<p>The problem is to solve an ODE.</p>
<p>Notice, since <span class="math-container">$\ell(\varepsilon,p)$</span> is a density function, the function <span class="math-container">$G(p):=\int_{p_{min}}^p\ell(\varepsilon,x)dx$</span> is absolutely continuous. So, by the Fundamental Theorem of Calculus, <span class="math-container">$G'(p):=g(p)=\ell(\varepsilon,p)$</span>.</p>
<p>Therefore, the equation you state can be rewritten in terms of <span class="math-container">$G$</span> and <span class="math-container">$g$</span> as <span class="math-container">$$(1+\kappa_1 \bar{F}(p))g(p)=[h(\varepsilon)+\kappa_1G(p)]f(p)$$</span>
We can solve this directly, <span class="math-container">$$\begin{align} (1+\kappa_1 \bar{F}(p))g(p)&amp;=[h(\varepsilon)+\kappa_1G(p)]f(p) \\
(1+\kappa_1 \bar{F}(p))g(p)-\kappa_1f(p)G(p)&amp;=h(\varepsilon)f(p) \\
\frac{d}{dp}[(1+\kappa_1\bar{F}(p))G(p)] &amp;= h(\varepsilon)f(p) \\
(1+\kappa_1\bar{F}(p))G(p) &amp;= h(\varepsilon)F(p) \\
G(p) &amp;= \frac{h(\varepsilon)F(p)}{1+\kappa_1\bar{F}(p)} \end{align}$$</span></p>
<p>Differentiation gives, <span class="math-container">$$\begin{align} g(p) &amp;= G'(p) \\
&amp;= \frac{h(\varepsilon)f(p)}{1+\kappa_1\bar{F}(p)}-\frac{h(\varepsilon)F(p)(-\kappa_1f(p))}{(1+\kappa_1\bar{F}(p))^2} \\
&amp;= \frac{1+\kappa_1\bar{F}(p)+\kappa_1F(p)}{(1+\kappa_1\bar{F}(p))^2}h(\varepsilon)f(p) \\
&amp;= \frac{1+\kappa_1}{(1+\kappa_1\bar{F}(p))^2}h(\varepsilon)f(p) \end{align}$$</span></p>
<p>Since <span class="math-container">$g(p)=\ell(\varepsilon,p)$</span>, we have proven that <span class="math-container">$$\ell(\varepsilon,p)=\frac{1+\kappa_1}{(1+\kappa_1\bar{F}(p))^2}h(\varepsilon)f(p)$$</span></p>

CC BY-SA 4.0
2024-07-05T12:10:35.523Z
2024-07-05T12:10:35.523Z
Joseph Basford
<p>In Postel-Vinay &amp; Robin 2002 they show an equation: <span class="math-container">$$\left\{\delta+\mu+\lambda_{1} \bar{F}(p)\right\} \ell(\varepsilon, p)=\left\{(\delta+\mu) h(\varepsilon)+\lambda_{1} \int_{p_{\min }}^{p} \ell(\varepsilon, x) d x\right\} f(p)$$</span> where <span class="math-container">$h$</span> is the pdf of <span class="math-container">$ε$</span>, and <span class="math-container">$\bar{F}(p)=1-F(p)$</span> is the inverse cdf of <span class="math-container">$p$</span>.</p>
<p>They then say that this solves as  <span class="math-container">$$\ell(\varepsilon, p)=\frac{1+\kappa_{1}}{\left[1+\kappa_{1} \bar{F}(p)\right]^{2}} h(\varepsilon) f(p)$$</span> where <span class="math-container">$\kappa_{1}=\lambda_{1} /(\delta+\mu)$</span>, but I can't see how.</p>
<p>Also note that at this time we don't know that if the joint distribution is independent or not because the authors integrate this result to get <span class="math-container">$\ell(p)=\frac{1+\kappa_{1}}{\left[1+\kappa_{1} \bar{F}(p)\right]^{2}} f(p)$</span>  and thus <span class="math-container">$\ell(\varepsilon, p)=h(\varepsilon) \ell(p)$</span> and finally argue that <span class="math-container">$ε$</span> and <span class="math-container">$p$</span> are uncorrelated.</p>

CC BY-SA 4.0
2022-07-13T13:19:08.037Z
Is your curly bracket $\{ \}$ means anything? or it has the same use as regular bracket $()$?
Redsbefall
CC BY-SA 4.0
2022-07-13T13:33:39.673Z
@Redsbefall Just a regular one.
Alalalalaki
CC BY-SA 4.0
2022-07-13T13:37:19.340Z
Can you share the link to the e-book?
Redsbefall
CC BY-SA 4.0
2022-07-13T13:48:03.937Z
@Redsbefall It's not a book it's a paper on econometrica. https://onlinelibrary.wiley.com/doi/10.1111/j.1468-0262.2002.00441.x
Alalalalaki
CC BY-SA 4.0
2022-07-19T12:23:20.980Z
Why dont you ask the author? (e-mail etc.) If it is from paper it is usually not elementary, also by looking at the equation.
Redsbefall
CC BY-SA 4.0
2022-07-19T14:05:51.890Z
@Redsbefall This is a pure math problem, i.e. solving the algebra. I am not sure it's a good idea to ask the authors for these kinds of questions. I think this site is more suitable.
Alalalalaki
CC BY-SA 4.0
2022-07-19T14:35:58.700Z
Then try ask at mathematics Stack exchange, or MathOverflow for research type question. Since this is math problem.
Redsbefall
<mathematical-economics><labor-economics>
Solve Joint Distribution From An Equation
Alalalalaki
# Solve Joint Distribution From An Equation

(Asked by: username_0 on 2022-06-22)

In Postel-Vinay & Robin 2002 they show an equation: $$\left\{\delta+\mu+\lambda\_{1} \bar{F}(p)\right\} \ell(\varepsilon, p)=\left\{(\delta+\mu) h(\varepsilon)+\lambda\_{1} \int\_{p\_{\min }}^{p} \ell(\varepsilon, x) d x\right\} f(p)$$ where $h$ is the pdf of $ε$, and $\bar{F}(p)=1-F(p)$ is the inverse cdf of $p$.

They then say that this solves as $$\ell(\varepsilon, p)=\frac{1+\kappa\_{1}}{\left[1+\kappa\_{1} \bar{F}(p)\right]^{2}} h(\varepsilon) f(p)$$ where $\kappa\_{1}=\lambda\_{1} /(\delta+\mu)$, but I can't see how.

Also note that at this time we don't know that if the joint distribution is independent or not because the authors integrate this result to get $\ell(p)=\frac{1+\kappa\_{1}}{\left[1+\kappa\_{1} \bar{F}(p)\right]^{2}} f(p)$ and thus $\ell(\varepsilon, p)=h(\varepsilon) \ell(p)$ and finally argue that $ε$ and $p$ are uncorrelated.

**Tags:** mathematical-economics, labor-economics
> **Comments:**
> **username_1**: Is your curly bracket $\{ \}$ means anything? or it has the same use as regular bracket $()$?  
> **username_0**: @Redsbefall Just a regular one.  
> **username_1**: Can you share the link to the e-book?  
> **username_0**: @Redsbefall It's not a book it's a paper on econometrica. https://onlinelibrary.wiley.com/doi/10.1111/j.1468-0262.2002.00441.x  
> **username_1**: Why dont you ask the author? (e-mail etc.) If it is from paper it is usually not elementary, also by looking at the equation.  
> **username_0**: @Redsbefall This is a pure math problem, i.e. solving the algebra. I am not sure it's a good idea to ask the authors for these kinds of questions. I think this site is more suitable.  
> **username_1**: Then try ask at mathematics Stack exchange, or MathOverflow for research type question. Since this is math problem.  

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## 1 answers

### Answer by username_2 on 2024-07-05. Score: 2 | Accepted answer

The problem is to solve an ODE.

Notice, since $\ell(\varepsilon,p)$ is a density function, the function $G(p):=\int\_{p\_{min}}^p\ell(\varepsilon,x)dx$ is absolutely continuous. So, by the Fundamental Theorem of Calculus, $G'(p):=g(p)=\ell(\varepsilon,p)$.

Therefore, the equation you state can be rewritten in terms of $G$ and $g$ as $$(1+\kappa\_1 \bar{F}(p))g(p)=[h(\varepsilon)+\kappa\_1G(p)]f(p)$$
We can solve this directly, $$\begin{align} (1+\kappa\_1 \bar{F}(p))g(p)&=[h(\varepsilon)+\kappa\_1G(p)]f(p) \\
(1+\kappa\_1 \bar{F}(p))g(p)-\kappa\_1f(p)G(p)&=h(\varepsilon)f(p) \\
\frac{d}{dp}[(1+\kappa\_1\bar{F}(p))G(p)] &= h(\varepsilon)f(p) \\
(1+\kappa\_1\bar{F}(p))G(p) &= h(\varepsilon)F(p) \\
G(p) &= \frac{h(\varepsilon)F(p)}{1+\kappa\_1\bar{F}(p)} \end{align}$$

Differentiation gives, $$\begin{align} g(p) &= G'(p) \\
&= \frac{h(\varepsilon)f(p)}{1+\kappa\_1\bar{F}(p)}-\frac{h(\varepsilon)F(p)(-\kappa\_1f(p))}{(1+\kappa\_1\bar{F}(p))^2} \\
&= \frac{1+\kappa\_1\bar{F}(p)+\kappa\_1F(p)}{(1+\kappa\_1\bar{F}(p))^2}h(\varepsilon)f(p) \\
&= \frac{1+\kappa\_1}{(1+\kappa\_1\bar{F}(p))^2}h(\varepsilon)f(p) \end{align}$$

Since $g(p)=\ell(\varepsilon,p)$, we have proven that $$\ell(\varepsilon,p)=\frac{1+\kappa\_1}{(1+\kappa\_1\bar{F}(p))^2}h(\varepsilon)f(p)$$


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