<p>Let <span class="math-container">$n \geq 2$</span>. Observe that
<span class="math-container">$$ \mathbb{E}[u_i] = \mathbb{E}[u_i|v_i &gt; r]P(v_i &gt; r) + \mathbb{E}[u_i|v_i \leq r]P(v_i \leq r)  = \mathbb{E}[u_i|v_i &gt; r]P(v_i &gt; r)$$</span>
since <span class="math-container">$\mathbb{E}[u_i|v_i \leq r] = 0$</span>. Moreover, <span class="math-container">$P(v_i &gt; r) = 1 - r$</span> if the values are standard uniform. So to compute the expected payoff <span class="math-container">$\mathbb{E}[u_i]$</span>, all that remains is to compute <span class="math-container">$\mathbb{E}[u_i|v_i &gt; r]$</span>.</p>
<p>Let <span class="math-container">$p$</span> denote the highest bid submitted by your opponents. We can then decompose
<span class="math-container">$$\mathbb{E}[u_i|v_i &gt; r] = \mathbb{E}[u_i|v_i &gt; r, p \leq r]P(p \leq r) + \mathbb{E}[u_i|v_i &gt; r, p &gt; r]P(p &gt; r)$$</span>
Starting with the first term,
<span class="math-container">$$ \mathbb{E}[u_i|v_i &gt; r, p \leq r] = \mathbb{E}[v_i - r|v_i &gt; r] = \frac{1+r}{2} - r = \frac{1-r}{2}$$</span>
More obviously, <span class="math-container">$P(p \leq r) = P(v_j \leq r)^{n-1} = r^{n-1}$</span> and so <span class="math-container">$P(p &gt; r) = 1 - r^{n-1}$</span>. Using symmetry of bidders and results on <a href="https://math.stackexchange.com/questions/2251604/expected-value-of-the-k-th-order-statistic-from-uniform-random-variables">order statistics</a>, one can see that</p>
<p><span class="math-container">\begin{equation}
\begin{split}
\mathbb{E}[u_i|v_i &gt; r, p &gt; r] &amp;&amp;= \mathbb{E}[u_i|\text{win}, v_i &gt; r, p &gt; r] P(\text{win}) \\
&amp;&amp;= \left(r + (1-r)\left(\frac{n}{n+1}\right) -  r - (1-r)\left(\frac{n-1}{n+1}\right) \right)\frac{1}{n} \\
&amp;&amp;= \frac{1-r}{n(n+1)}
\end{split}
\end{equation}</span>
Putting this all together, we have
<span class="math-container">$$ \mathbb{E}[u_i|v_i &gt; r] = \frac{1-r}{2} \times r^{n-1} + \frac{1-r}{n(n+1)} \times (1 - r^{n-1}) $$</span>
and so (multiplying by <span class="math-container">$1 - r$</span>)
<span class="math-container">$$ \mathbb{E}[u_i] = \frac{(1-r)^2 r^{n-1}}{2} + \frac{(1-r)^2 (1 - r^{n-1})}{n(n+1)} $$</span>
Observe that, if <span class="math-container">$r = 1$</span>, then <span class="math-container">$\mathbb{E}[u_i] = 0$</span> as one would expect. Moreover, if <span class="math-container">$r = 0$</span> (no reservation price), then one obtains the (hopefully familiar?) expression
<span class="math-container">$$ \mathbb{E}[u_i] = \frac{1}{n(n+1)} $$</span></p>

CC BY-SA 4.0
2021-11-02T13:44:18.800Z
I'm still not entirely sure if I get how $\mathbb{E}[u_i|v_i>r,p>r]$ becomes $(r+(1-r)\frac{n}{n+1}-r-(1-r)\frac{n-1}{n+1})\frac{1}{n}$, so I can't really figure out if there is a mistake in that part.

But I would think that $\mathbb{E}[u_i|v_i>r,p<r]=v_i-r$ since the bidder knows her own valuation and the reserve price, and doesn't have to take the expectation of $v_i$.
Sarah Plaug
CC BY-SA 4.0
2021-11-02T14:50:24.180Z
Sure, if you want to also condition on $v_i$. But I thought you wanted the unconditional expected utility? (In which case you need to average over all possible $v_i$ as I have done.)
PS I suspect that the error may indeed be with the first bit you mention, maybe you can have a go at that?
afreelunch
CC BY-SA 4.0
2021-11-02T15:18:58.967Z
OK mistake found! I somehow divided by $1-r$ at the end...
afreelunch
CC BY-SA 4.0
2021-11-02T15:21:33.630Z
Ahh, of course! you calculate the ex-ante expected utility, while I was originally thinking of the interim expected utility. Though on second thoughts, I do need both the interim and ex-ante expected utility. And the interim shouldn't be any problem for me to derive when I have the expression for the ex-ante expected utility.

But would you mind spelling out how you got $E[u_i|v_i>r,p>r]$? And great that you found the mistake :D
Sarah Plaug
CC BY-SA 4.0
2021-11-02T15:23:19.287Z
It's a bit complicated actually.... The $1/n$ term is because of the symmetry: each player is equally likely to win. The other term comes from thinking about the expected value of the maximum and highest highest of $n - 1$ uniform RVs on the interval $[r, 1]$.
afreelunch
CC BY-SA 4.0
2021-11-02T10:39:26.587Z
2021-11-02T15:21:50.467Z
2021-11-02T15:21:50.467Z
afreelunch
<p>I would like to know what bidder <em>i</em>'s expected payoff looks like in a second-price auction with <span class="math-container">$N=\{1,2,...,n\}$</span> bidders, where each bidder <span class="math-container">$i\in N$</span> has independent and uniform distributed valuations <span class="math-container">$v_i\sim U(0,1)$</span>, and the auctioneer sets a reserve price, <span class="math-container">$r$</span>. <br />
I assume that each bidder already knows <span class="math-container">$r$</span> like she knows her own true valuation, but only knows the distribution of the other bidders' valuations, and she assumes that all other bidders bid their true valuations <span class="math-container">$b_{-i}^*=v_{-i}$</span></p>
<p>Hence, bidder <em>i</em>'s payoff function is:
<span class="math-container">$u_i(b_i,b_{-i},r) = \left\{
    \begin{array}{lr} 
      v_i-\max\{b_{-i}\} &amp; \text{if } b_i&gt;\max\{b_{-i}\}&gt;r \\
      v_i-r &amp; \text{if } b_i&gt;r&gt;\max\{b_{-i}\}\\
      0 &amp; \text{if } b_i&lt;r \vee b_i&lt;\max\{b_{-i}\}
      \end{array}
      \right.$</span></p>
<p>I am unsure exactly what the expected payoff looks like, but I have assumed that it would be something like:
<span class="math-container">$ \mathbb{E}[u_i] = \mathbb{P}(b_i&gt;\max\{b_{-i}\}&gt;r)\cdot \mathbb{E}\left[  v_i-\max\{b_{-i}\} | b_i&gt;\max\{b_{-i}\}&gt;r  \right] \\
+\mathbb{P}(b_i&gt;r&gt;\max\{b_{-i}\})\cdot \mathbb{E}\left[ v_i-r | b_i&gt;r&gt;\max\{b_{-i}\} \right]$</span> <br />
but I don't know where to go from here..</p>
<p>Any help is highly appreciated!</p>

CC BY-SA 4.0
2021-11-02T08:33:00.403Z
Yes, so the reserve price r is set by the auctioneer to maximize his expected revenue. Thus, each bidder just sees the reserve price as if it was an additional bidder's bid. Then one could argue that it is still optimal for the bidder to bid their true values, which is not necessarily equal to the reserve price. But now there is a probability that the second-highest bidder bids below the reserve price, in which case the highest bidder doesn't pay the second-highest bid, but pays the reserve price instead.
If I understand your question correctly?
Sarah Plaug
CC BY-SA 4.0
2021-11-02T08:36:37.590Z
Yes, I meant to ask why you do not assume that $i$ bids her valuation. I mixed up the notations, sorry.
Giskard
<game-theory><expected-utility><auctions>
What is the expected payoff for a bidder in a second-price auction with N uniform distributed bidders, when the auctioneer sets a reserve price?
Sarah Plaug
# What is the expected payoff for a bidder in a second-price auction with N uniform distributed bidders, when the auctioneer sets a reserve price?

(Asked by: username_0 on 2021-11-01)

I would like to know what bidder *i*'s expected payoff looks like in a second-price auction with $N=\{1,2,...,n\}$ bidders, where each bidder $i\in N$ has independent and uniform distributed valuations $v\_i\sim U(0,1)$, and the auctioneer sets a reserve price, $r$.   
I assume that each bidder already knows $r$ like she knows her own true valuation, but only knows the distribution of the other bidders' valuations, and she assumes that all other bidders bid their true valuations $b\_{-i}^\*=v\_{-i}$

Hence, bidder *i*'s payoff function is:
$u\_i(b\_i,b\_{-i},r) = \left\{
\begin{array}{lr}
v\_i-\max\{b\_{-i}\} & \text{if } b\_i>\max\{b\_{-i}\}>r \\
v\_i-r & \text{if } b\_i>r>\max\{b\_{-i}\}\\
0 & \text{if } b\_i

I am unsure exactly what the expected payoff looks like, but I have assumed that it would be something like:
$ \mathbb{E}[u\_i] = \mathbb{P}(b\_i>\max\{b\_{-i}\}>r)\cdot \mathbb{E}\left[ v\_i-\max\{b\_{-i}\} | b\_i>\max\{b\_{-i}\}>r \right] \\
+\mathbb{P}(b\_i>r>\max\{b\_{-i}\})\cdot \mathbb{E}\left[ v\_i-r | b\_i>r>\max\{b\_{-i}\} \right]$   
but I don't know where to go from here..

Any help is highly appreciated!

**Tags:** game-theory, expected-utility, auctions
> **Comments:**
> **username_0**: Yes, so the reserve price r is set by the auctioneer to maximize his expected revenue. Thus, each bidder just sees the reserve price as if it was an additional bidder's bid. Then one could argue that it is still optimal for the bidder to bid their true values, which is not necessarily equal to the reserve price. But now there is a probability that the second-highest bidder bids below the reserve price, in which case the highest bidder doesn't pay the second-highest bid, but pays the reserve price instead. If I understand your question correctly?  
> **username_1**: Yes, I meant to ask why you do not assume that $i$ bids her valuation. I mixed up the notations, sorry.  

---

## 1 answers

### Answer by username_2 on 2021-11-02. Score: 3 | Accepted answer

Let $n \geq 2$. Observe that
$$ \mathbb{E}[u\_i] = \mathbb{E}[u\_i|v\_i > r]P(v\_i > r) + \mathbb{E}[u\_i|v\_i \leq r]P(v\_i \leq r) = \mathbb{E}[u\_i|v\_i > r]P(v\_i > r)$$
since $\mathbb{E}[u\_i|v\_i \leq r] = 0$. Moreover, $P(v\_i > r) = 1 - r$ if the values are standard uniform. So to compute the expected payoff $\mathbb{E}[u\_i]$, all that remains is to compute $\mathbb{E}[u\_i|v\_i > r]$.

Let $p$ denote the highest bid submitted by your opponents. We can then decompose
$$\mathbb{E}[u\_i|v\_i > r] = \mathbb{E}[u\_i|v\_i > r, p \leq r]P(p \leq r) + \mathbb{E}[u\_i|v\_i > r, p > r]P(p > r)$$
Starting with the first term,
$$ \mathbb{E}[u\_i|v\_i > r, p \leq r] = \mathbb{E}[v\_i - r|v\_i > r] = \frac{1+r}{2} - r = \frac{1-r}{2}$$
More obviously, $P(p \leq r) = P(v\_j \leq r)^{n-1} = r^{n-1}$ and so $P(p > r) = 1 - r^{n-1}$. Using symmetry of bidders and results on [order statistics](https://math.stackexchange.com/questions/2251604/expected-value-of-the-k-th-order-statistic-from-uniform-random-variables), one can see that

\begin{equation}
\begin{split}
\mathbb{E}[u\_i|v\_i > r, p > r] &&= \mathbb{E}[u\_i|\text{win}, v\_i > r, p > r] P(\text{win}) \\
&&= \left(r + (1-r)\left(\frac{n}{n+1}\right) - r - (1-r)\left(\frac{n-1}{n+1}\right) \right)\frac{1}{n} \\
&&= \frac{1-r}{n(n+1)}
\end{split}
\end{equation}
Putting this all together, we have
$$ \mathbb{E}[u\_i|v\_i > r] = \frac{1-r}{2} \times r^{n-1} + \frac{1-r}{n(n+1)} \times (1 - r^{n-1}) $$
and so (multiplying by $1 - r$)
$$ \mathbb{E}[u\_i] = \frac{(1-r)^2 r^{n-1}}{2} + \frac{(1-r)^2 (1 - r^{n-1})}{n(n+1)} $$
Observe that, if $r = 1$, then $\mathbb{E}[u\_i] = 0$ as one would expect. Moreover, if $r = 0$ (no reservation price), then one obtains the (hopefully familiar?) expression
$$ \mathbb{E}[u\_i] = \frac{1}{n(n+1)} $$

> **Comments:**
> **username_0**: I'm still not entirely sure if I get how $\mathbb{E}[u_i|v_i>r,p>r]$ becomes $(r+(1-r)\frac{n}{n+1}-r-(1-r)\frac{n-1}{n+1})\frac{1}{n}$, so I can't really figure out if there is a mistake in that part.  But I would think that $\mathbb{E}[u_i|v_i>r,p<r]=v_i-r$ since the bidder knows her own valuation and the reserve price, and doesn't have to take the expectation of $v_i$.  
> **username_2**: Sure, if you want to also condition on $v_i$. But I thought you wanted the unconditional expected utility? (In which case you need to average over all possible $v_i$ as I have done.) PS I suspect that the error may indeed be with the first bit you mention, maybe you can have a go at that?  
> **username_2**: OK mistake found! I somehow divided by $1-r$ at the end...  
> **username_0**: Ahh, of course! you calculate the ex-ante expected utility, while I was originally thinking of the interim expected utility. Though on second thoughts, I do need both the interim and ex-ante expected utility. And the interim shouldn't be any problem for me to derive when I have the expression for the ex-ante expected utility.  But would you mind spelling out how you got $E[u_i|v_i>r,p>r]$? And great that you found the mistake :D  
> **username_2**: It's a bit complicated actually.... The $1/n$ term is because of the symmetry: each player is equally likely to win. The other term comes from thinking about the expected value of the maximum and highest highest of $n - 1$ uniform RVs on the interval $[r, 1]$.  

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