Finding the Bayes-Nash Equilibrium for First-Price Auction with 2 bidders
$begingroup$
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?
game-theory economics algorithmic-game-theory
asked Dec 11 '18 at 21:51
user539807user539807
205
$endgroup$
add a comment |
$begingroup$
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?
game-theory economics algorithmic-game-theory
asked Dec 11 '18 at 21:51
user539807user539807
205
$endgroup$
add a comment |
$begingroup$
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?
game-theory economics algorithmic-game-theory
asked Dec 11 '18 at 21:51
user539807user539807
205
$endgroup$
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
game-theory economics algorithmic-game-theory
asked Dec 11 '18 at 21:51
user539807user539807
205
asked Dec 11 '18 at 21:51
user539807user539807
205
edited Dec 12 '18 at 0:48
user539807
asked Dec 11 '18 at 21:51
user539807user539807
205
asked Dec 11 '18 at 21:51
user539807user539807
205
asked Dec 11 '18 at 21:51
user539807user539807
205
205
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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).
answered Dec 17 '18 at 23:45
Herr K.Herr K.
5881617
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035881%2ffinding-the-bayes-nash-equilibrium-for-first-price-auction-with-2-bidders%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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).
answered Dec 17 '18 at 23:45
Herr K.Herr K.
5881617
$endgroup$
add a comment |
$begingroup$
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).
answered Dec 17 '18 at 23:45
Herr K.Herr K.
5881617
$endgroup$
add a comment |
$begingroup$
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).
answered Dec 17 '18 at 23:45
Herr K.Herr K.
5881617
$endgroup$
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).
answered Dec 17 '18 at 23:45
Herr K.Herr K.
5881617
answered Dec 17 '18 at 23:45
Herr K.Herr K.
5881617
answered Dec 17 '18 at 23:45
Herr K.Herr K.
5881617
answered Dec 17 '18 at 23:45
Herr K.Herr K.
5881617
5881617
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035881%2ffinding-the-bayes-nash-equilibrium-for-first-price-auction-with-2-bidders%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
