Finding the Bayes-Nash Equilibrium for First-Price Auction with 2 bidders












2












$begingroup$


Let $sigma_i$ be the strategy profile for bidder $i$ that indicates how they should bid based on their value. As we know, if there are 2 bidders both with their values $v_1,v_2$ on $U[0,1]$ in a First-Price Auction, they will bid $frac{v_1}{2}$ and $frac{v_2}{2}$ respectively, so $sigma_1(v_1)=frac{v_1}{2},sigma_2(v_2)=frac{v_2}{2}$.



However, if bidder 1's value is still $U[0,1]$ but bidder 2's value is $U[0,2]$, there does not exist a Bayes-Nash Equilibrium. Why is this? Specifically, why is the following not a BNE?



begin{equation*}
begin{split}
sigma_1(v_1) &= frac{v_1}{2} \
sigma_2(v_2)&= begin{cases} frac{v_2}{2} & mbox{ if }v_2in[0,1] \ frac{1}{2} & mbox{ if }v_2>1 end{cases}
end{split}
end{equation*}



Equilibrium says for all bidders $i$ and values $v_i$, $sigma_i(v_i)$ is the optimal bid. Is there a bidder and value for which this is not the case?










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$endgroup$

















    2












    $begingroup$


    Let $sigma_i$ be the strategy profile for bidder $i$ that indicates how they should bid based on their value. As we know, if there are 2 bidders both with their values $v_1,v_2$ on $U[0,1]$ in a First-Price Auction, they will bid $frac{v_1}{2}$ and $frac{v_2}{2}$ respectively, so $sigma_1(v_1)=frac{v_1}{2},sigma_2(v_2)=frac{v_2}{2}$.



    However, if bidder 1's value is still $U[0,1]$ but bidder 2's value is $U[0,2]$, there does not exist a Bayes-Nash Equilibrium. Why is this? Specifically, why is the following not a BNE?



    begin{equation*}
    begin{split}
    sigma_1(v_1) &= frac{v_1}{2} \
    sigma_2(v_2)&= begin{cases} frac{v_2}{2} & mbox{ if }v_2in[0,1] \ frac{1}{2} & mbox{ if }v_2>1 end{cases}
    end{split}
    end{equation*}



    Equilibrium says for all bidders $i$ and values $v_i$, $sigma_i(v_i)$ is the optimal bid. Is there a bidder and value for which this is not the case?










    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Let $sigma_i$ be the strategy profile for bidder $i$ that indicates how they should bid based on their value. As we know, if there are 2 bidders both with their values $v_1,v_2$ on $U[0,1]$ in a First-Price Auction, they will bid $frac{v_1}{2}$ and $frac{v_2}{2}$ respectively, so $sigma_1(v_1)=frac{v_1}{2},sigma_2(v_2)=frac{v_2}{2}$.



      However, if bidder 1's value is still $U[0,1]$ but bidder 2's value is $U[0,2]$, there does not exist a Bayes-Nash Equilibrium. Why is this? Specifically, why is the following not a BNE?



      begin{equation*}
      begin{split}
      sigma_1(v_1) &= frac{v_1}{2} \
      sigma_2(v_2)&= begin{cases} frac{v_2}{2} & mbox{ if }v_2in[0,1] \ frac{1}{2} & mbox{ if }v_2>1 end{cases}
      end{split}
      end{equation*}



      Equilibrium says for all bidders $i$ and values $v_i$, $sigma_i(v_i)$ is the optimal bid. Is there a bidder and value for which this is not the case?










      share|cite|improve this question











      $endgroup$




      Let $sigma_i$ be the strategy profile for bidder $i$ that indicates how they should bid based on their value. As we know, if there are 2 bidders both with their values $v_1,v_2$ on $U[0,1]$ in a First-Price Auction, they will bid $frac{v_1}{2}$ and $frac{v_2}{2}$ respectively, so $sigma_1(v_1)=frac{v_1}{2},sigma_2(v_2)=frac{v_2}{2}$.



      However, if bidder 1's value is still $U[0,1]$ but bidder 2's value is $U[0,2]$, there does not exist a Bayes-Nash Equilibrium. Why is this? Specifically, why is the following not a BNE?



      begin{equation*}
      begin{split}
      sigma_1(v_1) &= frac{v_1}{2} \
      sigma_2(v_2)&= begin{cases} frac{v_2}{2} & mbox{ if }v_2in[0,1] \ frac{1}{2} & mbox{ if }v_2>1 end{cases}
      end{split}
      end{equation*}



      Equilibrium says for all bidders $i$ and values $v_i$, $sigma_i(v_i)$ is the optimal bid. Is there a bidder and value for which this is not the case?







      game-theory economics algorithmic-game-theory






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      share|cite|improve this question













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      edited Dec 12 '18 at 0:48







      user539807

















      asked Dec 11 '18 at 21:51









      user539807user539807

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          $begingroup$

          It is easy to see why the strategy profile does not form a BNE. Since bidder 2 never bids above $frac12$, for a sufficiently high $v_1$, an alternative strategy $sigma_1'(v_1)=frac12+epsilon$ is a better response to $sigma_2$ than the proposed $sigma_1$.



          For example, suppose bidder 1 observes $v_1=1$. Following $sigma_1$, his probability of winning is $Pr(sigma_2le frac12)=frac12$ and so his expected payoff is $frac12(1-frac12)=frac14$. But if he bids slightly above $frac12$, his probability of winning jumps to $1$, and his expected payoff is just slightly below $frac12$, which is greater than $frac14$.



          In fact, there does exist a BNE in the asymmetric two-bidder first price auction where values are uniformly distributed on $[0,omega_i]$, with $omega_1ne omega_2$. The equilibrium bidding strategy is given by
          begin{equation}
          sigma_i(v_i)=frac{1}{k_iv_i}left(1-sqrt{1-k_iv_i^2}right),qquadtext{where }k_i=frac1{omega_i^2}-frac1{omega_j^2}.
          end{equation}

          For the details of derivation, I'd refer you to Section 4.3 of Krishna (2010).






          share|cite|improve this answer









          $endgroup$













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            $begingroup$

            It is easy to see why the strategy profile does not form a BNE. Since bidder 2 never bids above $frac12$, for a sufficiently high $v_1$, an alternative strategy $sigma_1'(v_1)=frac12+epsilon$ is a better response to $sigma_2$ than the proposed $sigma_1$.



            For example, suppose bidder 1 observes $v_1=1$. Following $sigma_1$, his probability of winning is $Pr(sigma_2le frac12)=frac12$ and so his expected payoff is $frac12(1-frac12)=frac14$. But if he bids slightly above $frac12$, his probability of winning jumps to $1$, and his expected payoff is just slightly below $frac12$, which is greater than $frac14$.



            In fact, there does exist a BNE in the asymmetric two-bidder first price auction where values are uniformly distributed on $[0,omega_i]$, with $omega_1ne omega_2$. The equilibrium bidding strategy is given by
            begin{equation}
            sigma_i(v_i)=frac{1}{k_iv_i}left(1-sqrt{1-k_iv_i^2}right),qquadtext{where }k_i=frac1{omega_i^2}-frac1{omega_j^2}.
            end{equation}

            For the details of derivation, I'd refer you to Section 4.3 of Krishna (2010).






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              It is easy to see why the strategy profile does not form a BNE. Since bidder 2 never bids above $frac12$, for a sufficiently high $v_1$, an alternative strategy $sigma_1'(v_1)=frac12+epsilon$ is a better response to $sigma_2$ than the proposed $sigma_1$.



              For example, suppose bidder 1 observes $v_1=1$. Following $sigma_1$, his probability of winning is $Pr(sigma_2le frac12)=frac12$ and so his expected payoff is $frac12(1-frac12)=frac14$. But if he bids slightly above $frac12$, his probability of winning jumps to $1$, and his expected payoff is just slightly below $frac12$, which is greater than $frac14$.



              In fact, there does exist a BNE in the asymmetric two-bidder first price auction where values are uniformly distributed on $[0,omega_i]$, with $omega_1ne omega_2$. The equilibrium bidding strategy is given by
              begin{equation}
              sigma_i(v_i)=frac{1}{k_iv_i}left(1-sqrt{1-k_iv_i^2}right),qquadtext{where }k_i=frac1{omega_i^2}-frac1{omega_j^2}.
              end{equation}

              For the details of derivation, I'd refer you to Section 4.3 of Krishna (2010).






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                It is easy to see why the strategy profile does not form a BNE. Since bidder 2 never bids above $frac12$, for a sufficiently high $v_1$, an alternative strategy $sigma_1'(v_1)=frac12+epsilon$ is a better response to $sigma_2$ than the proposed $sigma_1$.



                For example, suppose bidder 1 observes $v_1=1$. Following $sigma_1$, his probability of winning is $Pr(sigma_2le frac12)=frac12$ and so his expected payoff is $frac12(1-frac12)=frac14$. But if he bids slightly above $frac12$, his probability of winning jumps to $1$, and his expected payoff is just slightly below $frac12$, which is greater than $frac14$.



                In fact, there does exist a BNE in the asymmetric two-bidder first price auction where values are uniformly distributed on $[0,omega_i]$, with $omega_1ne omega_2$. The equilibrium bidding strategy is given by
                begin{equation}
                sigma_i(v_i)=frac{1}{k_iv_i}left(1-sqrt{1-k_iv_i^2}right),qquadtext{where }k_i=frac1{omega_i^2}-frac1{omega_j^2}.
                end{equation}

                For the details of derivation, I'd refer you to Section 4.3 of Krishna (2010).






                share|cite|improve this answer









                $endgroup$



                It is easy to see why the strategy profile does not form a BNE. Since bidder 2 never bids above $frac12$, for a sufficiently high $v_1$, an alternative strategy $sigma_1'(v_1)=frac12+epsilon$ is a better response to $sigma_2$ than the proposed $sigma_1$.



                For example, suppose bidder 1 observes $v_1=1$. Following $sigma_1$, his probability of winning is $Pr(sigma_2le frac12)=frac12$ and so his expected payoff is $frac12(1-frac12)=frac14$. But if he bids slightly above $frac12$, his probability of winning jumps to $1$, and his expected payoff is just slightly below $frac12$, which is greater than $frac14$.



                In fact, there does exist a BNE in the asymmetric two-bidder first price auction where values are uniformly distributed on $[0,omega_i]$, with $omega_1ne omega_2$. The equilibrium bidding strategy is given by
                begin{equation}
                sigma_i(v_i)=frac{1}{k_iv_i}left(1-sqrt{1-k_iv_i^2}right),qquadtext{where }k_i=frac1{omega_i^2}-frac1{omega_j^2}.
                end{equation}

                For the details of derivation, I'd refer you to Section 4.3 of Krishna (2010).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 17 '18 at 23:45









                Herr K.Herr K.

                5881617




                5881617






























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