# Burdett and Mortensen (1998), Equation (22), Integral by parts question

Suppose $F(x)$ and $H(x)$ are both cumulative prob dist functions on the support of $[b_0,b_1]$ and we know followings:

$$u(x|F)=\int_{b_0}^{x}\frac{m}{1+k[1-F(b)]} dH(b)$$

and from this

$$[1+k(1-F(x))]du(x|F)=mdH(x)$$

We also know that

$$G(w)(m-u(b_1|F))=\frac{k\int_{b_0}^w [F(w)-F(x)]du(x|F)}{1+k(1-F(w))}$$

Then in the paper, authors get following which is equation  in the paper

$$\frac{(m-u(b_1|F))dG(w)}{dF(w)}=\frac{kmH(w)}{[1+k[1-F(w)]]^2}$$

I really don't understand how the authors got this. Can somebody help me?

The authors apply in the second step the exact same rules they used in the first step, regarding the calculation of the total differential of an integral, taking also into account that the integral is taken over a function of the integrating variable and not just the integrating variable itself.

The first step is

$$d[u(x|F)]=d\left[\int_{b_0}^{x}\frac{m}{1+k[1-F(b)]} dH(b)\right]$$

$$=\frac{m}{1+k[1-F(x)]}dH(x) \tag{1}$$

and re-arranging we get the first result. Applying the same logic, we have

$$(m-u(b_1|F))\cdot d\big[G(w)\big]=kd\left[\frac{\int_{b_0}^w [F(w)-F(x)]du(x|F)}{1+k(1-F(w))}\right]$$

We are considering the differential with respect to $w$, since this is the argument of the $G$ function. To clearly indicate this we will write $d_w$. To clarify the picture write

$$kd_w\left[\frac{\int_{b_0}^w [F(w)-F(x)]du(x|F)}{1+k(1-F(w))}\right] \equiv k d_w\left(\frac{A(w)}{B(w)}\right) = k\frac{B(w)d_wA(w) - A(w)d_wB(w)}{[B(w)]^2}$$

We see that the denominator is exactly as the final expression of the authors, so it is of no concern. We turn to the numerator. One by one,

$$d_wA(w) = d_w\left[\int_{b_0}^w [F(w)-F(x)]du(x|F)\right]$$

$$= [F(w)-F(w)]du(w|F) + \left[\int_{b_0}^w d_w[F(w)-F(x)]du(x|F)\right]$$

$$= 0 + d_wF(w) \int_{b_0}^w du(x|F) + 0$$

We applied here Leibniz rule for differentiation w.r.t. a limit of integration under the integral when the integrand is also a function of the integrating limit. The second zero result comes from the fact that $d_wF(x) = 0$ ($F(x)$ does not include $w$, so as we change $w$ the change in $F(x)$ is zero). So

$$B(w)d_wA(w) = [1+k(1-F(w)]d_wF(w) \int_{b_0}^w du(x|F)$$

$$= d_wF(w)\int_{b_0}^w [1+k(1-F(w))]du(x|F) \tag{2}$$

where we have moved the term inside the integral since it does not depend on the integrating variable which is $x$.

Going to the second term, we have

$$d_wB(w) = -kd_wF(w)$$

So

$$-A(w)d_wB(w) = d_wF(w) \int_{b_0}^w k[F(w)-F(x)]du(x|F) \tag{3}$$

Therefore

$$B(w)d_wA(w) -A(w)d_wB(w) = d_wF(w)\int_{b_0}^w [1+k(1-F(w)]du(x|F) \\+ d_wF(w) \int_{b_0}^w k[F(w)-F(x)]du(x|F)$$

The $d_wF(w)$ is a common factor and the two integrals can be written as one due to linearity, so

$$B(w)d_wA(w) -A(w)d_wB(w)=\\ = d_wF(w)\int_{b_0}^w\Big([1+k(1-F(w)] + k[F(w)-F(x)]\Big)du(x|F)$$

The term $kF(w)$ in the integrand cancels off and we are left with

$$B(w)d_wA(w) -A(w)d_wB(w) = d_wF(w)\int_{b_0}^w[1+k(1-F(x))]du(x|F)$$

But using the re-arranged $(1)$ and substituting we get

$$B(w)d_wA(w) -A(w)d_wB(w) = d_wF(w)\int_{b_0}^wmdH(x)$$

$$= md_wF(w)\cdot \Big(H(x) \big|_{b_0}^w\Big) = md_wF(w)\cdot \Big(H(w) - H(b_0)\Big)$$

But since $H()$ is a CDF in $(b_0, b_1)$, it follows that $H(b_0) =0$.

And we are done. In short, don't be afraid to push through the mathematical calculations.

• That was amazing. I only don't follow the one part ( third integral after one by one where you get zero + ) where you used leibnitz rule. I looked it up here en.wikipedia.org/wiki/Leibniz_integral_rule but I still couldn't figure it out. Would you mind providing those details ? It's really appreciated. – mark leeds Sep 8 '18 at 20:16
• @markleeds A little care is needed because we apply Leibniz rule adjusted for differentials, not derivatives. The issue is the value of $$\int_{b_0}^w d[F(w)-F(x)]du(x|F)= \int_{b_0}^w dF(w)du(x|F)-\int_{b_0}^w dF(x)du(x|F)$$ $$=dF(w)\int_{b_0}^w du(x|F)-\int_{b_0}^w dF(x)du(x|F)$$ The first term is what is left in my post, so it must be the case that $$\int_{b_0}^w dF(x)du(x|F) = 0$$ But this holds becasue the diferential is taken with respect to $w$ specifically, and so $dF(x) =0$. I will add a subscript to my answer to make this relation to $w$ clear. – Alecos Papadopoulos Sep 9 '18 at 2:39
• Thanks so much for the detailed reply. I will go over carefully and hopefully follow it. – mark leeds Sep 9 '18 at 19:34
• @markleeds You're welcome. Since it appears that you are the only one interested in this post, perhaps you should consider upvoting this answer so that the the thread is removed from the "unanswered" queue. – Alecos Papadopoulos Sep 10 '18 at 1:35
• Hi Alecos: I'll try but I don't know if I'm allowed. Let me try. Also, I went through your latest explanation and it was beautiful. You should be ( maybe you are ? ) a graduate econometrics professor. there would be a lot more knowledgable econometricians out there !!! – mark leeds Sep 10 '18 at 4:15