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I am working through an economics paper and I need to take the derivative of the following function:

$h\left(\overline{\omega}\right) = \int^{\infty}_{\overline{\omega}} \omega \Phi \left(d\omega\right)$

Even though I don't understand it well, I can do the derivative for the case

$g\left(\overline{\omega}\right) = \int^{\overline{\omega}}_{0} \omega \Phi \left(d\omega\right)$

where the derivative is simply

$g'\left(\overline{\omega}\right) = \overline{\omega} \phi \left(\overline{\omega}\right)$

But for $h\left(\overline{\omega}\right)$ where the upper bound is $\infty$ I really have no idea of what to do.

Can anyone help me? Any explanation or even a pointer to where I can learn those things would be greatly appreciated.

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    I assume in the integral you have $d(\Phi(\omega))$ and not $\Phi(d(\omega))$2010-11-26
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    @Sivaram : The notation I use is the same as the paper (see page 895 of the linked file), but yeah, this is what I mean.2010-11-26
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    @Sivaram: $\Phi (d\omega )$ is basically the same as $d\Phi (\omega )$.2010-11-26

3 Answers 3

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If the integral is sufficiently nice by which I mean it doesn't blow up to infinity, we can write $h(\bar{\omega}) = \int_{l}^{\infty} \omega d(\Phi(w)) - \int_{l}^{\bar{\omega}} \omega d(\Phi(w))$.

where $l$ is some constant number. It could be $0$ or $\infty$ based on your problem.

Now the first integral is just a constant, and the second integral is similar to $g(\bar{\omega})$

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    Thanks, that makes a lot of sense. The lower bound is 0 rather than infinity, but you couldn't have known this :)2010-11-26
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    @Vivi: What TCL has written is also correct. $\int_{a}^{b} f(x)dx = - \int_{b}^{a} f(x)dx$ provided all these integral are finite.2010-11-26
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    @Sivaram : sorry, why not $1- \int _b^a f(x)dx$ ?2010-11-26
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    @Vivi: This comes from basic integration fact. $\int_{a}^{c} f(x) dx$ can be written as $\int_{a}^{b} f(x) dx + \int_{b}^{c} f(x) dx$. This holds true for all $a$,$b$ and $c$. If we take $c=a$, then we have $\int_{a}^{a} f(x) dx = \int_{a}^{b} f(x) dx + \int_{b}^{a} f(x) dx$. Now $\int_{a}^{a} f(x) dx = 0$. And hence $\int_{a}^{b} f(x) dx + \int_{b}^{a} f(x) dx = 0$, which implies $\int_{a}^{b} f(x) dx = - \int_{b}^{a} f(x) dx$2010-11-26
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    @Sivaram : I think I get it now... TCL was assuming that the expectation of omega over the whole interval was 0, when it is actually 1 (but I couldn't have expected him to know that), so it should be $1- \int _b^a f(x)dx$. Your explanation was very helpful :)2010-11-26
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    @Viva: No. TCL does not assume anything.To be clear, $h(\bar{\omega}) = \int_{\bar{\omega}}^{\infty} \omega d(\Phi(w)) = -\int_{\infty}^{\bar{\omega}} \omega d(\Phi(w)) = \int_{l}^{\infty} \omega d(\Phi(w)) - \int_{l}^{\bar{\omega}} \omega d(\Phi(w))$. All the answers are correct. TCL, Shai Covo are also correct.2010-11-26
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    @Sivaram: can't understand, sorry. The whole interval goes from 0 to $\infty$. Thus $\int_{0}^{\infty} \omega \Phi (d\omega) = \mathbb{E}[\omega]$. If $\int_{0}^{\infty}\omega\Phi (d\omega) = \int_{0}^{\overline{\omega}}\omega\Phi (d\omega) +\int_{\overline{\omega}}^{\infty}\omega\Phi (d\omega)$, then $\int_{\overline{\omega}}^{\infty} \omega \Phi (d\omega) = \mathbb{E}[\omega]-\int_{0}^{\overline{\omega}}\omega \Phi (d\omega) $. But in this case $\mathbb{E}[\omega] =1$, so $\int_{\overline{\omega}}^{\infty} \omega \Phi (d\omega) = 1 - \int_{0}^{\overline{\omega}}\omega\Phi(d\omega) $. :(2010-11-26
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    @Vivi: You are correct. However note that the $\textbf{lower limit}$ in TCL's answer is $\mathbf{\infty}$ and in your answer it is $\mathbf{0}$. TCL has $h(\bar{\omega}) = - \int_{\infty}^{\bar{\omega}} \omega d(\Phi(\omega))$ while you have $h(\bar{\omega}) = \mathbb{E}[\omega] - \int_{0}^{\bar{\omega}} \omega d(\Phi(\omega))$2010-11-26
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    @Vivi: You can use both to get your desired answer which is given by $h'(\bar{\omega}) = -\bar{\omega}\Phi(\bar{\omega})$.2010-11-26
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    @Sivaram: sorry, now I finally get it. It is $1-\int_0^{\overline{\omega}} \omega \Phi(\omega)$, but when I take the derivative the one drops out, so it is just $- \overline{\omega} \phi (\overline{\omega})$. Duh! Thanks so much again :)2010-11-26
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Note that $h(\bar \omega ) + g(\bar \omega )$ is constant, and use the result for $g(\bar \omega )$.

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Since $$h(\overline{\omega})=-\int_{\infty}^{\overline{\omega}}\omega\Phi(d\omega),$$ $h'(\overline{\omega})=-\overline{\omega}\phi(\overline{\omega})$.

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    I can't accept your answer yet (need to wait 4 minutes) but I will do it soon. Thanks heaps!2010-11-26