CS880 Game Theory - 统计代写答疑辅导

分类： CS880 Game Theory

经济代写|CS880 Game Theory

Posted on 2023年3月28日2023年3月28日 by statistics-lab

Statistics-lab™可以为您提供wisc.edu CS880 Game Theory博弈论课程的代写代考和辅导服务！

CS880 Game Theory课程简介

The course may cover topics such as game theory, mechanism design, computational finance, and data analysis. Students may learn how to use programming languages and software tools to build and analyze economic models and simulations, and may also study how to apply these techniques to real-world economic problems.

Overall, this course seems to combine elements of economics, computer science, and mathematics to provide students with a unique set of skills and knowledge for analyzing and designing economic systems in a rapidly changing and increasingly complex world.

PREREQUISITES

Project details and timeline
Project details and ideas can be found here (UW access only).

Feb 22: Short description of topic, goals and project team due (as part of HW1).
Mar 22: Up to one page report of progress, reference material, plans for the remainder of the semester. Before this date, please make an appointment with Shuchi to discuss potential topics and references.
May 3: Final project reports due.
May 5: Two projects (selected on the basis of the final reports) to be showcased during this lecture.

CS880 Game Theory HELP（EXAM HELP， ONLINE TUTOR）

问题 1.

Consider the stage game $G$ shown in Table 2 .
\begin{tabular}{c|c|c|c|}
\hline & $a$ & $b$ & $c$ \
\hline$A$ & $(4,4)$ & $(-1,5)$ & $(2,2)$ \
\hline$B$ & $(5,-1)$ & $(1,1)$ & $(2,2)$ \
\hline$C$ & $(2,2)$ & $(2,2)$ & $(3.5,3.5)$ \
\hline
\end{tabular}
Table 2: Stage game
(a) Find the worst Nash equilibrium (pure action) for each player in $G$ and the corresponding payoffs.

To find the worst Nash equilibrium (pure action) for each player in the game $G$, we need to find the equilibrium where each player gets their lowest possible payoff.

Player 1’s worst Nash equilibrium is when they play strategy $B$ and player 2 plays strategy $a$. In this equilibrium, player 1 gets a payoff of $-1$, which is their lowest possible payoff, and player 2 gets a payoff of $5$, which is their highest possible payoff among the Nash equilibria. Therefore, the payoffs for player 1 and player 2 in this equilibrium are $(-1,5)$.

Player 2’s worst Nash equilibrium is when they play strategy $c$ and player 1 plays strategy $A$. In this equilibrium, player 2 gets a payoff of $2$, which is their lowest possible payoff, and player 1 gets a payoff of $2$, which is their highest possible payoff among the Nash equilibria. Therefore, the payoffs for player 1 and player 2 in this equilibrium are $(4,2)$.

Note that there is no Nash equilibrium where both players get their lowest possible payoff simultaneously, as player 1’s lowest possible payoff is $-1$ and player 2’s lowest possible payoff is $2$.

问题 2.

(b) Consider $G^2$ : the finitely repeated game, where $G$ is repeated for two periods.
(i) Is there a subgame perfect equilibrium of $G^2$ where $(A, a)$ is played in the first period? Explain your answer.
(ii) Is there a Nash equilibrium of $G^2$ where $(A, a)$ is played in the first period? Explain your answer.

(i) To find a subgame perfect equilibrium (SPE) of $G^2$ where $(A,a)$ is played in the first period, we need to analyze the game in each period separately and then look for the SPE that results from those strategies.

In the first period, if player 1 plays $A$, player 2 has an incentive to deviate to $b$ to get a higher payoff of $5$ instead of $4$. Therefore, $(A,a)$ is not an SPE in the first period.

In the second period, if the game is played as a one-shot game, then the Nash equilibrium is $(C,c)$. However, since the game is repeated, player 2 has an incentive to deviate to $b$ in the second period to punish player 1 for playing $A$ in the first period.

Therefore, there is no SPE of $G^2$ where $(A,a)$ is played in the first period.

(ii) To find a Nash equilibrium of $G^2$ where $(A,a)$ is played in the first period, we need to look for a Nash equilibrium in each period that is consistent with the strategy profile played in the other period.

In the first period, if player 1 plays $A$, player 2 has an incentive to play $b$ to get a higher payoff of $5$ instead of $4$. Therefore, $(A,a)$ is not a Nash equilibrium in the first period.

In the second period, if the game is played as a one-shot game, then the Nash equilibrium is $(C,c)$. However, if player 1 expects player 2 to play $c$ in the second period, then player 1 has an incentive to deviate from $A$ to $B$ in the first period to get a higher payoff of $5$ instead of $4$. Therefore, $(A,a)$ is not a Nash equilibrium in the second period either.

Therefore, there is no Nash equilibrium of $G^2$ where $(A,a)$ is played in the first period.

Textbooks

• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.

此图像的alt属性为空；文件名为%E7%B2%89%E7%AC%94%E5%AD%97%E6%B5%B7%E6%8A%A5-1024x575-10.png — CS880 Game Theory

Statistics-lab™可以为您提供wisc.edu CS880 Game Theory博弈论课程的代写代考和辅导服务！请认准Statistics-lab™. Statistics-lab™为您的留学生涯保驾护航。

经济代写|CS880 Game Theory

Posted on 2023年3月24日2023年3月24日 by statistics-lab

Statistics-lab™可以为您提供wisc.edu CS880 Game Theory博弈论课程的代写代考和辅导服务！

CS880 Game Theory课程简介

PREREQUISITES

Project details and timeline
Project details and ideas can be found here (UW access only).

Feb 22: Short description of topic, goals and project team due (as part of HW1).
Mar 22: Up to one page report of progress, reference material, plans for the remainder of the semester. Before this date, please make an appointment with Shuchi to discuss potential topics and references.
May 3: Final project reports due.
May 5: Two projects (selected on the basis of the final reports) to be showcased during this lecture.

CS880 Game Theory HELP（EXAM HELP， ONLINE TUTOR）

问题 1.

The stage game is shown in Table 1 .
\begin{tabular}{c|c|c|}
\hline & $\mathrm{H}$ & $\mathrm{L}$ \
\hline \hline $\mathrm{H}$ & $(3,1)$ & $(0,0)$ \
\hline $\mathrm{L}$ & $(1,2)$ & $(5,3)$ \
\hline
\end{tabular}
Table 1: Stage game
Consider the infinite repetition of the game in Table 1 with discounted criterion to evaluate payoffs. Find a subgame perfect equilibrium of this game such that
(a) the equilibrium payoff of Players approach $(4,2)$ as $\delta \rightarrow 1$.
(b) the equilibrium payoff of Players approach $(3,2)$ as $\delta \rightarrow 1$.

(a) To find a subgame perfect equilibrium (SPE) that approaches a payoff of $(4,2)$ as $\delta \rightarrow 1$, we need to find a strategy for each player that is optimal at each stage of the game, given that the game will continue indefinitely with some probability $\delta \in [0,1)$. One possible SPE is as follows:

In the first stage, Player 1 plays H and Player 2 plays L. This yields a payoff of $(3,1)$ for Player 1 and $(0,0)$ for Player 2.
In all subsequent stages, both players play the following strategy:
- If the previous outcome was (H,L), play (H,L) again.
- If the previous outcome was (L,H), play (L,H) again.
- If the previous outcome was (H,H) or (L,L), play (L,H) with probability $p$ and (H,L) with probability $1-p$, where $p$ is the smallest value that satisfies the condition $\delta \geq \frac{1-p}{1+p}$.
This strategy ensures that both players punish deviations from the outcome (L,H) by playing (L,H) in the next stage, but also allows for occasional cooperation by playing (H,L) with some probability. The value of $p$ ensures that the expected discounted payoff from deviating to (H,H) or (L,L) is less than the payoff from playing (L,H), so there is no profitable deviation.

The equilibrium payoff under this strategy is $(4,2)$ when $\delta \rightarrow 1$, because the players play (L,H) with probability 1 as $\delta$ approaches 1. Note that this is not the only SPE, and there may be other equilibria that also approach $(4,2)$ as $\delta \rightarrow 1$.

问题 2.

If we repeat prisoner’s dilemma game for two periods, how many strategies does each player have in this repeated game?

In a repeated prisoner’s dilemma game, each player has multiple strategies that they can use. One common strategy is called “tit-for-tat,” where a player cooperates in the first period and then in subsequent periods does whatever the other player did in the previous period.

If we repeat the game for two periods, each player has four possible strategies:

Cooperate in both periods
Defect in both periods
Cooperate in the first period and then defect in the second period
Defect in the first period and then cooperate in the second period

It’s important to note that the number of possible strategies increases with each additional period in a repeated game, making it more difficult to predict the outcome of the game.

Textbooks

经济代写|CS880 Game Theory

Posted on 2023年3月18日2023年3月18日 by statistics-lab

Statistics-lab™可以为您提供wisc.edu CS880 Game Theory博弈论课程的代写代考和辅导服务！

CS880 Game Theory课程简介

PREREQUISITES

Project details and timeline
Project details and ideas can be found here (UW access only).

Feb 22: Short description of topic, goals and project team due (as part of HW1).
Mar 22: Up to one page report of progress, reference material, plans for the remainder of the semester. Before this date, please make an appointment with Shuchi to discuss potential topics and references.
May 3: Final project reports due.
May 5: Two projects (selected on the basis of the final reports) to be showcased during this lecture.

CS880 Game Theory HELP（EXAM HELP， ONLINE TUTOR）

问题 1.

Recall the MLA game as described on Slide 4 of Lecture 9 with its equilibria described on slides 6 and 11. This game involves three members of a legislative assembly (three MLAs) deciding how to vote on a pay raise.

Suppose now that each of the MLAs votes in sequence, using some pre-determined order (e.g., alphabetical). Let’s call the MLAs 1, 2 and 3: MLA 1 votes first; MLA 2 votes next knowing exactly how MLA 1 voted; and finally MLA 3 votes knowing the two previous votes.
(a) Draw the extensive form game tree for this sequential voting game. How will each MLA vote assuming that a subgame perfect equilibrium is played? What is the payoff for each of the MLAs? Explain.
(b) Describe one profile of Nash equilibrium strategies for this extensive form game that is not subgame perfect.
(c) Now suppose that the MLAs have seen a recent poll, which could potentially influence the payoff of MLA 3. Specifically, the new poll indicates a level of fiscal conservatism in the district of MLA 3 that makes it uncertain how voters will react if she votes for a pay raise. As such MLA 3 ‘s payoffs are now random variables. She will realize her original payoffs with probability 0.4 ; however, with probability 0.6 she will receive the following payoffs:

MLA 3 ‘s revised payoffs:
if the raise passes, and MLA 3 votes for: -1
if the raise passes, and MLA 3 votes against: 2
if the raise fails, and MLA 3 votes for: -3
if the raise fails, and MLA 3 votes against: 0
In other words, there is a $60 \%$ chance that her payoff when voting for a raise will be reduced by 2 . All three MLAs are aware of this poll and its implications.
Notice now that the payoffs for MLA 3 are random events. What is the matrix form of the game in which all three MLAs vote simultaneously, using the expected payoffs of MLA 3 ? Draw the extensive form game tree for this new game assuming they vote sequentially (in the same order as in part (a)). How will each MLA vote assuming that a subgame perfect equilibrium is played? What is the payoff for each of the MLAs? Explain.

(a) The extensive form game tree for this sequential voting game is as follows:

Assuming a subgame perfect equilibrium is played, each MLA will vote as follows:

MLA 1 will vote for the pay raise, as this gives her the highest payoff of 3, regardless of what the other MLAs do.
MLA 2 will vote against the pay raise, as this gives her the highest payoff of 2, given that MLA 1 voted for it.
MLA 3 will vote against the pay raise, as this gives her the highest expected payoff of -0.6, given that MLA 1 voted for it and MLA 2 voted against it.

The payoffs for each MLA are:

MLA 1: 3
MLA 2: 2
MLA 3: -0.6

(b) One profile of Nash equilibrium strategies that is not subgame perfect is for all MLAs to vote against the pay raise. This is a Nash equilibrium because no MLA can improve her payoff by changing her vote, given the votes of the other two MLAs. However, this strategy is not subgame perfect, because MLA 1 could improve her payoff by deviating and voting for the pay raise.

The extensive form game tree for this new game assuming they vote sequentially (in the same order as in part (a)) is:

Assuming a subgame perfect equilibrium is played, each MLA will vote as follows:

MLA 1 will vote for the pay raise, as this gives her the highest payoff of 2.4, regardless of what the other MLAs do.
MLA 2 will vote against the pay raise, as this gives her the highest payoff of 1.6, given that MLA 1 voted for it.
MLA 3 will vote against the pay raise, as this gives her the highest expected payoff of -0.

问题 2.

Consider the following game in matrix form with two players. Payoffs for the row player Izzy are indicated first in each cell, and payoffs for the column player Jack are second.
\begin{tabular}{c|c|c|c|}
\multicolumn{1}{c}{} & \multicolumn{1}{c}{$Y$} & \multicolumn{1}{c}{$Y$} & \multicolumn{1}{c}{$Z$} \
\cline { 2 – 4 }$S$ & 5,2 & 10,6 & 25,10 \
\cline { 2 – 4 }$T$ & 10,12 & 5,6 & 0,0 \
\cline { 2 – 4 } & & &
\end{tabular}
(a) This game has two pure strategy Nash equilibria. What are they (justify your answer)? Of the two pure equilibria, which would Izzy prefer? Which would Jack prefer?
(b) Suppose Izzy plays a strictly mixed strategy, where both $S$ and $T$ are chosen with positive probability. With what probability should Izzy choose $S$ and $T$ so that each of Jack’s three pure strategies is a best response to Izzy’s mixed strategy.
(c) Suppose Jack wants to play a mixed strategy in which he selects $X$ with probability 0.7. With what probability should Jack plays actions $Y$ and $Z$ so both of Izzy’s pure strategies is a best response to Jack’s mixed strategy? Explain your answer.
(d) Based on your responses above, describe a mixed strategy equilibrium for this game in which both Jack and Izzy play each of their actions (pure strategies) with positive probability. Explain why this is in fact a Nash equilibrium (you can rely on the quantities computed in the prior parts of this question).
(e) If we swap two of Izzy’s payoffs in this matrix – in other words, if we replace one of his payoffs $r$ in the matrix with another of his payoffs $t$ from the matrix, and replace $t$ with $r$ we can make one of his strategies dominant. What swap should we make, which strategy becomes dominant, and why is it now dominant?

(a) The pure strategy Nash equilibria are $(S,Y)$ and $(T,Z)$ because neither player has an incentive to deviate unilaterally. In the first equilibrium, Izzy prefers $S$ and Jack prefers $Y$. In the second equilibrium, Izzy prefers $T$ and Jack prefers $Z$.

(b) In order for Jack to be indifferent between playing $Y$ and $Z$, the expected payoff of playing $Y$ should be equal to the expected payoff of playing $Z$. Thus, we must have $0.5 \times 5 + 0.5 \times 10 = 0.5 \times 25$, which simplifies to $7.5 = 12.5p + 0p$, where $p$ is the probability of Izzy playing $S$. Solving for $p$, we get $p = 0.4$. Therefore, Izzy should play $S$ with probability $0.4$ and $T$ with probability $0.6$.

(c) In order for Izzy to be indifferent between playing $S$ and $T$, the expected payoff of playing $S$ should be equal to the expected payoff of playing $T$. Thus, we must have $0.7 \times 5 + y = 0.7 \times 10 + z$, where $y$ is the expected payoff of Jack playing $Y$ and $z$ is the expected payoff of Jack playing $Z$. Since Jack wants to play $X$ with probability $0.7$, he wants to make Izzy indifferent between playing $S$ and $T$. Thus, we must also have $0.3 \times 5 + y = 0.3 \times 10 + z$. Solving these two equations simultaneously, we get $y = 7$ and $z = 7.5$. Therefore, Jack should play $Y$ with probability $0.5$ and $Z$ with probability $0.5$.

(d) One mixed strategy equilibrium is for Izzy to play $S$ with probability $0.4$ and $T$ with probability $0.6$, and for Jack to play $Y$ with probability $0.5$ and $Z$ with probability $0.5$. This is a Nash equilibrium because neither player has an incentive to deviate unilaterally. If Izzy deviates by changing the probabilities of $S$ and $T$, then Jack’s best response is to continue playing $Y$ and $Z$ with equal probabilities. Similarly, if Jack deviates by changing the probabilities of $Y$ and $Z$, then Izzy’s best response is to continue playing $S$ and $T$ with the same probabilities as before.

(e) If we swap the payoff of $S$ and $T$, then $T$ becomes dominant. Specifically, if we replace $5$ with $10$ and replace $10$ with $5$, then the payoff matrix becomes:

\begin{tabular}{c|c|c|c|} \multicolumn{1}{c}{} & \multicolumn{1}{c}{$Y$} & \multicolumn{1}{c}{$Y$} & \multicolumn{1}{c}{$Z$} \ \cline { 2 – 4 }$S$ & 10,12 & 5,6 & 0,0 \ \cline { 2 – 4 }$T$ & 25,10 & 10,6 & 5,2 \ \cline { 2 – 4 } & & & \end{tabular}

Now, if Izzy plays

Textbooks

经济代写|CS880 Game Theory

Posted on 2023年3月15日2023年3月15日 by statistics-lab

Statistics-lab™可以为您提供wisc.edu CS880 Game Theory博弈论课程的代写代考和辅导服务！

CS880 Game Theory课程简介

PREREQUISITES

Project details and timeline
Project details and ideas can be found here (UW access only).

Feb 22: Short description of topic, goals and project team due (as part of HW1).
Mar 22: Up to one page report of progress, reference material, plans for the remainder of the semester. Before this date, please make an appointment with Shuchi to discuss potential topics and references.
May 3: Final project reports due.
May 5: Two projects (selected on the basis of the final reports) to be showcased during this lecture.

CS880 Game Theory HELP（EXAM HELP， ONLINE TUTOR）

问题 1.

A supplier is offering five different manufacturing widgets (W1, W2, W3, W4, W5) for sale to five different companies (companies A, B, C, D, E). The companies each have the following valuations for the different widgets based on how well they serve the companies needs:However, each company only needs one widget (so having two or more widgets is only as valuable as having the most valuable of those widgets; that is, they are “partial subsititutes”). The supplier decides to run a sequence of five second-price auctions, selling each of the widgets in turn. We assume each auction proceeds as follows: – In the first auction, all five companies bid. The winning company receives the widget being auctioned, then “leaves,” that is, it participates in none of the subsequent auctions. The remaining four companies participate in the second auction, and again the winner receives the widget and “leaves.” And so on. So five companies bid in the first auction, four companines bid in the second auction, three companines in the third, two in the fourth, and one in the fifth. – Each company participating in an auction bids using its dominant strategy for the second-price auction. In other words, it behaves in any auction-using its valuation for the widget being sold-as if it believed their would be no future opportunity to obtain a different widget. So they will act based on our understanding of second-price auctions for single items. (We sometimes call this “myopic” bidding.) – The supplier has a valuation of zero for all widgets.The supplier must now decide in which order to auction the items.

(a) Suppose the supplier auctions the items in the following order: W1, W2, W3, W4, W5. For each of the five auctions (call them A1, A2, etc.), state: – who participates in the auction; – what each participant bids; – who wins the auction; – the winner’s value for the widget; – the price paid for the widget; What is the total social welfare created through this sequence of auctions? What is the total revenue received by the supplier?

(b) If you knew the companies’ valuations, in what order would you recommend that the supplier auction the widgets so as to maximize social welfare? You don’t have to justify the choice of the ordering; but given your ordering, describe the participants, bids, winner and price for each of the five auctions as in part (a). What is the total social welfare created through this sequence of auctions? What is the total revenue received by the supplier?

(a) The following table summarizes the information for each auction:

Auction	Participants	Bids	Winner	Winner’s Value	Price Paid
A1	A, B, C, D, E	A: 40, B: 50, C: 60, D: 30, E: 20	C	60	50
A2	A, B, D, E	A: 40, B: 50, D: 30, E: 20	B	50	40
A3	A, D, E	A: 40, D: 30, E: 20	A	40	30
A4	A, E	A: 40, E: 20	A	40	20
A5	A	A: 40	A	40	0

The total social welfare created through this sequence of auctions is the sum of the values of the widgets to the winning companies, which is 60+50+40+40+40=230. The total revenue received by the supplier is the sum of the prices paid for each widget, which is 50+40+30+20+0=140.

(b) To maximize social welfare, the widgets should be auctioned in order of decreasing valuations. This means that the supplier should auction the widgets in the following order: W3, W2, W1, W4, W5. The rationale for this order is that it allocates the most valuable widgets to the companies that value them the most. The following table summarizes the information for each auction under this ordering:

Auction	Participants	Bids	Winner	Winner’s Value	Price Paid
A1	A, B, C, D, E	C: 60, B: 50, A: 40, D: 30, E: 20	C	60	50
A2	B, D, A, E	B: 50, A: 40, D: 30, E: 20	B	50	40
A3	A, D, E	A: 40, D: 30, E: 20	A	40	30
A4	E, A	E: 20, A: 40	A	40	20
A5	A	A: 40	A	40	0

问题 2.

A supplier is offering five different manufacturing widgets (W1, W2, W3, W4, W5) for sale to five different companies (companies A, B, C, D, E). The companies each have the following valuations for the different widgets based on how well they serve the companies needs:However, each company only needs one widget (so having two or more widgets is only as valuable as having the most valuable of those widgets; that is, they are “partial subsititutes”). The supplier decides to run a sequence of five second-price auctions, selling each of the widgets in turn. We assume each auction proceeds as follows: – In the first auction, all five companies bid. The winning company receives the widget being auctioned, then “leaves,” that is, it participates in none of the subsequent auctions. The remaining four companies participate in the second auction, and again the winner receives the widget and “leaves.” And so on. So five companies bid in the first auction, four companines bid in the second auction, three companines in the third, two in the fourth, and one in the fifth. – Each company participating in an auction bids using its dominant strategy for the second-price auction. In other words, it behaves in any auction-using its valuation for the widget being sold-as if it believed their would be no future opportunity to obtain a different widget. So they will act based on our understanding of second-price auctions for single items. (We sometimes call this “myopic” bidding.) – The supplier has a valuation of zero for all widgets.The supplier must now decide in which order to auction the items.(c) If you knew the companies’ valuations, in what order would you recommend that the supplier auction the widgets so as to maximize revenue received? You don’t have to justify the choice of the ordering; but given your ordering, describe the participants, bids, winner and price for each of the five auctions as in part (a). What is the total social welfare created through this sequence of auctions? What is the total revenue received by the supplier? (d) We assumed above that each company, once it receives a widget, leaves the auctioning process. Now suppose, instead, that a company that has obtained one widget will participate in a future auction if the widget being sold has greater value than the one they currently own. How should a company who owns a widget bid in an auction for a widget that is more valuable than the one they currently own? (As above, assume they are “myopic” in the sense that they do not anticipate further widgets being sold later.) Returning to the original ordering in part (a), how would this behavior impact the final allocation of widgets to companies? What would the resulting social welfare be? Justify your response. (e) Suppose instead of having value for one widget, companies A and B, being rather large, can actually each derive a value from two (specific) widgets that is greater than the sum of their individual values. Specifically, – If A wins W1 and W4, its value is the sum of the individual values of these widgets plus 10 (i.e., the widgets offer some synergies, so their value in combination is 10 greater than the sum of their individual values). – If B wins W2 and W3, its value is the sum of the individual values of these widgets plus 10.

(c) To maximize revenue, the supplier should auction the widgets in the following order: W5, W4, W3, W2, W1. This order sells the widgets in decreasing order of their total valuations, as follows:

W5: Valued at $60 by company E, who will bid their true value. Company D values it at $50, so will bid $50. Company C values it at $40, so will bid $40. Company B values it at $30, so will bid $30. Company A values it at $20, so will bid $20. Company E wins with a bid of $50, and pays the second-highest bid of $40.
W4: Valued at $50 by company A, who will bid their true value. Company E values it at $40, so will bid $40. Company D values it at $30, so will bid $30. Company C values it at $20, so will bid $20. Company B values it at $10, so will bid $10. Company A wins with a bid of $40, and pays the second-highest bid of $30.
W3: Valued at $40 by company B, who will bid their true value. Company E values it at $30, so will bid $30. Company D values it at $20, so will bid $20. Company C values it at $10, so will bid $10. Company B wins with a bid of $30, and pays the second-highest bid of $20.
W2: Valued at $30 by company C, who will bid their true value. Company E values it at $20, so will bid $20. Company D values it at $10, so will bid $10. Company C wins with a bid of $30, and pays the second-highest bid of $20.
W1: Valued at $20 by company D, who will bid their true value. Company E values it at $10, so will bid $10. Company D wins with a bid of $20, and pays the second-highest bid of $10.

The total revenue received by the supplier is the sum of the second-highest bids for each auction: $40+$30+$20+$20+$10=$120.

The total social welfare created through this sequence of auctions is the sum of the values of the widgets for the winning companies: $50+$40+$30+$30+$20=$170.

(d) If a company that has obtained one widget will participate in a future auction if the widget being sold has greater value than the one they currently own, then they will bid their true value for the new widget only if it is higher than the value of the widget they already own. Otherwise, they will not participate in the auction.

In the original ordering, if a company wins a widget with lower value than the one they already own, they will not participate in future auctions. This means that the final allocation of widgets to companies will be the same as in the myopic case, and the total social welfare will also be the same: $170.

(e) In this case, the supplier should still auction the widgets in the same order as in part (c), because this order maximizes the total value of the widgets for the winning companies. The only difference is that companies A and B will bid more for the widgets that they need to achieve their synergies.

For example, in the auction for W1, company A will bid $20 because that is their true value for W1. However, if they also win W4, their combined value for the two widgets is $70 ($50+$20+10), which

Textbooks

经济代写|CS880 Game Theory

Posted on 2023年3月9日2023年3月9日 by statistics-lab

Statistics-lab™可以为您提供wisc.edu CS880 Game Theory博弈论课程的代写代考和辅导服务！

CS880 Game Theory课程简介

PREREQUISITES

Project details and timeline
Project details and ideas can be found here (UW access only).

Feb 22: Short description of topic, goals and project team due (as part of HW1).
Mar 22: Up to one page report of progress, reference material, plans for the remainder of the semester. Before this date, please make an appointment with Shuchi to discuss potential topics and references.
May 3: Final project reports due.
May 5: Two projects (selected on the basis of the final reports) to be showcased during this lecture.

CS880 Game Theory HELP（EXAM HELP， ONLINE TUTOR）

问题 1.

Consider the following game in matrix form with two players. Payoffs for the row player Izzy are indicated first in each cell, and payoffs for the column player Jack are second.

(a) This game has two pure strategy Nash equilibria. What are they (justify your answer)? Of the two pure equilibria, which would Izzy prefer? Which would Jack prefer?
(b) Suppose Izzy plays a strictly mixed strategy, where both $S$ and $T$ are chosen with positive probability. With what probability should Izzy choose $S$ and $T$ so that each of Jack’s three pure strategies is a best response to Izzy’s mixed strategy.
(c) Suppose Jack wants to play a mixed strategy in which he selects $X$ with probability 0.7. With what probability should Jack plays actions $Y$ and $Z$ so both of Izzy’s pure strategies is a best response to Jack’s mixed strategy? Explain your answer.
(d) Based on your responses above, describe a mixed strategy equilibrium for this game in which both Jack and Izzy play each of their actions (pure strategies) with positive probability. Explain why this is in fact a Nash equilibrium (you can rely on the quantities computed in the prior parts of this question).
(e) If we swap two of Izzy’s payoffs in this matrix – in other words, if we replace one of his payoffs $r$ in the matrix with another of his payoffs $t$ from the matrix, and replace $t$ with $r$ we can make one of his strategies dominant. What swap should we make, which strategy becomes dominant, and why is it now dominant?

(a) There are two pure strategy Nash equilibria in this game. The first equilibrium is for Izzy to play $S$ and for Jack to play $X$, resulting in payoffs of (3,1). Neither player can improve their payoff by unilaterally deviating from this strategy. The second equilibrium is for Izzy to play $T$ and for Jack to play $Y$, resulting in payoffs of (2,2). Again, neither player can improve their payoff by unilaterally deviating from this strategy.

Izzy would prefer the first equilibrium because she receives a higher payoff of 3 compared to 2 in the second equilibrium. Jack would prefer the second equilibrium because he also receives a payoff of 2, compared to 1 in the first equilibrium.

(b) If Izzy plays a strictly mixed strategy where both $S$ and $T$ are chosen with positive probability, then the probability of playing $S$ must be such that Jack is indifferent between playing $X$ and $Y$. Similarly, the probability of playing $T$ must be such that Jack is indifferent between playing $Y$ and $Z$.

Let $p$ be the probability of playing $S$ and $1-p$ be the probability of playing $T$. Then we have the following system of equations:

$3p + 2(1-p) = 1$
$1p + 2(1-p) = 2$ $0 \leq p \leq 1$

Solving this system of equations, we get $p = 0.25$ and $1-p = 0.75$. Therefore, Izzy should play $S$ with probability 0.25 and $T$ with probability 0.75.

(c) If Jack wants to play a mixed strategy where he selects $X$ with probability 0.7, then the probability of playing $Y$ and $Z$ must be such that Izzy is indifferent between playing $S$ and $T$.

Let $q$ be the probability of playing $Y$ and $1-q$ be the probability of playing $Z$. Then we have the following system of equations:

$3(0.7) + 1(1-0.7) = 2q + 1(1-q)$
$2(0.7) + 0(1-0.7) = 3q + 0(1-q)$ $0 \leq q \leq 1$

Solving this system of equations, we get $q = 0.4$ and $1-q = 0.6$. Therefore, Jack should play $X$ with probability 0.7, $Y$ with probability 0.4, and $Z$ with probability 0.6.

(d) A mixed strategy equilibrium for this game in which both players play each of their actions with positive probability can be constructed as follows: Izzy plays $S$ with probability 0.25 and $T$ with probability 0.75, and Jack plays $X$ with probability 0.7, $Y$ with probability 0.4, and $Z$ with probability 0.6.

To see that this is a Nash equilibrium, we need to check that neither player can improve their payoff by unilaterally deviating from their strategy. Suppose Izzy deviates by playing $S$ with higher probability than 0.25. Then Jack would prefer to play $Y$ instead of $X$, since he would receive a payoff of 3 instead of 1. Similarly, if Izzy deviates by playing $T$ with higher probability than 0.75, then

问题 2.

Suppose a seller runs a second-price, sealed-bid auction for a painting. There are two bidders with independent, private values. The seller does not know their precise valuations, but knows: (a) each bidder $i$ has one of three values, $v_i=2, v_i=4$ or $v_i=8$; and (b) each of these values is equally likely (i.e., occurs with probability $\frac{1}{3}$ ). When running the auction, if the two bids are tied (say, at $x$ ), the winner is chosen at random (and pays $x$ ). The seller has no value for the painting (i.e., her valuation is 0 ).
(a) Assume both bidders use their dominant strategies for bidding in a second-price auction. What is the seller’s expected revenue in this auction? Please explain your answer.
(b) Now the seller decides to set a reserve price of $r$-as discussed in class, this means that if the highest bid is at least $r$, then the painting will go to the highest bidder, and the winner will pay the maximum of $r$ and the second-highest bid.
Suppose the reserve price is set to $r=4$. Assume both bidders use their dominant strategies. What is the seller’s expected revenue in this auction? Please explain your answer. If the expected revenue increases or decreases relative to your answer in part (a), give a qualitative explanation for why this change occurs.
(c) Is there a better reserve price than $r=4$ (i.e., that will provide more revenue for the seller)? Give a brief justification for your response.

(a) Since both bidders are using their dominant strategy, they will bid their true valuations. Let $v_1$ and $v_2$ denote the values of bidder 1 and bidder 2, respectively. There are nine possible cases, and in each case, the highest bid wins the auction and pays the second-highest bid. Therefore, the seller’s expected revenue is the sum of the probabilities of each case times the second-highest bid in that case: \begin{align*} \text{Expected revenue} &= \frac{1}{9}\cdot 2 + \frac{4}{9}\cdot 2 + \frac{4}{9}\cdot 4 + \frac{1}{9}\cdot 4 + \frac{1}{9}\cdot 2 + \frac{4}{9}\cdot 2 + \frac{4}{9}\cdot 4 + \frac{1}{9}\cdot 4 + \frac{1}{9}\cdot 2\ &= \frac{20}{9} \end{align*} Therefore, the expected revenue for the seller in this auction is $\frac{20}{9}$.

(b) If the reserve price is set at $r=4$, then bidder 3 (i.e., the hypothetical bidder who values the painting at $v_3=8$) will not participate, since she would not be willing to pay more than $r$. Therefore, only bidder 1 and bidder 2 will participate, and the winner will pay at most $4$. Let $p_1$ and $p_2$ denote the probabilities that bidder 1 and bidder 2, respectively, win the auction. The seller’s expected revenue is then: \begin{align*} \text{Expected revenue} &= p_1\cdot 2 + p_2\cdot 2 + (1-p_1-p_2)\cdot 4\ &= 2(p_1+p_2) + 4(1-p_1-p_2) \end{align*} To find the values of $p_1$ and $p_2$, we can solve the following system of equations: \begin{align*} p_1 &= \frac{1}{3}\cdot \left(\mathbb{P}(v_2\leq x<v_1)+\mathbb{P}(x=v_1)\right)\ p_2 &= \frac{1}{3}\cdot \left(\mathbb{P}(v_1\leq x<v_2)+\mathbb{P}(x=v_2)\right) \end{align*} where $\mathbb{P}(v_2\leq x<v_1)$ is the probability that the highest bid is between $v_2$ and $v_1$, and $\mathbb{P}(x=v_1)$ and $\mathbb{P}(x=v_2)$ are the probabilities that the highest bid is equal to $v_1$ and $v_2$, respectively. Note that if $v_1=v_2$, then $\mathbb{P}(x=v_1)=\mathbb{P}(x=v_2)=\frac{1}{6}$, and $\mathbb{P}(v_2\leq x<v_1)=0$. If $v_1>v_2$, then $\mathbb{P}(v_2\leq x<v_1)=\frac{1}{3}$, and $\mathbb{P}(x=v_1)=\frac{1}{6}$ and $\mathbb{P