<p>For notational simplicity, let me define the distribution <span class="math-container">$G(s) = F^{N-1}(s)$</span> with density <span class="math-container">$g(s)$</span>. Let <span class="math-container">$\underline{s} = 0$</span> (for simplicity).</p>
<p>We have
<span class="math-container">$$
b'(s)G(s)+b(s)g(s)=s g(s)
$$</span>
Integrating to <span class="math-container">$x$</span> gives us
<span class="math-container">$$
\int_0^x b'(s)G(s)+b(s)g(s) ds = \int_0^x s g(s)ds
$$</span>
Notice that <span class="math-container">$\frac{\partial }{\partial s}(b(s)G(s)) =  b'(s)G(s)+b(s)g(s)$</span>, so
<span class="math-container">$$
 b(x)G(x) = \int_0^x s g(s) ds
$$</span>
where we use the boundary condition that <span class="math-container">$b(0) = 0$</span>.</p>
<p>Now,
<span class="math-container">$$
 b(x) = \frac{1}{G(x)}\int_0^x s g(s) ds
$$</span></p>
<p>Finally, apply integration by parts to <span class="math-container">$\int_0^x s g(s)$</span> to get
<span class="math-container">$$
 b(x) = \frac{1}{G(x)}\left[ G(x) x - \int_0^xG(y) dy\right]= x - \int_0^x\frac{G(y)}{G(x)} dy
$$</span></p>

CC BY-SA 4.0
2021-03-16T19:26:21.810Z
2021-03-16T19:26:21.810Z
Walrasian Auctioneer
<p>I want to solve a differential equation of the first-price auction. In particular, from Jonathan Levin's October 2004 <a href="https://web.stanford.edu/%7Ejdlevin/Econ%20286/Auctions.pdf" rel="nofollow noreferrer">lecture notes</a>, we have the following differential equation:</p>
<p><span class="math-container">$$b'(s) = (s-b(s))(n-1)\frac{f(s)}{F(s)}$$</span></p>
<p>Solving this, we should get</p>
<p><span class="math-container">$$b(s) = s - \frac{1}{F^{N-1}} \int_\bar{s}^{s_i} F^{N-1}(\tilde{s}) \,{\rm d}\tilde{s}$$</span></p>
<p>I would like to understand how to get this equation.</p>
<p>I found a November 2011 <a href="https://vinci.cs.uiowa.edu/%7Ehjp/download/HoCE.pdf" rel="nofollow noreferrer">paper</a> by Timothy P. Hubbard &amp; 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:</p>
<p><span class="math-container">$$y=\frac{1}{\mu(x)}\int_{x_0}^x \mu(u)q(u) {\,\rm d}u + k$$</span></p>
<p>Could you help how to solve this equation to get <span class="math-container">$b(s)$</span>?</p>

<auctions><ode>
Differential equation for first-price auction
hrkshr
# Differential equation for first-price auction

(Asked by: username_0 on 2021-03-16)

I want to solve a differential equation of the first-price auction. In particular, from Jonathan Levin's October 2004 [lecture notes](https://web.stanford.edu/%7Ejdlevin/Econ%20286/Auctions.pdf), 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](https://vinci.cs.uiowa.edu/%7Ehjp/download/HoCE.pdf) 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 $b(s)$?

**Tags:** auctions, ode

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## 1 answers

### Answer by username_1 on 2021-03-16. Score: 7 | Accepted answer

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
$$
b(x) = \frac{1}{G(x)}\left[ G(x) x - \int\_0^xG(y) dy\right]= x - \int\_0^x\frac{G(y)}{G(x)} dy
$$


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