<p>To find expected revenue of the seller in a second price auction with reserved price consisting of two bidders who bid their valuations in equilibrium, we do the following :</p>

<p>Given that valuations are i.i.d with pdf $f$ and CDF $F$ over $[0, \hat{v}]$, the revenue that the seller gets for different realizations of players' valuations are as indicated in the graph below:
<a href="https://i.sstatic.net/SwvIY.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/SwvIY.png" alt="enter image description here"></a></p>

<p>Therefore, expected revenue of the seller is :
\begin{eqnarray*}  &amp;&amp;\int_r^\hat{v}\int_r^{v_2} v_1 f(v_1)f(v_2)dv_1dv_2 + \int_r^\hat{v}\int_{v_2}^\hat{v} v_2 f(v_1)f(v_2)dv_1dv_2  \\ &amp;&amp;+\int_0^r\int_r^\hat{v} r f(v_1)f(v_2)dv_1dv_2  + \int_r^\hat{v}\int_0^r r f(v_1)f(v_2)dv_1dv_2 \\ &amp;=&amp;  \int_r^\hat{v}v_1 (1-F(v_1))f(v_1)dv_1 + \int_r^\hat{v} v_2 (1-F(v_2))f(v_2)dv_2 + 2rF(r)(1-F(r)) \\ &amp;=&amp; 2\int_r^\hat{v} v_2 (1-F(v_2))f(v_2)dv_2 + 2rF(r)(1-F(r))\end{eqnarray*}</p>

CC BY-SA 4.0
2018-05-20T10:04:48.830Z
2018-05-20T14:02:31.297Z
2018-05-20T14:02:31.297Z
Amit
<p>I think that the first part of your question must ask for a conditional probability. In other words, for $v_1$ and $v_2$ representing valuations of player 1 and player 2, we should have been asked for the derivation (namely, the density) of the following probability:</p>

<p>$$Pr( v_1&gt;x \land v_2&gt;x \;|\; r&lt;v_1 \land r&lt;v_2) $$
Thus it is equal to:
$$\frac{d\left( \frac{(1-F(x))^2}{(1-F(r))^2} \right)}{dx} = \frac{2f(x)(1-F(x))}{(1-F(r))^2}$$</p>

<p>From your words, however, it is deduced that what it asks for is the derivation (so the density) of the following probability:</p>

<p>$$Pr(v_1&gt;max(r,x) \;\land\; v_2&gt;max(x,r))$$ 
But, assuming wlog x>r, this is just equal to $\frac{d\left( (1-F(x))^2 \right)}{dx} = 2f(x)(1-F(x))$, so there is no place for $r$ since if $v_1$ and $v_2$ are grater than $x$, and $x$ is greater than $r$, then $v_i$>$x$ implies $v_i$>$r$ for $i=1,2$. So no need to bother with $r$ in this case. But I don't think that your question asks for this.</p>

<p>For the last part of your question, I'm again suspicious whether there is no requirement for $v_1$ and $v_2$ to be higher than $x$. If the only constraint for them is to be higher than the reserve price $r$, then  Amit above has given a very well answer except that $v_1$ or $v_2$ can't get any value less than $r$ in equilibrium, so in his calculations the integrals $\int_0^r\int_r^\hat{v} r f(v_1)f(v_2)dv_1dv_2$  and $ \int_r^\hat{v}\int_0^r r f(v_1)f(v_2)dv_1dv_2 $ are useless. When we remove these two integrals, we get the answer $2\int_r^\hat{v} v_2 (1-F(v_2))f(v_2)dv_2 $ (you can replace variable $v_2$ by $x$ here to get an identical result with your answer). </p>

CC BY-SA 4.0
2018-05-20T15:30:53.913Z
2018-05-20T15:30:53.913Z
Osman Bulut
<p>The situation:</p>
<p>There is a second price auction with 2 players. Their valuations of the object at auction are independently and identically distributed with pdf <span class="math-container">$f$</span> and cdf <span class="math-container">$F$</span> over <span class="math-container">$[0,\hat v]$</span>. Assume <span class="math-container">$f$</span> is continuous and positive over <span class="math-container">$[0,\hat v]$</span>.</p>
<p>A reservation bid <span class="math-container">$r$</span> is now implemented - the winner pays the second of the highest bids including the reservation price, or if both bid lower no-one wins. I want to find the pdf that both bids are above <span class="math-container">$r$</span> and above some <span class="math-container">$x$</span>, and add this to an equation calculating expected revenue for the auctioneer.</p>
<p>I have already found the pdf for both bids being above some value of <span class="math-container">$x$</span>: <span class="math-container">$2f(x)(1-F(x))$</span>. The pdf for both being above <span class="math-container">$r$</span> is <span class="math-container">$(1-F(r))^2$</span>.</p>
<p>I've had a look at an answer for the problem, and it suggests that the combined pdf is <span class="math-container">$\frac{2f(x)(1-F(x))}{(1-F(r))^2}$</span>. Could someone explain to me how this is so?</p>
<p>Then, when calculating expected revenue for the auctioneer, we have for the case where both bids are above <span class="math-container">$r$</span>: <span class="math-container">$(1-F(r))^2\int_r^\hat v{\frac{2f(x)(1-F(x))}{(1-F(r))^2}}dx$</span>. I'm also quite confused why we multiply by <span class="math-container">$(1-F(r))^2$</span>.</p>

<auctions>
Second Price Auction - adjusting PDF for reservation price
Chaerephon
# Second Price Auction - adjusting PDF for reservation price

(Asked by: username_0 on 2018-05-20)

The situation:

There is a second price auction with 2 players. Their valuations of the object at auction are independently and identically distributed with pdf $f$ and cdf $F$ over $[0,\hat v]$. Assume $f$ is continuous and positive over $[0,\hat v]$.

A reservation bid $r$ is now implemented - the winner pays the second of the highest bids including the reservation price, or if both bid lower no-one wins. I want to find the pdf that both bids are above $r$ and above some $x$, and add this to an equation calculating expected revenue for the auctioneer.

I have already found the pdf for both bids being above some value of $x$: $2f(x)(1-F(x))$. The pdf for both being above $r$ is $(1-F(r))^2$.

I've had a look at an answer for the problem, and it suggests that the combined pdf is $\frac{2f(x)(1-F(x))}{(1-F(r))^2}$. Could someone explain to me how this is so?

Then, when calculating expected revenue for the auctioneer, we have for the case where both bids are above $r$: $(1-F(r))^2\int\_r^\hat v{\frac{2f(x)(1-F(x))}{(1-F(r))^2}}dx$. I'm also quite confused why we multiply by $(1-F(r))^2$.

**Tags:** auctions

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## 2 answers

### Answer by username_1 on 2018-05-20. Score: 4

To find expected revenue of the seller in a second price auction with reserved price consisting of two bidders who bid their valuations in equilibrium, we do the following :

Given that valuations are i.i.d with pdf $f$ and CDF $F$ over $[0, \hat{v}]$, the revenue that the seller gets for different realizations of players' valuations are as indicated in the graph below:
[![enter image description here](https://i.sstatic.net/SwvIY.png)](https://i.sstatic.net/SwvIY.png)

Therefore, expected revenue of the seller is :
\begin{eqnarray\*} &&\int\_r^\hat{v}\int\_r^{v\_2} v\_1 f(v\_1)f(v\_2)dv\_1dv\_2 + \int\_r^\hat{v}\int\_{v\_2}^\hat{v} v\_2 f(v\_1)f(v\_2)dv\_1dv\_2 \\ &&+\int\_0^r\int\_r^\hat{v} r f(v\_1)f(v\_2)dv\_1dv\_2 + \int\_r^\hat{v}\int\_0^r r f(v\_1)f(v\_2)dv\_1dv\_2 \\ &=& \int\_r^\hat{v}v\_1 (1-F(v\_1))f(v\_1)dv\_1 + \int\_r^\hat{v} v\_2 (1-F(v\_2))f(v\_2)dv\_2 + 2rF(r)(1-F(r)) \\ &=& 2\int\_r^\hat{v} v\_2 (1-F(v\_2))f(v\_2)dv\_2 + 2rF(r)(1-F(r))\end{eqnarray\*}


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### Answer by username_2 on 2018-05-20. Score: 3

I think that the first part of your question must ask for a conditional probability. In other words, for $v\_1$ and $v\_2$ representing valuations of player 1 and player 2, we should have been asked for the derivation (namely, the density) of the following probability:

$$Pr( v\_1>x \land v\_2>x \;|\; r

From your words, however, it is deduced that what it asks for is the derivation (so the density) of the following probability:

$$Pr(v\_1>max(r,x) \;\land\; v\_2>max(x,r))$$
But, assuming wlog x>r, this is just equal to $\frac{d\left( (1-F(x))^2 \right)}{dx} = 2f(x)(1-F(x))$, so there is no place for $r$ since if $v\_1$ and $v\_2$ are grater than $x$, and $x$ is greater than $r$, then $v\_i$>$x$ implies $v\_i$>$r$ for $i=1,2$. So no need to bother with $r$ in this case. But I don't think that your question asks for this.

For the last part of your question, I'm again suspicious whether there is no requirement for $v\_1$ and $v\_2$ to be higher than $x$. If the only constraint for them is to be higher than the reserve price $r$, then Amit above has given a very well answer except that $v\_1$ or $v\_2$ can't get any value less than $r$ in equilibrium, so in his calculations the integrals $\int\_0^r\int\_r^\hat{v} r f(v\_1)f(v\_2)dv\_1dv\_2$ and $ \int\_r^\hat{v}\int\_0^r r f(v\_1)f(v\_2)dv\_1dv\_2 $ are useless. When we remove these two integrals, we get the answer $2\int\_r^\hat{v} v\_2 (1-F(v\_2))f(v\_2)dv\_2 $ (you can replace variable $v\_2$ by $x$ here to get an identical result with your answer).


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