I'm trying to solve followed first-price auction problem.

Bidder's pdf is $$ f(v_i)= \begin{cases} \dfrac{1}{8}v_i, & \text{if} & 0\leq v_i\leq4\\ 0, & \text{if} & \text{otherwise}\\ \end{cases} $$

Bidders only know their own values. One gets 0 if lose the auction and ($v_i-b_i$) if win the auction game.

The question asks the symmetric bidders' bid under Bayesian Nash equilibrium and expected payoff of bidders and seller.

So, one bidder will maximize as follows:


If we take the derivative with respect to $b_1$ we get

$$b(v_1)=\dfrac{\int v_1f(v_1)\,dv_1}{F(v_1)}$$

We know from question $f(v_1)=\dfrac{1}{8}v_i$ and we can find $F(v_1)=\dfrac{1}{16}v_i^2$. If we replace them, we get

$$b(v_1)=\dfrac{\int v_1\dfrac{1}{8}v_1\,dv_1}{\dfrac{1}{16}v_1^2} = \dfrac{2}{3}v_1$$

So this is the bidding of bidder 1. However, I couldn't find the expected payoff bidder 1 and bidder 2. My question is how can I use PDF or CDF to find the expected payoffs of the bidders.

  • 1
    $\begingroup$ Plug in the equilibrium bid into the equation you were attempting to maximise, and you're done! $\endgroup$ – Walrasian Auctioneer May 12 at 2:38
  • $\begingroup$ Thank you! I plugged the $2/3v_1$ into maximization problem. Now, what I should write in the maximization problem instead of $\Pr(b_1>b_2)$? $\endgroup$ – ali May 12 at 8:17
  • $\begingroup$ Total probability is 1/2 with this pdf...? $\endgroup$ – VARulle May 12 at 9:03
  • $\begingroup$ @VARulle you're right. I tried to make my own example. I have edited the question. Sorry for the mistake. $\endgroup$ – ali May 12 at 9:09
  • $\begingroup$ Use the fact that equilibrium bids are strictly monotone. So $b(v_1) > b(v_2)$ whenever $v_1 > v_2$. What does this tell you about $Pr(b_1 > b_2)$? $\endgroup$ – Walrasian Auctioneer May 12 at 21:40

Let me answer here putting together all the hints in the comments.

You've figured out that the equilibrium bid for a type $v$ bidder is $b(v) = \frac{2}{3}v$.

The bids are strictly increasing in $v$, so bidder $i$ wins whenever $b(v_i) \geq b(v_j)$ or $v_i \geq v_j$.

The expected payoff for a type $v$ bidder is thus $$ (v -b(v))P(b_1 \geq b_2) $$ or $$ (v - \frac{2}{3}v)P(v \geq v_2) = \frac{1}{3}P(v \geq v_2) $$ So what is the probability that $v \geq v_2$?

By definition of CDF $P(x \leq v) = F(v) = \int_0^v f(x) dx$

So $P(v_2 \leq v) = F(v)$.

Thus expected utility for a type $v$ bidder is: $$ u(v) = \frac{v}{3} F(v) $$

| improve this answer | |
  • $\begingroup$ Thank you! Now, $\int_0^vf(v_1)\,dv_2=\frac{1}{8}v_1^2$ and the result is $\frac{v_1}{3}\frac{v_1^2}{8}=\frac{v_1^3}{24}$. I was expecting to find the result with number. I mean that the result will not include any variable $\endgroup$ – ali May 13 at 12:21
  • $\begingroup$ Well the answer is the interim payoff of a type $v$ bidder (which is usually what matters). If you're interested in the ex-ante payoff (so even before types are realised), $\int \frac{v^3}{24} f(v) dv$ would give you the ex-ante payoff that wont have any variables. $\endgroup$ – Walrasian Auctioneer May 13 at 17:21

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