xi$ is independent of $v_2$. You can see this explained in Cameron and Trivedi's Microeconometrics on p. 551 or in "Adult" Wooldridge on p. 803. Neither book describes this as shared property of some family of bivariate distributions, and I am not aware of any such family. This might actually make a very good separate question.

In the FIML version, the fact that $\mathbb E (U_1 | V_2=v_2) = \gamma v_2$ is a consequence of joint normality of the errors, their mean being zero, and their correlation being nonzero. If

$$ \begin{pmatrix} U_1 \\ V_2 \end{pmatrix} \sim \mathcal{N} \left( \begin{pmatrix} \mu_1 \\ \mu_2 \end{pmatrix} , \begin{pmatrix} \sigma^2_1 & \rho \sigma_1 \sigma_2 \\ \rho \sigma_1 \sigma_2 & \sigma^2_2 \end{pmatrix} \right) $$

then the conditional mean is $$\mathbb E (U_1 | V_2=v_2) = \mu_1 + \rho \frac{\sigma_1}{\sigma_2}(v_2 - \mu_2).$$ The $\mu$s are zero by assumption and they drop out, so your $\gamma=\rho \frac{\sigma_1}{\sigma_2}$. The $\sigma$s are both positive since they are standard deviations, so the conditional mean really depends on the sign of $\rho$, which is the correlation between $u$ and $v$ and is somewhere between -1 and 1. But what does that correlation mean?

Let's take a concrete example to fix the idea. In the case of a purchase on a durable good, the behavioral interpretation of the conditional mean is that the expected error term in the expenditure equation is a multiple of the error in the purchase decision equation. Is that multiple positive? Is it greater than 1?

We might argue that $\rho$ is likely to be positive in the durable goods case, so the multiple is positive. The errors are all the things the econometrician doesn't get to observe, like the fact that your mother's birthday is coming up or that the kids need braces. These are the random components of utility, which we need to explain why seemingly identical people make different purchase decisions. If you really want to buy something (big $v_2>0$), it makes sense that you may be willing to pay more for it (big $u_1>0$).

The magnitude of the effect is a bit more ambiguous. It depends on the ratio of the standard deviations. Usually, $\sigma_2$ is normalized to $1$ to make things easier, so if the variance of the unobserved traits that make people pay more is large, then $\gamma$ is the product of a number between $0$ and $1$ and large positive number, so we might expect $\gamma$ to be greater than one.

Obviously, in another context, the sign and the magnitude may be different.

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  • $\begingroup$ Thanks Dimitriy. The example is very useful. I was under the impression that bivariate normality was just one distribution which satisfies the assumption that $E({U}_{1}|{V}_{2})=\gamma {v}_{2}$? So we assume bivariate normality for the case when we want to estimate the model by ML, but it isn't an assumption we necessarily make for 2-step estimation? $\endgroup$ – singlepeaked Nov 12 '14 at 9:07
  • $\begingroup$ @adsh13ab I should have read your question more carefully. I edited the answer. $\endgroup$ – Dimitriy V. Masterov Nov 12 '14 at 19:36
  • $\begingroup$ @adsh13ab Flipping through Continuous Bivariate Distributions by Balakrishnan and Lai, I don't see an obvious pattern about when the expectation is linear like that. The classical t, F, triangular, some Gammas, $\chi^2$, exponential, and normal are examples. $\endgroup$ – Dimitriy V. Masterov Nov 12 '14 at 22:37
  • $\begingroup$ @Dimitriy V. Masterov Wooldridge, in his book Introductory Econometrics, in deriving Heckman model starts from the assumption that explanatory variables are independent of both error terms. How is that possible? Can errors be mutually correlated and, at the same time, explanatory variables be independent of error terms? Don't we use Heckman when errors are correlated and explanatory variables dependent of error terms? $\endgroup$ – Quirik Jul 3 '16 at 10:13
  • $\begingroup$ The selection process is what creates the dependence. $\endgroup$ – Dimitriy V. Masterov Jul 3 '16 at 15:04

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