-1
$\begingroup$

We find expected revenue in first bid auction by following method.

let us say $V_1$ and $V_2$ denotes maximum amount that player 1 and player 2 willing to pay.

$V_1,V_2 \in [0,1]$

In case when we have only 2 players.

Nash equilibrium is $(\frac{V_1}{2},\frac{V_2}{2})$.

Expected revenue = $P(V_1>V_2)\frac{V_1}{2} + P(V_2>V_1) \frac{V_2}{2}$

we replace $V_1$ by $\frac{1+V_2}{2}$ (mean in range $V_2$ to 1).

Now Expected revenue = $\frac{V_2^{2}}{2} +(1-V_2)\frac{1+V_2}{4}$ = $\frac{1+V_2^{2}}{4}$

Now by integrating from 0 to 1 we get $\frac{1}{3}$ answer.

I did not Understand why we take mean.

Can anyone please explain.

$\endgroup$
5
  • $\begingroup$ Do you understand basis statistics and how to calculate expected value? Looking at how you write the equation for expected revenue suggests to me that you should review expected value, since the equation doesn't make sense. $\endgroup$
    – Calvin Lin
    Jun 26, 2013 at 6:47
  • $\begingroup$ We are saying Expected value of $V_1$ in range $V_2$ to 1 is $\frac{V_2+1}{2}$,because it is mean. But I did not understand why we are multiplying by expected value. $\endgroup$
    – TLE
    Jun 26, 2013 at 6:51
  • $\begingroup$ As Calvin says, your expected revenue formula is not quite correct. $\endgroup$
    – Michael
    Jun 26, 2013 at 6:54
  • $\begingroup$ Note that you currently have a constant on the LHS, and a variable on the RHS, which doesn't make sense. This applies to both of your equations. Fix that, by using the correct equation. $\endgroup$
    – Calvin Lin
    Jun 26, 2013 at 6:54
  • $\begingroup$ kk thanks for explanation. $\endgroup$
    – TLE
    Jun 26, 2013 at 6:57

1 Answer 1

1
$\begingroup$

Your formula is not quite correct. The expected revenue for the auctioneer is (using the fact that valuations are independently drawn)

$$ \int_{v_1 \geq v_2} \frac{v_1}{2} + \int_{v_2 > v_1} \frac{v_2}{2} = \int_0^1 \int_0^{v_1}\frac{v_1}{2} dv_2 dv_1 + \int_0^1 \int_0^{v_2}\frac{v_2}{2} dv_1 dv_2 = \frac{1}{3}. $$

$\endgroup$
2
  • $\begingroup$ How did we get this formula. $\endgroup$
    – TLE
    Jun 26, 2013 at 7:25
  • $\begingroup$ @TLE this follows directly from the definition of expected value, and is exactly what you are trying to say. Review that concept first, esp in the case of continuous random variables and joint distributions. $\endgroup$
    – Calvin Lin
    Jun 26, 2013 at 7:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .