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I am studying a paper where I have the following indicator function: $\text{I}(i) = \mathbf{1} \{ \eta^* - \eta(\beta(i)) + u(i) > 0 \}$ where $\beta(i)$ is taken from uniform distribution $U[\chi-\epsilon, \ \chi+\epsilon]$ and $\eta(i)=\frac{1}{h}\psi^{\frac{-1}{\beta (i)(1+\beta (i))}}$.

The study states that the marginal effect is given by: $$\frac{\partial \text{I}(i)}{\partial \beta(i)} = \frac{1}{2\varepsilon} \left|\frac{d\beta(i)}{d\eta(i)}\right|$$ Would you please explain how to derive this result and give me an intuition of how it works? Thanks in advance.

The paper is "Patience and Comparative Development, The Review of Economic Studies, Volume 89, Issue 5, October 2022" .

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    $\begingroup$ A reference to the paper would be helpful. $\endgroup$ Jan 14 at 23:14
  • $\begingroup$ What is the 1 ? Is the probability of ... ? What is $\eta(\beta(i))$? Without these info, it is impossible to answer! Btw, the $1/2\epsilon$ suggets that the 1 is the probability of something, since is derived from the CdF of a uniformly distributed random variable. $\endgroup$ Jan 15 at 8:37
  • $\begingroup$ @Dario, could you please tell us which paper is this to take a look? $\endgroup$ Jan 15 at 15:30
  • $\begingroup$ The paper is "Patience and Comparative Development, The Review of Economic Studies, Volume 89, Issue 5, October 2022" . You will find this expression at section 5.1 $\endgroup$
    – Dario
    Jan 17 at 20:58

1 Answer 1

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As a matter of mathematical trivia, the derivative of an indicator function is zero, except at the threshold where it does not exist.

But \begin{align} \mathbb E[\text{I}(i)] &= \mathbb E\big[\mathbf{1} \{ \eta^* - \eta(\beta(i)) + u(i) > 0 \}\big]\\ &=\Pr\big[\eta^* - \eta(\beta(i)) + u(i) > 0\big]\\ &= \Pr\big[ \eta(\beta(i)) \leq \eta^* + u(i)\big]\\ &=\Pr\big[ \beta(i) \leq \eta^{-1}(\eta^* + u(i))\big]. \end{align} ...assuming invertibility.

This now is the distribution function of the Uniform random variable in question, $$\Pr\big[ \beta(i) \leq \eta^{-1}(\eta^* + u(i))\big] = \frac{\eta^{-1}(\eta^* + u(i))- (\chi-\epsilon)}{\chi+\epsilon-(\chi-\epsilon)} = \frac{\eta^{-1}(\eta^* + u(i))- (\chi-\epsilon)}{2\epsilon}.$$

Then

$$\frac{\partial \mathbb E[\text{I}(i)]}{ \partial \beta(i)} = \frac 1 {2\epsilon}\cdot \frac{d\eta^{-1}}{d\beta(i)},$$

and by applying the inverse function theorem (it appears so, but why and how it is applicable here?), we get $$\frac{\partial \mathbb E[\text{I}(i)]}{ \partial \beta(i)} = \frac 1 {2\epsilon}\cdot \frac{1}{\frac{d\eta}{d\beta(i)}}= \frac 1 {2\epsilon}\cdot \frac{d\beta(i)}{d\eta}.$$

So it could be the "marginal effect" on the expected value of the indicator function.

All these appear, and are, a bit too informal though, and the details in the paper are crucial. So I also urge the OP to provide these details or at least a full reference to the paper.

ADDENDUM

As reported by the OP,

The paper is "Patience and Comparative Development, The Review of Economic Studies, Volume 89, Issue 5, October 2022"

Looking at the paper, the expression in question is eq. $(5.11)$. Just before it the authors write

Therefore, we represent the empirical analogue of the decision to become skilled as a linear probability model..."

and it is clear that

a) This indicator function is the backbone of a standard binary choice / conditional probability model

b) Any marginal effects are on the probability/expected value of indicator of the event being realized.

The authors refer for details to their on-line Appendix "G". Going there (Section $G1$), we read

"The marginal effect of an increase in patience on the propensity of becoming skilled can then be expressed as"

...and then they use the symbol $\partial I /\partial \beta$. But "the propensity" is the probability, the expected value of, and not the 0-1 indicator function itself. So, it appears, that this is just a notational mistake, and what I wrote above holds.

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  • $\begingroup$ Thanks a lot! The paper is "Patience and Comparative Development, The Review of Economic Studies, Volume 89, Issue 5, October 2022" . You will find this expression at section 5.1. If you find out more you are more than welcomed to share. $\endgroup$
    – Dario
    Jan 17 at 20:59
  • $\begingroup$ @Dario Added some things. $\endgroup$ Jan 17 at 22:36

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