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I am having a hard time figuring out a problem. In a first price auction with a reserve price R and values of the bidders are U[0,1], how do we find expected revenue given the strategy of both of them is V/2. I am thinking of it like this if v < r or v = r revenue is zero. So what we are really looking at is scenarios where v>r for bidder 1 and v < r for bidder 2 (and vice versa) or we have a situation where we have both bidders v > r and so we'll have to take the max of that. Am I on the right track here? Why is this not bayes-nash equilibrium? I having been trying to figure this out forever. Help!

Martin Van der Linden
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guest
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  • With respect to your last question, why do you think it is not a Bayesian-Nash equilibrium? For the expected revenue, you are on the right track. Now you "just" need to compute the expected value of this random variable you just described. Is this something you feel comfortable doing or do you have any question regarding the way this expectation should be computed? As far as the expected value of the max of two random variables is concerned, see http://math.stackexchange.com/questions/197299/expected-value-of-maximum-of-two-random-variables-from-uniform-distribution/573833#573833 if needed. – Martin Van der Linden Feb 10 '14 at 02:42
  • So you mean my solution would just be the expected value of this random variable or the expected value of this random variable plus 2(r(1-r)r)? The question in my assignment asks, why this isnt a bayes nash equil. So I am a little confused. – guest Feb 10 '14 at 03:38

1 Answers1

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For your future questions, you might wanna know that it is good practice on MathSE to tag assignment or homework question with the homework tag.

Regarding the strategy being a Bayesian-Nash equilibrium, my comment was misleading. I did not take into account the fact that you have a reservation price $R$. The strategy profile $(V/2,V/2)$ is a Bayesian-Nash equilibrium only for $R =0$.

To convince yourself that $(V/2,V/2)$ is not a Bayesian-Nash equilibrium for other values of $R$, consider the example $R = 1/4$. The strategy profile prescribes that players play $V/2$ whatever the actual realization of $V$. Now ask yourself whether this is really optimal when, for instance, $V$ happens to be $1/3$. In this case, does playing $1/6$ really maximizing your expected utility given that the value of the other bidder is distributed as a $U[0,1]$ and that you know that she will always play $V/2$? Once you get some intuition from this example, try to generalize it to any $R>0$.

As for the expected revenue, you do not need to bother about equilibrium considerations. You are being told that the agents play $(V/2,V/2)$, no matter that it be an equilibrium.

Now I realize my comment was maybe misleading again : the random variable I referred to was not $U[0,1]$ but the random variable you describe, namely the random variable which takes values

  • $V_1$ when $V_1 > R$ and $V_2 \leq R$
  • $V_2$ when $V_2 > R$ and $V_1 \leq R$
  • $\max \{V_1,V_2\}$ when $V_1>R$ and $V_2 >R$
  • and $0$ otherwise.

This is really the random variable you want to compute the expected value of.

Martin Van der Linden
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