1
$\begingroup$

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:

$$\max_{b_1}\,\,(v_1-b_1)\Pr(b_1>b_2)$$

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.

$\endgroup$
8
  • 1
    $\begingroup$ Plug in the equilibrium bid into the equation you were attempting to maximise, and you're done! $\endgroup$ May 12 '20 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 '20 at 8:17
  • $\begingroup$ Total probability is 1/2 with this pdf...? $\endgroup$
    – VARulle
    May 12 '20 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 '20 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 $Pr(b_1 > b_2)$? $\endgroup$ May 12 '20 at 21:40
1
$\begingroup$

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: $$ u(v) = \frac{v}{3} F(v) $$

$\endgroup$
2
  • $\begingroup$ 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 '20 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. $\endgroup$ May 13 '20 at 17:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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