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Suppose that the observations $(y_t, x_t, k_t)_{t=1}^N$ satisfy the linear regression equation:

\begin{equation} \begin{split} y_t = \begin{cases} x_t \beta + e_t & w.p. \; \theta \\ k_t \gamma + e_t & w.p. \; 1-\theta \end{cases} \end{split} \end{equation} with $\mathbb{E}(e_t|x_t, k_t)=0$ and other usual assumptions.

Can I estimate $\beta*\theta$ and $\gamma*(1-\theta)$?

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  • $\begingroup$ What does "w.p." mean?? Are the apparent error terms "$e_t$" the same quantities in both cases or not? $\endgroup$
    – whuber ♦
    Commented Sep 1, 2022 at 17:49
  • $\begingroup$ w.p. is standard notation for 'with probability,' it denotes a random variable. Not sure how the error term being the same is relevant, the error term is defined as the difference between $y_t$ and the conditional mean. Thanks. $\endgroup$
    – Luca Gi
    Commented Sep 1, 2022 at 19:14
  • $\begingroup$ It wouldn't make sense for $\theta$ to be a random variable. Would you mean that there is a lurking random Bernoulli$(\theta)$ variable independent of $e_t$ and that $y_t = kx_t\beta + e_t$ conditional on $U=1$ and otherwise $y_t = k_t\gamma + e_t$ when $U=0$? $\endgroup$
    – whuber ♦
    Commented Sep 1, 2022 at 20:17
  • $\begingroup$ Yes, that's what I mean. Thanks. $\endgroup$
    – Luca Gi
    Commented Sep 1, 2022 at 20:19

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The conditional mean of $y_t$ is $\mathbb{E}(y_t|x_t, k_t) = x_t \beta \theta + k_t \gamma (1-\theta)$, so parameters can be identified with the regression $$y_t = a_1 x_t + a_2 k_t + e_t$$

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