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:

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


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

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!

  • $\begingroup$ 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. $\endgroup$
    – Oliv
    Aug 6, 2015 at 11:21
  • $\begingroup$ 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. $\endgroup$
    – ml0105
    Aug 6, 2015 at 12:42
  • 3
    $\begingroup$ 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. $\endgroup$
    – Oliv
    Aug 6, 2015 at 14:17
  • 2
    $\begingroup$ 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 (repository.cmu.edu/cgi/…) points to private correspondence with Cheng. I've emailed Cheng to see if he could send me the derivation. $\endgroup$
    – ml0105
    Aug 6, 2015 at 21:40

1 Answer 1


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

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


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

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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Oliv
    Aug 7, 2015 at 9:01
  • $\begingroup$ That's a neat approach! Thanks for pointing that out :-) $\endgroup$
    – ml0105
    Aug 7, 2015 at 15:44
  • $\begingroup$ You're welcome! Please don't hesitate to report here if you find a general solution. $\endgroup$
    – Oliv
    Aug 7, 2015 at 16:09
  • $\begingroup$ Will do. Hopefully Professor Cheng gets back to me. The only citation for the general case is private correspondence with him. $\endgroup$
    – ml0105
    Aug 7, 2015 at 16:21

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