Timeline for need help from theorists: proof in Cole, Mailath, and Postlewaite (2001)
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Oct 16, 2018 at 21:05 | comment | added | mark leeds | probably not in this lifetime aside from trying to follow the integration that's done on this list !!!!!! All the best. | |
| Oct 16, 2018 at 21:00 | comment | added | Ubiquitous | @markleeds if you need any more help with integration, you can get specialist answers over at math stack exchange. | |
| Oct 16, 2018 at 20:59 | comment | added | mark leeds | It's really appreciated. I will print out and go over carefully. Thanks for taking the time for the detailed explanation. I've noticed that people on this list are quite generous and helpful. | |
| Oct 15, 2018 at 20:27 | comment | added | Ubiquitous | @markleeds When we differentiate, we are asking how does a (small) change in $\epsilon$ change the value of the integral. It does so through two effects. Firstly, the limits of the integral change, which is the first two terms. Secondly, the integrand itself changes (that's the $\partial h/\partial \epsilon$ bit), and we have to apply this change along the entire length of the interval over which we are integrating, which is why the $\partial h/\partial \epsilon$ appears inside an integral. | |
| Oct 15, 2018 at 20:18 | comment | added | Ubiquitous | @markleeds: The first term has no minus sign: when the lower limit gets smaller (because $\epsilon$ increases) the value of the integral increases. In essence, there would be a minus sign that comes from the fact that an increase in the lower limit causes the value of the integral to shrink, and a second minus sign that comes from the fact that the lower limit is decreasing in $\epsilon$. The two minus signs cancel. This is basically just the chain rule: $$\frac{\partial}{\partial x}\int_{f(x)}^{c} g(z) dz=f'(x)\frac{\partial}{\partial f(x)}\int_{f(x)}^{c} g(z) dz=-f'(x)g(f(x)).$$ | |
| Oct 15, 2018 at 20:05 | comment | added | mark leeds | That was great but my calculus is rusty. In your original expression for the integral, shouldn't the first term have a minus in front of it ? and where does the third term come from ? I thought that when you take the derivative of the integral, you just evaluate it at both limits. I'm not familiar with the derivative term. Or maybe I used to be and forgot. Thanks. | |
| Oct 15, 2018 at 15:52 | history | edited | Ubiquitous | CC BY-SA 4.0 |
added 11 characters in body
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| S Oct 15, 2018 at 15:51 | history | suggested | Kenneth Rios | CC BY-SA 4.0 |
fixed typo in the last limit
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| Oct 15, 2018 at 14:34 | review | Suggested edits | |||
| S Oct 15, 2018 at 15:51 | |||||
| Oct 15, 2018 at 12:27 | history | edited | Ubiquitous | CC BY-SA 4.0 |
added 14 characters in body
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| Oct 15, 2018 at 12:21 | history | answered | Ubiquitous | CC BY-SA 4.0 |