I have one question in the proof for section 4.1. in Cole, Mailath, and Postlewaite (2001). $$\lim_{\varepsilon \to 0}\frac{1}{2\varepsilon}\int_{\overline{l}-\varepsilon}^{\overline{l}+\varepsilon} v(\beta(i) + \delta, \sigma(i+\beta^{-1}(\beta(i) +\delta) - \overline{l}-\varepsilon))-v(\beta(i), s(i))di$$

They say the above limit is equal to : $$v(\beta(\overline{l}+\delta, \sigma(\tilde{l}))-v(\beta(\overline{l}, s(\overline{l})),$$ for $\tilde{l} = \beta^{-1}(\beta(\overline{l})+\delta).$

I am not sure how do they this. I think if $\varepsilon \to 0$, the term inside the integral goes to 0. But, the answer they give us is to substitute $\overline{l}$ for $i$ in the integral. I appreciate if you give some help to figure this out.


This can be achieved with L'Hopital's rule, which says $$\lim_{x\rightarrow0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow0}\frac{f'(x)}{g'(x)}.$$

The result from Cole, Mailath, and Postlewaite holds fairly generally and is not special to the particular functions in their integrand.

Let $$f(\epsilon)=\int_{l-\epsilon}^{l+\epsilon}h(x,\epsilon)dx \iff f'(\epsilon)=h(l-\epsilon,\epsilon)+h(l+\epsilon,\epsilon)+\int_{l-\epsilon}^{l+\epsilon}\frac{\partial h(x,\epsilon)}{\partial \epsilon}dx$$

and $$g(\epsilon)=2\epsilon\iff g'(\epsilon)=2.$$

Then, using L'Hopital's rule: $$\lim_{\epsilon\rightarrow0}\frac{f(\epsilon)}{g(\epsilon)}=\lim_{\epsilon\rightarrow0}\frac{f'(\epsilon)}{g'(\epsilon)}=\frac{h(l,0)+h(l,0)}{2}=h(l,0).$$

A final word on 'intuition'. You reason, correctly, that the integral goes to zero with $\epsilon$. But the problem is that the denominator, $2\epsilon$ also goes to zero. So, to know what the overall expression does as $\epsilon$ approaches zero, we need to know whether the numerator or the denominator approaches zero faster. In other words, for $\epsilon\rightarrow0$ are we dividing a small number by a very very small number, or are we dividing a very very small number by a small number? This, roughly speaking, is the logic behind L'Hopital's rule.

  • $\begingroup$ 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. $\endgroup$
    – mark leeds
    Oct 15 '18 at 20:05
  • $\begingroup$ @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: $$\frac{\partial}{\partial x}\int_{f(x)}^{c} g(z) dz=f'(x)\frac{\partial}{\partial f(x)}\int_{f(x)}^{c} g(z) dz=-f'(x)g(f(x)).$$ $\endgroup$
    – Ubiquitous
    Oct 15 '18 at 20:18
  • $\begingroup$ @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. $\endgroup$
    – Ubiquitous
    Oct 15 '18 at 20:27
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    $\begingroup$ 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. $\endgroup$
    – mark leeds
    Oct 16 '18 at 20:59
  • 1
    $\begingroup$ probably not in this lifetime aside from trying to follow the integration that's done on this list !!!!!! All the best. $\endgroup$
    – mark leeds
    Oct 16 '18 at 21:05

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