integrating value function by parts

This is a derivation from "Exercises in Recursive Macroeconomic Theory preliminary and incomplete by Stijn Van Nieuwerburgh Pierre-Olivier Weill Lars Ljungqvist Thomas J. Sargent"

which uses integration by parts. Can someone explain how this derivation works? I have tried to replicate the result but to no avail. The text says that they use integration by parts on the second integral on the second line [the integral from w to B of w'dF(w') ]


The integral of interest is $\int_{w^*}^Bw'dF(w')$, which we get after bringing $\frac{1}{1-\beta}$ out from the last integral on the second line. I assume that $F(B)=1$. First, we have:

$$ \int_{w^*}^Bw'dF(w')=w'F(w')\vert_{w^*}^B - \int^B_{w^*} F(w')dw' $$ from integration by parts with $u=w'$ and $dv=dF(w')$. We can evaluate the first term on the right-hand side and both add and subtract $B-w^*=\int^B_{w^*}dw'$ to get:

$$ \int_{w^*}^Bw'dF(w')= [B - w^*F(w^*)]-[B-w^*]+\int^B_{w^*}(1-F(w')dw'. $$

Note that I used my initial assumption here, and that I brought $B-w^*=\int^B_{w^*}dw'$ under the pre-existing integral.It should be clear after a bit of algebra that:

$$ \int_{w^*}^Bw'dF(w')= w^*(1 - F(w^*))+\int^B_{w^*}(1-F(w')dw'. $$

The rest should be easy after multipling both sides by the $\frac{\beta}{1-\beta}$ factor.

  • $\begingroup$ Thanks so much! I wonder why textbooks so often neglect to mention they are using algebraic 'tricks' like adding and subtracting B-w* in this case? $\endgroup$ – MHall Nov 1 '16 at 21:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.