I want to solve a differential equation of the first-price auction. In particular, from Jonathan Levin's October 2004 lecture notes, we have the following differential equation:

$$b'(s) = (s-b(s))(n-1)\frac{f(s)}{F(s)}$$

Solving this, we should get

$$b(s) = s - \frac{1}{F^{N-1}} \int_\bar{s}^{s_i} F^{N-1}(\tilde{s}) \,{\rm d}\tilde{s}$$

I would like to understand how to get this equation.

I found a November 2011 paper by Timothy P. Hubbard & Harry J. Paarsch discussing this part. On page 6, although it is a regular expression of the differential equation, I should solve the following equation:

$$y=\frac{1}{\mu(x)}\int_{x_0}^x \mu(u)q(u) {\,\rm d}u + k$$

Could you help how to solve this equation to get $b(s)$?


1 Answer 1


For notational simplicity, let me define the distribution $G(s) = F^{N-1}(s)$ with density $g(s)$. Let $\underline{s} = 0$ (for simplicity).

We have $$ b'(s)G(s)+b(s)g(s)=s g(s) $$ Integrating to $x$ gives us $$ \int_0^x b'(s)G(s)+b(s)g(s) ds = \int_0^x s g(s)ds $$ Notice that $\frac{\partial }{\partial s}(b(s)G(s)) = b'(s)G(s)+b(s)g(s)$, so $$ b(x)G(x) = \int_0^x s g(s) ds $$ where we use the boundary condition that $b(0) = 0$.

Now, $$ b(x) = \frac{1}{G(x)}\int_0^x s g(s) ds $$

Finally, apply integration by parts to $\int_0^x s g(s)$ to get $$ b(x) = \frac{1}{G(x)}\left[ G(x) x - \int_0^xG(y) dy\right]= x - \int_0^x\frac{G(y)}{G(x)} dy $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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