# First Price Auction (Expected Payoff)

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.

• Plug in the equilibrium bid into the equation you were attempting to maximise, and you're done! May 12 '20 at 2:38
• Thank you! I plugged the $2/3v_1$ into maximization problem. Now, what I should write in the maximization problem instead of $\Pr(b_1>b_2)$?
– ali
May 12 '20 at 8:17
• Total probability is 1/2 with this pdf...? May 12 '20 at 9:03
• @VARulle you're right. I tried to make my own example. I have edited the question. Sorry for the mistake.
– ali
May 12 '20 at 9:09
• 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 $Pr(b_1 > b_2)$? May 12 '20 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: $$u(v) = \frac{v}{3} F(v)$$

• 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
– ali
May 13 '20 at 12:21
• 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. May 13 '20 at 17:21