3
$\begingroup$

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 $(1-F(r))^2$.

$\endgroup$

2 Answers 2

4
$\begingroup$

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

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*}

$\endgroup$
3
$\begingroup$

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:

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

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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