I am studing a little bit of auction theory. I found the optimal bid value in the Milgrom paper for the first price auction that is $$ P=v \frac{n-1}{n} $$ where $P$ is the optimal bid, $v$ is the true value and $n$ is the number of bidders. Now I am wondering if there exist an equivalent formula for reverse auctions. What is the formula for the optimal bidding in reverse auctions?
$\begingroup$
$\endgroup$
4
-
1$\begingroup$ Exactly the same method works. Where did you run into trouble? $\endgroup$– Steven LandsburgFeb 8 '15 at 20:25
-
$\begingroup$ I would like to see the resulting formula or some references. I have some doubt in the convergence of the distribution. $\endgroup$– emanueleFeb 8 '15 at 20:31
-
$\begingroup$ What is reverse auction? What are the rules? Are the rules different from those of a first price auction? $\endgroup$– Herr K.Feb 9 '15 at 0:51
-
$\begingroup$ Reverse auction are auctions where the bidder is the final seller of the goods. Tendering is a reverse auction. $\endgroup$– emanueleFeb 9 '15 at 3:37
Add a comment
|
$\begingroup$
$\endgroup$
12
A first price standard and reverse auction are formally equivalent to each other, and the same method can be used to solve both:
First Price Auction
In a first price auction, $n$ bidders choose their bid, $b_i$, as a function of their value $v_i$ (distributed according to $F$. They seek to maximise their expected payoff: $$[v_i-b_i(v_i)]\Pr(b_i\geq\max_j b_j).$$ If we look for a symmetric equilibrium in which bidders with higher values bid more then the probability that I have the highest bid is just the probability I have the highest value: $$\Pr(b_i\geq\max_j b_j)=F(v_i)^{n-1}.$$
A useful trick is to think of a bidder as choosing what kind of value to "pretend to have". If we call their choice $\widetilde{v}$ then a bidder's probablem is to $$\max_{\widetilde{v}}[v_i-b(\widetilde{v})]F(\widetilde{v})^{n-1}.$$
Differentiation yields the first-order condition: $$(n-1)F'(\widetilde{v})F(\widetilde{v})^{n-2}[v_i-b(\widetilde{v})]-F(\widetilde{v})^{n-1}b'(\widetilde{v})=0.$$
In equilibrium, we know that a bidder should not want to pretend to be anyone other than his true type, so this first-order condition must hold at $\widetilde{v}=v_i$: $$(n-1)F'(v_i)F(v_i)^{n-2}[v_i-b(v_i)]-F(v_i)^{n-1}b'(v_i)=0.$$ This is a differential equation that can be solved for the equilibrium bid function, $b(v_i)$. In, particular, if values are uniformly distributed on $[0,1]$, we have $F(v_i)=v_i$ and the equation becomes $$(n-1)v_i^{n-2}[v_i-b(v_i)]-v_i^{n-1}b'(v_i)=0,$$ which can be simplified to $$(n-1)\frac{1}{v_i}[v_i-b(v_i)]=b'(v_i).$$
The solution to this differential equation is $$b(v_i)=v_i\frac{n-1}{n},$$ which is what you have in your question.
Reverse Auction
In a reverse auction, $n$ bidders choose their bid, $b_i$, as a function of their cost $c_i$ (distributed according to $F$. They seek to maximise their expected payoff: $$[b_i(c_i)-c_i]\Pr(b_i\leq\min_j b_j).$$
If we look for a symmetric equilibrium in which bidders with higher costs bid more then the probability that I have the lowest bid is just the probability I have the lowest cost: $$\Pr(b_i\leq\min_j b_j)=[1-F(c_i)]^{n-1}.$$
A useful trick is to think of a bidder as choosing what kind of cost to "pretend to have". If we call their choice $\widetilde{c}$ then a bidder's probblem is to $$\max_{\widetilde{c}}[b(\widetilde{c})-c_i][1-F(\widetilde{c})]^{n-1}.$$
Differentiation yields the first-order condition: $$-(n-1)F'(\widetilde{c})[1-F(\widetilde{c})]^{n-2}[b(\widetilde{c})-c_i]+[1-F(\widetilde{c})]^{n-1}b'(\widetilde{c})=0.$$
In equilibrium, we know that a bidder should not want to pretend to be anyone other than his true type, so this first-order condition must hold at $\widetilde{c}=c_i$: $$-(n-1)F'(c_i)[1-F(c_i)]^{n-2}[b(c_i)-c_i]+[1-F(c_i)]^{n-1}b'(c_i)=0.$$ This is a differential equation that can be solved for the equilibrium bid function, $b(c_i)$. In, particular, if costs are uniformly distributed on $[0,1]$, we have $F(c_i)=c_i$ and the differential equation can be simplified to $$-(n-1)[1-c_i]^{n-2}[b(c_i)-c_i]+[1-c_i]^{n-1}b'(c_i)=0.$$ $$(n-1)\frac{1}{1-c_i}[b(c_i)-c_i]=b'(c_i).$$ The solution to this differential equation is $$b(c_i)=\frac{1+c_i(n-1)}{n}.$$
Comparing the two
The two solutions, $$b(v_i)=v_i\frac{n-1}{n},\qquad b(c_i)=\frac{1+c_i(n-1)}{n}$$ are equivalent in the sense that a bidder in the first price auction shades (bids below his true valuation) by exactly the same amount as a bidder in a procurement auction bids above his true cost. You can confirm this by, for example, plotting these two bid functions for some value of $n$. Here's an example with $n=2$:
-
$\begingroup$ The second solution cannot be true. The value of the offer must be smaller by increasing the number of the bidders, otherwise public tendering would be meaningles. $\endgroup$– emanueleFeb 9 '15 at 13:53
-
$\begingroup$ If I am the only seller,i can set the price i like. If am in a crowded market i must lowering my price. So your solution has an error somewhere. $\endgroup$– emanueleFeb 9 '15 at 13:56
-
$\begingroup$ @emanuele There's no error: differentiating $b(c_i)$ with respect to $n$, one obtains $b'(c_i)=(c_i-1)/n^2<0$ so the solution does predict that sellers will offer to perform the task for less if the number of bidders is large—just as your intuition suggests it should. $\endgroup$ Feb 9 '15 at 14:11
-
$\begingroup$ @Ubiquitus sorry but i don't understand. $b'(c_i)<0$ if and only if $c_i<1$ $\endgroup$– emanueleFeb 9 '15 at 14:36
-
$\begingroup$ ok. i get it, but why cost are limited in $[0,1]$ range? $\endgroup$– emanueleFeb 9 '15 at 14:42