<p>As a matter of mathematical trivia, the derivative of an indicator function is zero, except at the threshold where it does not exist.</p>
<p>But
<span class="math-container">\begin{align}
\mathbb E[\text{I}(i)] &amp;= \mathbb E\big[\mathbf{1} \{ \eta^* - \eta(\beta(i)) + u(i) &gt; 0 \}\big]\\
&amp;=\Pr\big[\eta^* - \eta(\beta(i)) + u(i) &gt; 0\big]\\
&amp;= \Pr\big[ \eta(\beta(i)) \leq \eta^* + u(i)\big]\\
&amp;=\Pr\big[ \beta(i) \leq \eta^{-1}(\eta^* + u(i))\big].
\end{align}</span>
...assuming invertibility.</p>
<p>This now is the distribution function of the Uniform random variable in question,
<span class="math-container">$$\Pr\big[ \beta(i) \leq \eta^{-1}(\eta^* + u(i))\big] = \frac{\eta^{-1}(\eta^* + u(i))- (\chi-\epsilon)}{\chi+\epsilon-(\chi-\epsilon)} = \frac{\eta^{-1}(\eta^* + u(i))- (\chi-\epsilon)}{2\epsilon}.$$</span></p>
<p>Then</p>
<p><span class="math-container">$$\frac{\partial \mathbb E[\text{I}(i)]}{ \partial \beta(i)} = \frac 1 {2\epsilon}\cdot \frac{d\eta^{-1}}{d\beta(i)},$$</span></p>
<p>and by applying the inverse function theorem (it appears so, but why and how it is applicable here?), we get
<span class="math-container">$$\frac{\partial \mathbb E[\text{I}(i)]}{ \partial \beta(i)} = \frac 1 {2\epsilon}\cdot \frac{1}{\frac{d\eta}{d\beta(i)}}= \frac 1 {2\epsilon}\cdot \frac{d\beta(i)}{d\eta}.$$</span></p>
<p>So it <em>could</em> be the &quot;marginal effect&quot; on the <em>expected value</em> of the indicator function.</p>
<p>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.</p>
<p><strong>ADDENDUM</strong></p>
<p>As reported by the OP,</p>
<blockquote>
<p>The paper is &quot;Patience and Comparative Development, The Review of
Economic Studies, Volume 89, Issue 5, October 2022&quot;</p>
</blockquote>
<p>Looking at the paper, the expression in question is eq. <span class="math-container">$(5.11)$</span>. Just before it the authors write</p>
<blockquote>
<p><em>Therefore, we represent the empirical analogue of the decision to
become skilled as a linear probability model...&quot;</em></p>
</blockquote>
<p>and it is clear that</p>
<p><strong>a)</strong> This indicator function is the backbone of a standard binary choice / conditional probability model</p>
<p><strong>b)</strong> Any marginal effects are on the probability/expected value of indicator of the event being realized.</p>
<p>The authors refer for details to their on-line Appendix &quot;G&quot;. Going there (Section <span class="math-container">$G1$</span>),  we read</p>
<blockquote>
<p><em>&quot;The marginal effect of an increase in patience on the <strong>propensity</strong> of
becoming skilled can then be expressed as&quot;</em></p>
</blockquote>
<p>...and then they use the symbol <span class="math-container">$\partial I /\partial \beta$</span>. But &quot;the propensity&quot; 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.</p>

CC BY-SA 4.0
2024-01-17T20:59:24.353Z
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.
Dario
CC BY-SA 4.0
2024-01-17T22:36:10.803Z
@Dario Added some things.
Alecos Papadopoulos
CC BY-SA 4.0
2024-01-15T21:42:16.177Z
2024-01-17T22:35:56.813Z
2024-01-17T22:35:56.813Z
Alecos Papadopoulos
<p>I am studying a paper where I have the following indicator function: <span class="math-container">$\text{I}(i) = \mathbf{1} \{ \eta^* - \eta(\beta(i)) + u(i) &gt; 0 \}$</span> where <span class="math-container">$\beta(i)$</span> is taken from uniform distribution <span class="math-container">$U[\chi-\epsilon, \ \chi+\epsilon]$</span> and <span class="math-container">$\eta(i)=\frac{1}{h}\psi^{\frac{-1}{\beta (i)(1+\beta (i))}}$</span>.</p>
<p>The study states that the marginal effect is given by: <span class="math-container">$$\frac{\partial \text{I}(i)}{\partial \beta(i)} = \frac{1}{2\varepsilon} \left|\frac{d\beta(i)}{d\eta(i)}\right|$$</span>
Would you please explain how to derive this result and give me an intuition of how it works?
Thanks in advance.</p>
<p>The paper is &quot;Patience and Comparative Development, The Review of Economic Studies, Volume 89, Issue 5, October 2022&quot; .</p>

CC BY-SA 4.0
2024-01-14T23:14:33.437Z
A reference to the paper would be helpful.
Michael Greinecker
CC BY-SA 4.0
2024-01-15T08:37:46.857Z
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.
Matteo Bulgarelli
CC BY-SA 4.0
2024-01-15T15:30:32.753Z
@Dario, could you please tell us which paper is this to take a look?
Oliver Queen
CC BY-SA 4.0
2024-01-17T20:58:29.453Z
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
Dario
<econometrics><mathematical-economics><statistics>
"Derivative" of an indicator function
Dario
# "Derivative" of an indicator function

(Asked by: username_0 on 2024-01-14)

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" .

**Tags:** econometrics, mathematical-economics, statistics
> **Comments:**
> **username_1**: A reference to the paper would be helpful.  
> **username_2**: 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.  
> **username_3**: @Dario, could you please tell us which paper is this to take a look?  
> **username_0**: 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  

---

## 1 answers

### Answer by username_4 on 2024-01-15. Score: 11 | Accepted answer

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.

> **Comments:**
> **username_0**: 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.  
> **username_4**: @Dario Added some things.  

---

