Game Theory and Strategy · Quiz 1 · Sep 2024

Rows are player 1 bids; columns are player 2 bids. Each pair is (player 1 payoff, player 2 payoff).\[\begin{array}{c|ccc} &0&1&2\\ 0&(3/2,5/2)&(0,4)&(0,3)\\ 1&(2,0)&(1,2)&(0,3)\\ 2&(1,0)&(1,0)&(1/2,3/2)\end{array}\] **1. Calculate the payoffs:** A higher bidder earns value minus bid; the loser earns zero. A tie at b gives expected payoff \((v_i-b)/2\), producing the displayed matrix. **2. Check pure-strategy strict dominance:** For player 2, bid 2 gives payoffs (3,3,1.5), strictly above bid 0’s (2.5,0,0). For player 1, no single alternative bid is strictly better than another at every column: bid 1 only weakly dominates 0, while bids 1 and 2 trade off gains. **3. State the convention-sensitive answer:** Under elimination by another pure strategy, player 1 has no strictly dominated strategy in the original matrix. **Answer: A — Player 1.** **Convention note:** The answer uses pure-strategy dominance, as the supplied key intends. If domination by mixed strategies is allowed, player 1’s bid 0 is strictly dominated by a mixture of 0.6 on bid 1 and 0.4 on bid 2: its payoffs (1.6,1,0.2) exceed (1.5,0,0). Under that broader definition, neither player lacks a strictly dominated strategy and option C would apply. The question does not state this restriction explicitly. A: **Correct:** Under pure-strategy comparisons, no row is strictly worse than another row in every column. B: **Incorrect:** Player 2’s bid 0 is strictly dominated by bid 2. C: **Incorrect:** Under the intended pure-strategy convention, player 1 does have the requested property.

Rows are player 1 bids; columns are player 2 bids. Each pair is (player 1 payoff, player 2 payoff).\[\begin{array}{c|ccc} &0&1&2\\ 0&(3/2,5/2)&(0,4)&(0,3)\\ 1&(2,0)&(1,2)&(0,3)\\ 2&(1,0)&(1,0)&(1/2,3/2)\end{array}\] **1. Remove player 2’s bid 0:** Against player 1 bids 0,1,2, player 2’s bid 2 earns (3,3,1.5), strictly exceeding bid 0’s (2.5,0,0). **2. Remove player 1’s bid 0:** In remaining columns 1 and 2, player 1’s bid 0 gives (0,0), while bid 2 gives (1,0.5), both strictly greater. **3. Continue in the reduced game:** Against remaining rows 1 and 2, player 2’s bid 2 gives (3,1.5), strictly exceeding bid 1’s (2,0). With only column 2 left, player 1 prefers bid 2’s 0.5 to bid 1’s 0. **4. Identify survivors:** Each player is left with bid 2. The surviving profile is (2,2), with expected payoffs (0.5,1.5). **Answer: C — (2, 2).** A: **Incorrect:** Both bid-0 strategies are eliminated during the procedure. B: **Incorrect:** Bid 1 is eliminated for both players as the game is reduced. C: **Correct:** Iterated strict elimination leaves only bid 2 for each player. D: **Incorrect:** There is a surviving strategy for each player.

Rows are police choices; columns are dealer choices, with S=street and P=park. Each pair is (police payoff, dealer payoff).\[\begin{array}{c|cc}&S&P\\ S&(80,20)&(0,100)\\ P&(10,90)&(60,40)\end{array}\] **1. Find the officer’s best responses:** If the dealer chooses street, the officer prefers street (80>10). If the dealer chooses park, the officer prefers park (60>0). **2. Find the dealer’s best responses:** Against street patrol, the dealer prefers park (100>20). Against park patrol, the dealer prefers street (90>40). **3. Check mutual best responses:** In every cell at least one player benefits by switching. Therefore no pure-strategy Nash equilibrium exists. **Answer: C — There is no Nash equilibrium in this game in pure strategy.** **Convention note:** Literally zero is a finite number, so B is imprecise rather than logically disjoint from C. C states the exact conclusion and is the intended single best answer; B should say “a positive finite number” to avoid overlap. A: **Incorrect:** No cell is a mutual best response, so there is no unique pure equilibrium. B: **Incorrect:** The intended contrast is a positive finite number of pure equilibria; here the number is zero. C: **Correct:** Every pure profile permits a profitable unilateral deviation. D: **Incorrect:** The explicit no-pure-equilibrium alternative applies.

Rows are police choices; columns are dealer choices, with S=street and P=park. Each pair is (police payoff, dealer payoff).\[\begin{array}{c|cc}&S&P\\ S&(80,20)&(0,100)\\ P&(10,90)&(60,40)\end{array}\] **1. Make the dealer indifferent:** Let alpha be the officer’s probability of street patrol. The dealer’s street payoff is \(20\alpha+90(1-\alpha)=90-70\alpha\); its park payoff is \(100\alpha+40(1-\alpha)=40+60\alpha\). Equality gives \(\alpha=5/13\), hence park patrol probability 8/13. **2. Make the officer indifferent:** Let beta be the dealer’s probability of street. The officer’s street payoff is \(80\beta\); park payoff is \(10\beta+60(1-\beta)=60-50\beta\). Equality gives \(\beta=6/13\), hence dealer park probability 7/13. **3. Establish uniqueness:** There is no pure equilibrium, all best responses to pure actions are strict, and each indifference equation has a unique interior solution. Thus the unique equilibrium has officer mix (5/13,8/13) and dealer mix (6/13,7/13). **Answer: A — There is a unique Nash equilibrium in this game.** **Convention note:** The options overlap literally: a unique equilibrium is also finitely many. A is the more precise single best answer intended by the source. This equilibrium is in mixed, not pure, strategies. A: **Correct:** The two indifference equations identify exactly one mixed-strategy equilibrium. B: **Incorrect:** One is finite, so this is literally true but weaker than the explicit uniqueness statement; it is not the intended alternative. C: **Incorrect:** A mixed equilibrium exists even though no pure equilibrium does. D: **Incorrect:** The uniqueness alternative gives the exact result.

Rows are police choices; columns are dealer choices, with S=street and P=park. Each pair is (police payoff, dealer payoff).\[\begin{array}{c|cc}&S&P\\ S&(80,20)&(0,100)\\ P&(10,90)&(60,40)\end{array}\] **1. Make the dealer indifferent:** Let alpha be the officer’s probability of street patrol. The dealer’s street payoff is \(20\alpha+90(1-\alpha)=90-70\alpha\); its park payoff is \(100\alpha+40(1-\alpha)=40+60\alpha\). Equality gives \(\alpha=5/13\), hence park patrol probability 8/13. **2. Make the officer indifferent:** Let beta be the dealer’s probability of street. The officer’s street payoff is \(80\beta\); park payoff is \(10\beta+60(1-\beta)=60-50\beta\). Equality gives \(\beta=6/13\), hence dealer park probability 7/13. **3. Establish uniqueness:** There is no pure equilibrium, all best responses to pure actions are strict, and each indifference equation has a unique interior solution. Thus the unique equilibrium has officer mix (5/13,8/13) and dealer mix (6/13,7/13). **4. Read the requested numerator:** The officer chooses street with probability alpha=5/13, so the requested numerator p is 5. **Answer: 5.**

Rows are police choices; columns are dealer choices, with S=street and P=park. Each pair is (police payoff, dealer payoff).\[\begin{array}{c|cc}&S&P\\ S&(80,20)&(0,100)\\ P&(10,90)&(60,40)\end{array}\] **1. Make the dealer indifferent:** Let alpha be the officer’s probability of street patrol. The dealer’s street payoff is \(20\alpha+90(1-\alpha)=90-70\alpha\); its park payoff is \(100\alpha+40(1-\alpha)=40+60\alpha\). Equality gives \(\alpha=5/13\), hence park patrol probability 8/13. **2. Make the officer indifferent:** Let beta be the dealer’s probability of street. The officer’s street payoff is \(80\beta\); park payoff is \(10\beta+60(1-\beta)=60-50\beta\). Equality gives \(\beta=6/13\), hence dealer park probability 7/13. **3. Establish uniqueness:** There is no pure equilibrium, all best responses to pure actions are strict, and each indifference equation has a unique interior solution. Thus the unique equilibrium has officer mix (5/13,8/13) and dealer mix (6/13,7/13). **4. Read the requested numerator:** The officer chooses park with probability 1-alpha=8/13, so the requested numerator q is 8. **Answer: 8.**

Rows are police choices; columns are dealer choices, with S=street and P=park. Each pair is (police payoff, dealer payoff).\[\begin{array}{c|cc}&S&P\\ S&(80,20)&(0,100)\\ P&(10,90)&(60,40)\end{array}\] **1. Make the dealer indifferent:** Let alpha be the officer’s probability of street patrol. The dealer’s street payoff is \(20\alpha+90(1-\alpha)=90-70\alpha\); its park payoff is \(100\alpha+40(1-\alpha)=40+60\alpha\). Equality gives \(\alpha=5/13\), hence park patrol probability 8/13. **2. Make the officer indifferent:** Let beta be the dealer’s probability of street. The officer’s street payoff is \(80\beta\); park payoff is \(10\beta+60(1-\beta)=60-50\beta\). Equality gives \(\beta=6/13\), hence dealer park probability 7/13. **3. Establish uniqueness:** There is no pure equilibrium, all best responses to pure actions are strict, and each indifference equation has a unique interior solution. Thus the unique equilibrium has officer mix (5/13,8/13) and dealer mix (6/13,7/13). **4. Read the requested numerator:** The dealer chooses street with probability beta=6/13, so the requested numerator r is 6. **Answer: 6.**

Rows are police choices; columns are dealer choices, with S=street and P=park. Each pair is (police payoff, dealer payoff).\[\begin{array}{c|cc}&S&P\\ S&(80,20)&(0,100)\\ P&(10,90)&(60,40)\end{array}\] **1. Make the dealer indifferent:** Let alpha be the officer’s probability of street patrol. The dealer’s street payoff is \(20\alpha+90(1-\alpha)=90-70\alpha\); its park payoff is \(100\alpha+40(1-\alpha)=40+60\alpha\). Equality gives \(\alpha=5/13\), hence park patrol probability 8/13. **2. Make the officer indifferent:** Let beta be the dealer’s probability of street. The officer’s street payoff is \(80\beta\); park payoff is \(10\beta+60(1-\beta)=60-50\beta\). Equality gives \(\beta=6/13\), hence dealer park probability 7/13. **3. Establish uniqueness:** There is no pure equilibrium, all best responses to pure actions are strict, and each indifference equation has a unique interior solution. Thus the unique equilibrium has officer mix (5/13,8/13) and dealer mix (6/13,7/13). **4. Read the requested numerator:** The dealer chooses park with probability 1-beta=7/13, so the requested numerator s is 7. **Answer: 7.**

Payoffs are (player 1, player 2). Each downward arrow continues play; each rightward arrow ends it. All three decision nodes are singleton information sets.\[\begin{array}{rcl}P_1&\xrightarrow{A}&(2,0)\\ {\scriptstyle B}\downarrow&&\\ P_2&\xrightarrow{C}&(3,1)\\ {\scriptstyle D}\downarrow&&\\ P_1&\xrightarrow{E}&(0,0)\\ {\scriptstyle F}\downarrow&&\\ (1,2)&&\end{array}\] **1. Identify what each player knows:** Player 2 moves only after B, and player 1’s final decision occurs only after the history BD. Each moving player can identify the exact decision node. **2. Classify information:** Each information set contains a single decision node. The game therefore has perfect information. **Answer: A — Perfect information.** A: **Correct:** Every decision maker knows the full history when acting. B: **Incorrect:** No indistinguishable decision nodes or hidden past moves are specified. C: **Incorrect:** The game fits the perfect-information category. D: **Incorrect:** The sequence and observations determine the classification.

Payoffs are (player 1, player 2). Each downward arrow continues play; each rightward arrow ends it. All three decision nodes are singleton information sets.\[\begin{array}{rcl}P_1&\xrightarrow{A}&(2,0)\\ {\scriptstyle B}\downarrow&&\\ P_2&\xrightarrow{C}&(3,1)\\ {\scriptstyle D}\downarrow&&\\ P_1&\xrightarrow{E}&(0,0)\\ {\scriptstyle F}\downarrow&&\\ (1,2)&&\end{array}\] **1. List the complete terminal histories:** The game can end after A, after BC, after BDE, or after BDF. **2. Count leaves:** These four histories end at four separate terminal nodes, with payoffs (2,0),(3,1),(0,0),(1,2). **Answer: C — 4.** A: **Incorrect:** This counts the three decision nodes, not the terminal leaves. B: **Incorrect:** The final E/F choices are not the only ways the game can end. C: **Correct:** The four terminal histories are A, BC, BDE and BDF. D: **Incorrect:** There is no fifth terminal history in the tree.

Payoffs are (player 1, player 2). Each downward arrow continues play; each rightward arrow ends it. All three decision nodes are singleton information sets.\[\begin{array}{rcl}P_1&\xrightarrow{A}&(2,0)\\ {\scriptstyle B}\downarrow&&\\ P_2&\xrightarrow{C}&(3,1)\\ {\scriptstyle D}\downarrow&&\\ P_1&\xrightarrow{E}&(0,0)\\ {\scriptstyle F}\downarrow&&\\ (1,2)&&\end{array}\] **1. Locate decision nodes:** There is an initial player 1 node, a player 2 node after B, and a later player 1 node after BD. **2. Count information sets:** Under perfect information each decision node is its own singleton information set, giving 3. Terminal nodes are not decision information sets. **Answer: B — 3.** A: **Incorrect:** This counts players rather than information sets; player 1 has two distinct decision situations. B: **Correct:** Each of the three decision nodes forms a separate information set. C: **Incorrect:** Four is the number of terminal nodes, not information sets. D: **Incorrect:** All decision nodes cannot share one information set across different players and histories.

Payoffs are (player 1, player 2). Each downward arrow continues play; each rightward arrow ends it. All three decision nodes are singleton information sets.\[\begin{array}{rcl}P_1&\xrightarrow{A}&(2,0)\\ {\scriptstyle B}\downarrow&&\\ P_2&\xrightarrow{C}&(3,1)\\ {\scriptstyle D}\downarrow&&\\ P_1&\xrightarrow{E}&(0,0)\\ {\scriptstyle F}\downarrow&&\\ (1,2)&&\end{array}\] **1. Use a complete contingent plan:** A pure strategy specifies what player 1 would do initially and what they would do later after BD, even if the later node is never reached. **2. Multiply independent choices:** There are 2 initial actions and 2 later actions, giving \(2\times2=4\) strategies: AE, AF, BE and BF. **Answer: D — 4.** A: **Incorrect:** Player 1 has choices at two separate decision nodes. B: **Incorrect:** This counts actions at a single node rather than complete plans. C: **Incorrect:** AE and AF remain distinct full strategies even though both stop play immediately. D: **Correct:** The four complete plans are AE, AF, BE and BF.

Payoffs are (player 1, player 2). Each downward arrow continues play; each rightward arrow ends it. All three decision nodes are singleton information sets.\[\begin{array}{rcl}P_1&\xrightarrow{A}&(2,0)\\ {\scriptstyle B}\downarrow&&\\ P_2&\xrightarrow{C}&(3,1)\\ {\scriptstyle D}\downarrow&&\\ P_1&\xrightarrow{E}&(0,0)\\ {\scriptstyle F}\downarrow&&\\ (1,2)&&\end{array}\] **1. Locate player 2’s choice:** Player 2 acts at only one information set, after player 1 chooses B. **2. List complete plans:** Choosing C or choosing D are the two possible pure strategies. No additional contingency for player 2 is needed. **Answer: B — 2.** A: **Incorrect:** Player 2 has two available actions, so there is more than one strategy. B: **Correct:** The complete strategy set is {C,D}. C: **Incorrect:** Player 2 has no third action or extra information set. D: **Incorrect:** Four is player 1’s strategy count, not player 2’s.

Payoffs are (player 1, player 2). Each downward arrow continues play; each rightward arrow ends it. All three decision nodes are singleton information sets.\[\begin{array}{rcl}P_1&\xrightarrow{A}&(2,0)\\ {\scriptstyle B}\downarrow&&\\ P_2&\xrightarrow{C}&(3,1)\\ {\scriptstyle D}\downarrow&&\\ P_1&\xrightarrow{E}&(0,0)\\ {\scriptstyle F}\downarrow&&\\ (1,2)&&\end{array}\] Rows are complete player 1 plans (initial action, later action); columns are player 2 plans. Each pair is (player 1 payoff, player 2 payoff).\[\begin{array}{c|cc}&C&D\\ AE&(2,0)&(2,0)\\ AF&(2,0)&(2,0)\\ BE&(3,1)&(0,0)\\ BF&(3,1)&(1,2)\end{array}\] **1. Find player 1’s best responses:** Against C, BE and BF earn 3, exceeding AE/AF’s 2. Against D, AE and AF earn 2, exceeding BE’s 0 and BF’s 1. **2. Find player 2’s best responses:** Against AE or AF, C and D both yield 0. Against BE, C is better (1>0). Against BF, D is better (2>1). **3. Intersect the best responses:** The pure equilibria are (AE,D), (AF,D), and (BE,C), giving exactly three. **Answer: C — 3.** A: **Incorrect:** This counts only one equilibrium and misses two other mutual best responses. B: **Incorrect:** There are two equilibria with initial A plus one with initial B. C: **Correct:** Exactly (AE,D), (AF,D) and (BE,C) are mutual best responses. D: **Incorrect:** (BF,C) is not an equilibrium because player 2 gains by switching to D.

Payoffs are (player 1, player 2). Each downward arrow continues play; each rightward arrow ends it. All three decision nodes are singleton information sets.\[\begin{array}{rcl}P_1&\xrightarrow{A}&(2,0)\\ {\scriptstyle B}\downarrow&&\\ P_2&\xrightarrow{C}&(3,1)\\ {\scriptstyle D}\downarrow&&\\ P_1&\xrightarrow{E}&(0,0)\\ {\scriptstyle F}\downarrow&&\\ (1,2)&&\end{array}\] **1. Solve the last decision:** After BD, player 1 chooses F because 1>0. Any subgame-perfect strategy must prescribe F there, including off the equilibrium path. **2. Solve player 2’s decision:** Anticipating F, choosing D yields 2 while C yields 1, so player 2 chooses D. **3. Solve the initial decision:** Player 1 compares A’s payoff 2 with the continuation BF payoff 1 and chooses A. The full subgame-perfect strategy profile is (AF,D); the realized outcome is A with payoffs (2,0). **Answer: B — (AF, D).** A: **Incorrect:** Prescribing E after BD is not optimal in that subgame, since F gives player 1 a larger payoff. B: **Correct:** Every prescribed decision is optimal in its continuation subgame. C: **Incorrect:** After D, E is not credible; backward induction selects F and then D instead of C. D: **Incorrect:** Given the continuation F, player 2 prefers D to C, so this profile is not even Nash.

Payoffs are (player 1, player 2). Each downward arrow continues play; each rightward arrow ends it. All three decision nodes are singleton information sets.\[\begin{array}{rcl}P_1&\xrightarrow{A}&(2,0)\\ {\scriptstyle B}\downarrow&&\\ P_2&\xrightarrow{C}&(3,1)\\ {\scriptstyle D}\downarrow&&\\ P_1&\xrightarrow{E}&(0,0)\\ {\scriptstyle F}\downarrow&&\\ (1,2)&&\end{array}\] **1. Apply the subgame definition:** In this perfect-information tree, each decision node can start a subgame consisting of that node and all its successors. **2. List all roots:** They are the initial player 1 node, the player 2 node after B, and the player 1 node after BD. Counting the whole game gives 3 subgames; excluding it would leave 2 proper subgames. **Answer: C — 3.** **Convention note:** The standard count here includes the entire game, as the supplied answer does. The question does not ask specifically for proper subgames. A: **Incorrect:** This counts only the entire game and misses two continuation subgames. B: **Incorrect:** This is the number of proper subgames, excluding the whole game. C: **Correct:** There are three subgames when the whole game is included. D: **Incorrect:** Terminal nodes alone are not counted as strategic subgames in this convention.

There are four players, one per driver, each minimizing their own travel time (equivalently maximizing its negative). Initially let X denote A-X-B and Y denote A-Y-B. If n drivers choose X, the route costs for routes in use are:\[\begin{array}{c|cc}n&T_X(n)&T_Y(4-n)\\ 0&\text{unused}&38\\ 1&26&34\\ 2&29.9&30\\ 3&33.8&26\\ 4&37.7&\text{unused}\end{array}\] **1. Form the strategic game:** Each driver chooses X or Y simultaneously. For k>=1, \(T_X(k)=6+3(k-1)+20+0.9(k-1)=26+3.9(k-1)\), while \(T_Y(k)=26+4(k-1)\). **2. Check the balanced profile:** With two drivers on each route, X takes 29.9 minutes and Y takes 30. An X-driver switching to Y would face three drivers there and take 34; a Y-driver switching to X would take 33.8. Neither gains. **3. Rule out other load splits:** If X has three drivers, one can switch and reduce 33.8 to 30; if Y has three, one can switch and reduce 34 to 29.9. A four-to-zero split likewise allows a switch to the empty route taking 26. Hence every pure equilibrium has a 2-2 split. **4. Choose the complete alternative:** Both the two-per-route statement and the 29.9-or-30 travel-time statement hold. **Answer: C — In Nash equilibrium two people take each route and For the NE action profile, .** **Convention note:** As intended by the alternatives, “Nash equilibrium” here refers to pure route-choice profiles. There are six labelled pure equilibria, one for each choice of which two drivers use X. Mixed-strategy equilibria also exist and need not realize a 2-2 split on every trip. A: **Incorrect:** This is true, but option C includes this and the additional true time statement. B: **Incorrect:** This is true at the equilibrium split, but option C states both correct conclusions. C: **Correct:** This combines the correct equilibrium load split with its correct travel times. D: **Incorrect:** The two statements in C are both valid.

Four drivers minimize travel time. Label routes X=A-X-B, Y=A-Y-B, Z=A-X-Y-B and W=A-Y-X-B. An edge used by n>=1 drivers has cost:\[\begin{array}{c|c}\text{edge}&\text{minutes}\\ AX,\ YB&6+3(n-1)\\ AY&20+(n-1)\\ XB&20+0.9(n-1)\\ XY&7+(n-1)\end{array}\]The XY road is shared by both directions, as in the stated common-road congestion rule. **1. Remove the long reverse route:** A driver switching from W to X keeps using XB, drops AY and XY (a combined cost at least 27), and adds AX (at most 15 after the switch). Thus W is always strictly worse than X and cannot be used in a pure equilibrium. **2. Calculate route costs from the load counts:** Let a,b,c be the numbers choosing X,Y,Z, with a+b+c=4. For a route that is used, the costs are \(T_X=22.1+3.9a+3c\), \(T_Y=22+4b+3c\), and \(T_Z=24+4c\). These include shared-edge congestion. **3. Find the only possible pure-equilibrium split:** If c=0, the original 2-2 split is the only candidate among X and Y, but a driver can switch to Z for 28 minutes. If c>0, Z-users’ no-deviation inequalities require \(1+c\le3.9a\) and \(1+c\le4b\), so a,b>=1 and c<=2. With c=1, the doubled X or Y route has a driver who gains by moving to Z. Thus c=2 and a=b=1. **4. Check all deviations at (a,b,c)=(1,1,2):** Every used route takes 32 minutes. An X-user’s alternatives Y,Z,W take 36,36,50; a Y-user’s alternatives X,Z,W take 35.9,36,49.9; a Z-user’s alternatives X,Y,W take 32.9,33,49.9. None improves. **5. Select the statement:** One driver uses X, one uses Y, two use Z, and none uses W. The common time is 32, so the statement claiming 36 minutes is false. **Answer: A — In any Nash equilibrium, one person takes A–X–B, two people take A–X–Y–B, .** **Convention note:** The route-count claim is about pure-strategy equilibrium, as the source intends. Mixed strategies permit randomized realized counts. A unilateral deviation must recompute the congestion loads; holding the old edge times fixed would give an incorrect equilibrium test. A: **Correct:** This is the unique pure-equilibrium route-load pattern, up to relabelling drivers. B: **Incorrect:** The equilibrium costs are 12+20=32 on X, 20+12=32 on Y, and 12+8+12=32 on Z. C: **Incorrect:** Its load pattern is correct, but the added claim of 36 minutes is false. D: **Incorrect:** Option A correctly gives the pure-equilibrium route counts.

**1. Optimize player i’s own effort:** Holding ej fixed, \(u_i=(a+e_j)e_i-e_i^2\). The derivative with respect to ei is \(a+e_j-2e_i\), and the second derivative is -2. **2. Enforce the nonnegative constraint:** The strictly concave objective has unique constrained maximizer \(BR_i(e_j)=\max\{0,(a+e_j)/2\}\). **3. Match the supplied alternatives:** Under the intended assumption a>=0 and the stated ej>=0, the interior formula is already nonnegative. Hence \(BR_i(e_j)=(a+e_j)/2\), option C. **Answer: C — Best response (a+ej)/2, under a nonnegative.** **Convention note:** The prompt does not state the sign of a. Option C and the source key assume a>=0. For an unrestricted real a, the complete best response is max{0,(a+ej)/2}; when a+ej<0 none of the displayed formulas gives the correct boundary solution. The wording “for all ei>=0” should refer to opponent effort ej>=0. A: **Incorrect:** The first-order condition contains no square of a+ej. B: **Incorrect:** It both squares the expression and uses the player’s own effort as the best-response argument. C: **Correct:** For a>=0, this is the unique nonnegative maximizer for every opponent effort ej>=0. D: **Incorrect:** A best response is a function of the opponent effort ej, not the already chosen own effort ei.

**1. Solve mutual best responses:** Under a>=0 the equations are \(2e_1=a+e_2\) and \(2e_2=a+e_1\). Subtracting gives \(3(e_1-e_2)=0\), so the efforts are equal. **2. Determine the common effort:** Substitution gives \(2e=a+e\), hence \(e_1=e_2=a\). The solution satisfies nonnegativity. **3. Establish uniqueness:** Each player’s objective is strictly concave in own effort, and these two best-response equations have exactly one solution. At a=0 the only solution is also (0,0). Both the equilibrium formula and uniqueness statement are therefore true. **Answer: C — Unique equilibrium with e1=e2=a, assuming a is nonnegative.** **Convention note:** The source omits the sign restriction on a. With the full constrained best response, the unique equilibrium for any real a is e1=e2=max{a,0}. Thus if a<0, the printed e1=e2=a claim is infeasible; only the uniqueness statement in B remains true. C follows the nonnegative-a convention used by the source. A: **Incorrect:** This is true under the intended a>=0 assumption, but C includes the uniqueness conclusion as well. B: **Incorrect:** Uniqueness is true, but C also gives the equilibrium effort levels. C: **Correct:** Both players choose a, and this is the unique equilibrium for a>=0. D: **Incorrect:** Under the intended parameter assumption, C gives both correct conclusions.