<p>I followed Oliv's suggestion, which was quite fruitful. So we have the differential equations:</p>

<p>$$g_{1}^{\prime}(b) = \dfrac{1}{2} \cdot \dfrac{g_{1}(b) - 1}{g_{2}(b) - b}$$</p>

<p>And:</p>

<p>$$g_{2}^{\prime}(b) = \dfrac{3}{2} \cdot \dfrac{1}{g_{1}(b) - b}$$</p>

<p>With the boundary conditions $g_{2}(0) = 0$ and $g_{1}(\overline{b}) = g_{2}(\overline{b}) = 3$, where $\overline{b}$ is the maximum bid. </p>

<p>Now we guess that $g_{1}(b) = \alpha b + \gamma$ and $g_{2}(b) = \delta b + \lambda$. Applying $g_{2}(b) = 0$ yields that $\lambda = 0$. </p>

<p>Next, I substitute $g_{1}(b)$ into $g_{2}^{\prime}(b)$ to obtain:</p>

<p>$$g_{2}^{\prime}(b) = \dfrac{3}{2} \cdot \dfrac{1}{(\alpha - 1)b + \gamma}$$</p>

<p>Integrating $g_{2}^{\prime}(b)$ yields </p>

<p>$$g_{2}(b) = \dfrac{3}{2(\alpha - 1)} ln( (\alpha - 1)b + \gamma)$$</p>

<p>We note there is no constant when integrating, as $g_{2}(b) = \delta b$. We now apply $g_{2}(0) = 0$ again, concluding that $ln( \gamma) = 0$. And so $\gamma = 1$. Thus, $g_{1}(b) = \alpha b + 1$.</p>

<p>We now solve:</p>

<p>$$g_{2}(\overline{b}) = 3 = \dfrac{3}{2(\alpha - 1)} ln( (\alpha - 1)\overline{b} + \gamma)$$</p>

<p>From this and noting that $g_{1}(\overline{b}) = 3 = \alpha \overline{b} + 1$, we obtain:</p>

<p>$$e^{2(\alpha - 1)} = 3 - \overline{b} \implies \overline{b} = 3 - e^{2(\alpha - 1)}$$</p>

<p>Plugging this into $g_{1}(b)$ yields:</p>

<p>$$g_{1}(\overline{b}) = \alpha(3 - e^{2(\alpha - 1)}) + 1 = 3$$ </p>

<p>Which implies that the solution $\alpha = 1$. Thus, $\overline{b} = 2$.</p>

<p>So $\delta = \dfrac{3}{2}$. </p>

<p>Thus, $g_{1}(b) = b + 1 \implies \beta_{1}(v) = v - 1$; and $g_{2}(b) = \dfrac{3}{2}b$ implies $\beta_{2}(v) = \dfrac{2}{3}v$ as desired.</p>

CC BY-SA 3.0
2015-08-07T09:01:05.333Z
You can obtain your result more efficiently (without integrating) by deriving $g_2'(b)=\delta$, and therefore $\delta = \frac{3}{2} \frac{1}{(\alpha-1)b+\gamma}$. Since this must be true for all $b$, this yields $\alpha=1$ and $2 \delta \gamma =3$. Then you can do the same for the other equation and solve for all values of $\alpha,\gamma,\delta,\beta$ that way.
Oliv
CC BY-SA 3.0
2015-08-07T15:44:25.620Z
That's a neat approach! Thanks for pointing that out :-)
ml0105
CC BY-SA 3.0
2015-08-07T16:09:23.033Z
You're welcome! Please don't hesitate to report here if you find a general solution.
Oliv
CC BY-SA 3.0
2015-08-07T16:21:22.077Z
Will do. Hopefully Professor Cheng gets back to me. The only citation for the general case is private correspondence with him.
ml0105
CC BY-SA 3.0
2015-08-06T23:17:38.560Z
2015-08-07T00:32:52.870Z
2015-08-07T00:32:52.870Z
ml0105
<p>I am working on a problem in my Auction Theory textbook regarding a two-player asymmetric first price auction. Assume the bidders are risk neutral. The problem statement is as follows:</p>

<blockquote>
  <blockquote>
    <p>Suppose that bidder $1$'s value $X_{1}$ is distributed according to $F_{1}(x) = \frac{1}{4}(x-1)^{2}$ over $[1, 3]$, and bidder $2$'s value is distributed according to $\text{exp}(\frac{2}{3}x - 2)$ over $[0, 3]$. Show that $\beta_{1}(x) = x - 1$ and $\beta_{2}(x) = \frac{2}{3}x$ constitute equilibrium bidding strategies in a first price auction.</p>
  </blockquote>
</blockquote>

<p>I am trying to work on deriving $\beta_{1}$ and $\beta_{2}$. Unfortunately, my knowledge of differential equations isn't terribly strong. Would someone be able to double check my work and let me know if I have logic errors? I have derived the correct bidding functions, but am not entirely confident my work is sound.</p>

<p>First, suppose the equilibrium bidding functions $\beta_{1} : [1, 3] \to \mathbb{R}_{+}, \beta_{2} : [0, 3] \to \mathbb{R}_{+}$ are strictly increasing and differentiable. Define $g_{1}(x) = \beta_{1}^{-1}(x)$ and $g_{2}(x) = \beta_{2}^{-1}(x)$. </p>

<p>Player $i$ with valuation $v$ can only vary his bid, so he seeks to find the optimal bid given by the optimization problem below.</p>

<p>$$\max_{b} F_{-i}(g_{-i}(b)) \cdot (v - b)$$</p>

<p>We consider the First Order Conditions:</p>

<p>$$F_{-i}(g_{-i}(b)) = \dfrac{f_{-i}(g_{-i}(b))}{\beta_{-i}^{\prime}(g_{-i}(b))} \cdot (v-b)$$</p>

<p>At equilibrium, $v = g_{i}(b)$. Applying this and noting $\dfrac{1}{\beta_{-i}^{\prime}(g_{-i}(b))} = (g_{-i}(b))^{\prime}$, we have:</p>

<p>$$(g_{-i}(b))^{\prime} = \dfrac{F_{-i}(g_{-i}(b))}{f_{-i}(g_{-i}(b))} \cdot \dfrac{1}{g_{i}(b) - b}$$</p>

<p>Plugging in each $F_{i}$, we obtain:</p>

<p>$$g_{2}^{\prime}(b) = \dfrac{3}{2} \cdot \dfrac{1}{g_{1}(b) - b}$$</p>

<p>And:</p>

<p>$$g_{1}^{\prime}(b) = \dfrac{1}{2} \cdot \dfrac{g_{1}(b) - 1}{g_{2}(b) - b}$$</p>

<p>At equilibrium, we have $\beta_{1}(3) = \beta_{2}(3)$. By individual rationality, $\beta_{2}(0) = 0 \implies g_{2}(0) = 0$.</p>

<p>While I could obviously use the problem statement that $\beta_{1}(x) = x - 1$ to conclude that $g_{1}(0) = 1$, I don't know how to justify this boundary condition independently. Does anyone have any insights into this?</p>

<p>Assuming this boundary condition though, I note:</p>

<p>$$g_{2}^{\prime}(0) = \dfrac{3}{2} \cdot \dfrac{1}{1 - 0} = \dfrac{3}{2}$$ </p>

<p>From here, I can wave my hand and guess that $g_{2}^{\prime}(b) = \dfrac{3}{2}$, which would imply $g_{2}(b) = \dfrac{3}{2}b$. I'm not sure how to formally derive this though. Would anyone have insights into this?</p>

<p>Once I have $g_{2}(b) = \dfrac{3}{2}b$, I can plug into $g_{1}^{\prime}(b)$ to get:</p>

<p>$$g_{1}^{\prime}(b) = \dfrac{1}{2} \cdot \dfrac{g_{1}(b) - 1}{\dfrac{3}{2}b - b} = \dfrac{g_{1}(b) - 1}{b}$$</p>

<p>Which is a first order linear differential equation, whose solution is:</p>

<p>$g_{1}(b) = b + 1 \implies \beta_{1}(v) = v - 1$.</p>

<p>And we have $\beta_{2}(v) = \dfrac{2}{3}v$.</p>

<p>My work is certainly a little hand-wavy. I would greatly appreciate any help in solidifying the details. Thank you in advance for any help!</p>

CC BY-SA 3.0
2015-08-06T11:21:23.810Z
Why are you trying to derive $\beta_1$ and $\beta_2$? It seems that you are only asked to show that these strategies constitute an equilibrium (i.e. that no deviations are profitable), not that this equilibrium is unique.
Oliv
CC BY-SA 3.0
2015-08-06T12:42:08.930Z
Academic interest. I'd like to know how to obtain the solutions without knowing them a priori. I'm not even trying to show uniqueness, just a more axiomatic derivation.
ml0105
CC BY-SA 3.0
2015-08-06T14:17:23.693Z
It does not completely answer your question, but a way to obtain some solutions without knowing them is to "guess and verify" by postulating some functional forms. For instance, you can assume $g_1(b)=\alpha b + \beta$ and $g_2(b)=\gamma b + \delta$ and solve for the values of the coefficients that satisfy your system of differential equations. There is no general method to find all solutions. You can for instance express $g_1$ as a function of $g_2'$ given your first equation and plug this into the second to obtain an equation in $g_2$ only, but you loose tractability doing so.
Oliv
CC BY-SA 3.0
2015-08-06T21:40:05.570Z
Thank you for your help! That looks like a promising idea! I am going to play around with it some. In digging, it looks like this class of distribution pairs has a closed form solution which was derived by Professor Harrison Cheng. The citation in this paper for the class C2 (http://repository.cmu.edu/cgi/viewcontent.cgi?article=1137&context=tepper) points to private correspondence with Cheng. I've emailed Cheng to see if he could send me the derivation.
ml0105
<game-theory><auctions>
System of Differential Equations- Asymmetric First-Price Auction
ml0105
# System of Differential Equations- Asymmetric First-Price Auction

(Asked by: username_0 on 2015-08-06)

I am working on a problem in my Auction Theory textbook regarding a two-player asymmetric first price auction. Assume the bidders are risk neutral. The problem statement is as follows:

> > Suppose that bidder $1$'s value $X\_{1}$ is distributed according to $F\_{1}(x) = \frac{1}{4}(x-1)^{2}$ over $[1, 3]$, and bidder $2$'s value is distributed according to $\text{exp}(\frac{2}{3}x - 2)$ over $[0, 3]$. Show that $\beta\_{1}(x) = x - 1$ and $\beta\_{2}(x) = \frac{2}{3}x$ constitute equilibrium bidding strategies in a first price auction.

I am trying to work on deriving $\beta\_{1}$ and $\beta\_{2}$. Unfortunately, my knowledge of differential equations isn't terribly strong. Would someone be able to double check my work and let me know if I have logic errors? I have derived the correct bidding functions, but am not entirely confident my work is sound.

First, suppose the equilibrium bidding functions $\beta\_{1} : [1, 3] \to \mathbb{R}\_{+}, \beta\_{2} : [0, 3] \to \mathbb{R}\_{+}$ are strictly increasing and differentiable. Define $g\_{1}(x) = \beta\_{1}^{-1}(x)$ and $g\_{2}(x) = \beta\_{2}^{-1}(x)$.

Player $i$ with valuation $v$ can only vary his bid, so he seeks to find the optimal bid given by the optimization problem below.

$$\max\_{b} F\_{-i}(g\_{-i}(b)) \cdot (v - b)$$

We consider the First Order Conditions:

$$F\_{-i}(g\_{-i}(b)) = \dfrac{f\_{-i}(g\_{-i}(b))}{\beta\_{-i}^{\prime}(g\_{-i}(b))} \cdot (v-b)$$

At equilibrium, $v = g\_{i}(b)$. Applying this and noting $\dfrac{1}{\beta\_{-i}^{\prime}(g\_{-i}(b))} = (g\_{-i}(b))^{\prime}$, we have:

$$(g\_{-i}(b))^{\prime} = \dfrac{F\_{-i}(g\_{-i}(b))}{f\_{-i}(g\_{-i}(b))} \cdot \dfrac{1}{g\_{i}(b) - b}$$

Plugging in each $F\_{i}$, we obtain:

$$g\_{2}^{\prime}(b) = \dfrac{3}{2} \cdot \dfrac{1}{g\_{1}(b) - b}$$

And:

$$g\_{1}^{\prime}(b) = \dfrac{1}{2} \cdot \dfrac{g\_{1}(b) - 1}{g\_{2}(b) - b}$$

At equilibrium, we have $\beta\_{1}(3) = \beta\_{2}(3)$. By individual rationality, $\beta\_{2}(0) = 0 \implies g\_{2}(0) = 0$.

While I could obviously use the problem statement that $\beta\_{1}(x) = x - 1$ to conclude that $g\_{1}(0) = 1$, I don't know how to justify this boundary condition independently. Does anyone have any insights into this?

Assuming this boundary condition though, I note:

$$g\_{2}^{\prime}(0) = \dfrac{3}{2} \cdot \dfrac{1}{1 - 0} = \dfrac{3}{2}$$

From here, I can wave my hand and guess that $g\_{2}^{\prime}(b) = \dfrac{3}{2}$, which would imply $g\_{2}(b) = \dfrac{3}{2}b$. I'm not sure how to formally derive this though. Would anyone have insights into this?

Once I have $g\_{2}(b) = \dfrac{3}{2}b$, I can plug into $g\_{1}^{\prime}(b)$ to get:

$$g\_{1}^{\prime}(b) = \dfrac{1}{2} \cdot \dfrac{g\_{1}(b) - 1}{\dfrac{3}{2}b - b} = \dfrac{g\_{1}(b) - 1}{b}$$

Which is a first order linear differential equation, whose solution is:

$g\_{1}(b) = b + 1 \implies \beta\_{1}(v) = v - 1$.

And we have $\beta\_{2}(v) = \dfrac{2}{3}v$.

My work is certainly a little hand-wavy. I would greatly appreciate any help in solidifying the details. Thank you in advance for any help!

**Tags:** game-theory, auctions

> **Comments:**
> **username_1**: Why are you trying to derive $\beta_1$ and $\beta_2$? It seems that you are only asked to show that these strategies constitute an equilibrium (i.e. that no deviations are profitable), not that this equilibrium is unique.  
> **username_0**: Academic interest. I'd like to know how to obtain the solutions without knowing them a priori. I'm not even trying to show uniqueness, just a more axiomatic derivation.  
> **username_1**: It does not completely answer your question, but a way to obtain some solutions without knowing them is to "guess and verify" by postulating some functional forms. For instance, you can assume $g_1(b)=\alpha b + \beta$ and $g_2(b)=\gamma b + \delta$ and solve for the values of the coefficients that satisfy your system of differential equations. There is no general method to find all solutions. You can for instance express $g_1$ as a function of $g_2'$ given your first equation and plug this into the second to obtain an equation in $g_2$ only, but you loose tractability doing so.  
> **username_0**: Thank you for your help! That looks like a promising idea! I am going to play around with it some. In digging, it looks like this class of distribution pairs has a closed form solution which was derived by Professor Harrison Cheng. The citation in this paper for the class C2 (http://repository.cmu.edu/cgi/viewcontent.cgi?article=1137&context=tepper) points to private correspondence with Cheng. I've emailed Cheng to see if he could send me the derivation.  

---

## 1 answers

### Answer by username_0 on 2015-08-06. Score: 3 | Accepted answer

I followed username_1's suggestion, which was quite fruitful. So we have the differential equations:

$$g\_{1}^{\prime}(b) = \dfrac{1}{2} \cdot \dfrac{g\_{1}(b) - 1}{g\_{2}(b) - b}$$

And:

$$g\_{2}^{\prime}(b) = \dfrac{3}{2} \cdot \dfrac{1}{g\_{1}(b) - b}$$

With the boundary conditions $g\_{2}(0) = 0$ and $g\_{1}(\overline{b}) = g\_{2}(\overline{b}) = 3$, where $\overline{b}$ is the maximum bid.

Now we guess that $g\_{1}(b) = \alpha b + \gamma$ and $g\_{2}(b) = \delta b + \lambda$. Applying $g\_{2}(b) = 0$ yields that $\lambda = 0$.

Next, I substitute $g\_{1}(b)$ into $g\_{2}^{\prime}(b)$ to obtain:

$$g\_{2}^{\prime}(b) = \dfrac{3}{2} \cdot \dfrac{1}{(\alpha - 1)b + \gamma}$$

Integrating $g\_{2}^{\prime}(b)$ yields

$$g\_{2}(b) = \dfrac{3}{2(\alpha - 1)} ln( (\alpha - 1)b + \gamma)$$

We note there is no constant when integrating, as $g\_{2}(b) = \delta b$. We now apply $g\_{2}(0) = 0$ again, concluding that $ln( \gamma) = 0$. And so $\gamma = 1$. Thus, $g\_{1}(b) = \alpha b + 1$.

We now solve:

$$g\_{2}(\overline{b}) = 3 = \dfrac{3}{2(\alpha - 1)} ln( (\alpha - 1)\overline{b} + \gamma)$$

From this and noting that $g\_{1}(\overline{b}) = 3 = \alpha \overline{b} + 1$, we obtain:

$$e^{2(\alpha - 1)} = 3 - \overline{b} \implies \overline{b} = 3 - e^{2(\alpha - 1)}$$

Plugging this into $g\_{1}(b)$ yields:

$$g\_{1}(\overline{b}) = \alpha(3 - e^{2(\alpha - 1)}) + 1 = 3$$

Which implies that the solution $\alpha = 1$. Thus, $\overline{b} = 2$.

So $\delta = \dfrac{3}{2}$.

Thus, $g\_{1}(b) = b + 1 \implies \beta\_{1}(v) = v - 1$; and $g\_{2}(b) = \dfrac{3}{2}b$ implies $\beta\_{2}(v) = \dfrac{2}{3}v$ as desired.

> **Comments:**
> **username_1**: You can obtain your result more efficiently (without integrating) by deriving $g_2'(b)=\delta$, and therefore $\delta = \frac{3}{2} \frac{1}{(\alpha-1)b+\gamma}$. Since this must be true for all $b$, this yields $\alpha=1$ and $2 \delta \gamma =3$. Then you can do the same for the other equation and solve for all values of $\alpha,\gamma,\delta,\beta$ that way.  
> **username_0**: That's a neat approach! Thanks for pointing that out :-)  
> **username_1**: You're welcome! Please don't hesitate to report here if you find a general solution.  
> **username_0**: Will do. Hopefully Professor Cheng gets back to me. The only citation for the general case is private correspondence with him.  

---

