<p>Let me answer here putting together all the hints in the comments.</p>

<p>You've figured out that the equilibrium bid for a type <span class="math-container">$v$</span> bidder is <span class="math-container">$b(v) = \frac{2}{3}v$</span>.</p>

<p>The bids are strictly increasing in <span class="math-container">$v$</span>, so bidder <span class="math-container">$i$</span> wins whenever <span class="math-container">$b(v_i) \geq b(v_j)$</span> or <span class="math-container">$v_i \geq v_j$</span>.</p>

<p>The expected payoff for a type <span class="math-container">$v$</span> bidder is thus
<span class="math-container">$$
(v -b(v))P(b_1 \geq b_2)
$$</span> 
or
<span class="math-container">$$
(v - \frac{2}{3}v)P(v \geq v_2) = \frac{1}{3}P(v \geq v_2)
$$</span>
So what is the probability that <span class="math-container">$v \geq v_2$</span>?</p>

<p>By definition of CDF <span class="math-container">$P(x \leq v) =  F(v) = \int_0^v f(x) dx$</span></p>

<p>So <span class="math-container">$P(v_2 \leq v) = F(v)$</span>.</p>

<p>Thus expected utility for a type <span class="math-container">$v$</span> bidder is:
<span class="math-container">$$
u(v) = \frac{v}{3} F(v)
$$</span></p>

CC BY-SA 4.0
2020-05-13T12:21:24.180Z
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
CC BY-SA 4.0
2020-05-13T17:21:06.667Z
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.
Walrasian Auctioneer
CC BY-SA 4.0
2020-05-13T01:42:39.837Z
2020-05-13T01:42:39.837Z
Walrasian Auctioneer
<p>I'm trying to solve followed first-price auction problem.</p>

<p>Bidder's pdf is
<span class="math-container">$$
f(v_i)= \begin{cases}
\dfrac{1}{8}v_i, &amp; \text{if} &amp; 0\leq v_i\leq4\\
0, &amp; \text{if} &amp;  \text{otherwise}\\
\end{cases}
$$</span></p>

<p>Bidders only know their own values. One gets 0 if lose the auction and (<span class="math-container">$v_i-b_i$</span>) if win the auction game.</p>

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

<p>So, one bidder will maximize as follows:</p>

<p><span class="math-container">$$\max_{b_1}\,\,(v_1-b_1)\Pr(b_1&gt;b_2)$$</span></p>

<p>If we take the derivative with respect to <span class="math-container">$b_1$</span> we get</p>

<p><span class="math-container">$$b(v_1)=\dfrac{\int v_1f(v_1)\,dv_1}{F(v_1)}$$</span></p>

<p>We know from question <span class="math-container">$f(v_1)=\dfrac{1}{8}v_i$</span> and we can find <span class="math-container">$F(v_1)=\dfrac{1}{16}v_i^2$</span>. If we replace them, we get</p>

<p><span class="math-container">$$b(v_1)=\dfrac{\int v_1\dfrac{1}{8}v_1\,dv_1}{\dfrac{1}{16}v_1^2} = \dfrac{2}{3}v_1$$</span></p>

<p>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.</p>

CC BY-SA 4.0
2020-05-12T02:38:24.917Z
Plug in the equilibrium bid into the equation you were attempting to maximise, and you're done!
Walrasian Auctioneer
CC BY-SA 4.0
2020-05-12T08:17:18.677Z
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
CC BY-SA 4.0
2020-05-12T09:03:35.560Z
Total probability is 1/2 with this pdf...?
VARulle
CC BY-SA 4.0
2020-05-12T09:09:56.833Z
@VARulle you're right. I tried to make my own example. I have edited the question. Sorry for the mistake.
ali
CC BY-SA 4.0
2020-05-12T21:40:53.607Z
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 $Pr(b_1 > b_2)$?
Walrasian Auctioneer
CC BY-SA 4.0
2020-05-12T22:20:05.113Z
@WalrasianAuctioneer it says the probability of the bidder 1's bid is greater than the bidder 2. But when I try to calculate the probability $\int f(v_i)$ with $2/3v_i$, I get the result a function which includes $v_i$ and could not be able to find the expected payoff.
ali
CC BY-SA 4.0
2020-05-12T22:30:10.500Z
Fix bidder type as $v_1$. $P(b_1 > b_2) = P(b_1(v_1) > b_2(v_2)) = P(v_1 > v_2)$. Can you now figure out what it is now?
Walrasian Auctioneer
CC BY-SA 4.0
2020-05-12T22:43:31.290Z
I got this point but I need how can I practise this result on pdf.
ali
<game-theory><auctions><bayesian-game>
First Price Auction (Expected Payoff)
ali
# First Price Auction (Expected Payoff)

(Asked by: username_0 on 2020-05-11)

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.

**Tags:** game-theory, auctions, bayesian-game
> **Comments:**
> **username_1**: Plug in the equilibrium bid into the equation you were attempting to maximise, and you're done!  
> **username_0**: Thank you! I plugged the $2/3v_1$ into maximization problem. Now, what I should write in the maximization problem instead of $\Pr(b_1>b_2)$?  
> **username_2**: Total probability is 1/2 with this pdf...?  
> **username_0**: @VARulle you're right. I tried to make my own example. I have edited the question. Sorry for the mistake.  
> **username_1**: 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 $Pr(b_1 > b_2)$?  
> **username_0**: @WalrasianAuctioneer it says the probability of the bidder 1's bid is greater than the bidder 2. But when I try to calculate the probability $\int f(v_i)$ with $2/3v_i$, I get the result a function which includes $v_i$ and could not be able to find the expected payoff.  
> **username_1**: Fix bidder type as $v_1$. $P(b_1 > b_2) = P(b_1(v_1) > b_2(v_2)) = P(v_1 > v_2)$. Can you now figure out what it is now?  
> **username_0**: I got this point but I need how can I practise this result on pdf.  

---

## 1 answers

### Answer by username_1 on 2020-05-13. Score: 1 | Accepted answer

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

> **Comments:**
> **username_0**: 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  
> **username_1**: 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.  

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