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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.

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  • $\begingroup$ Is your curly bracket $\{ \}$ means anything? or it has the same use as regular bracket $()$? $\endgroup$
    – Redsbefall
    Commented Jul 13, 2022 at 13:19
  • $\begingroup$ @Redsbefall Just a regular one. $\endgroup$
    – Alalalalaki
    Commented Jul 13, 2022 at 13:33
  • $\begingroup$ Can you share the link to the e-book? $\endgroup$
    – Redsbefall
    Commented Jul 13, 2022 at 13:37
  • $\begingroup$ @Redsbefall It's not a book it's a paper on econometrica. onlinelibrary.wiley.com/doi/10.1111/j.1468-0262.2002.00441.x $\endgroup$
    – Alalalalaki
    Commented Jul 13, 2022 at 13:48
  • $\begingroup$ 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. $\endgroup$
    – Redsbefall
    Commented Jul 19, 2022 at 12:23

1 Answer 1

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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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