109 年 國立臺灣大學經濟學系碩士班《個體經濟學》

📄 試題原卷 免費註冊後即可對照原始考卷 PDF免費註冊

第 1 題

Multiple-answer questions. Choose correct answer(s). The number of correct answers for each question could be from 0 to 5. For each wrong answer in a question, you are deducted 4 points (from 10 points) until you get 0 point in that question. For instance, suppose the correct answers are abc, but your answers are acd, then you get a right, miss b, get c right, wrongly include d. Thus for this question there are two wrong answers (miss b and wrongly include d), hence you get 10-4-4=2 points. If you leave this question blank, you wrongly miss abc and correctly miss de, hence you get 0 point because 10-4-4-4-2<0.

  1. Rush hour is a board game for a kid. There are vehicles parking on a 6*6 surface squares. Among them is a red vehicle. There is an exit hole and the kid's goal is to move the red vehicle through the exit hole of the board by moving the other vehicles out of its way. The game ends when the kid achieves the goal. Each vehicle is one square wide and is either two squares long or three squares long. Each square is labeled by an alphabet and a number. For instance, the northwestern corner is labeled as A1. The exit is next to C6.

Consider the following example shown in the picture. In this specific example there are five vehicles on the board. The red vehicle is on squares C1C2. Other vehicles are blue (on A4B4C4), orange (on C5D5), yellow (on E4E5E6) and white (on F4F5). Vehicles can only move forward or backward on the surface squares. They cannot move sideways. Hence, for instance, the blue vehicle, as it is facing down, can only move on column 4. Similarly, the orange vehicle can only move on column 5, the white vehicle can only move on row F, the yellow vehicle can only move on row E and the red vehicle can only move on row C. No lifting the vehicles off the surface squares is allowed. Two vehicles cannot occupy the same space. Apply what you have learned from game theory to this, you will conclude

(a) This game can last forever.
(b) The idea of backward induction suggests that during the process that the red vehicle passes through the exit, the yellow vehicle must have slid towards the left at least once.
(c) In equilibrium, the red vehicle does not pass through the exit.
(d) The idea of backward induction suggests that the orange vehicle must move to A5B5.
(e) In equilibrium, there could be a vehicle which never moves.

  1. Continue from above.
    (a) In a general game, increasing the number of vehicles makes the red vehicle pass through the exit less likely in equilibrium.
    (b) In a general game, the red vehicle passes through the exit only when the number of vehicles is odd.
    (c) To make the exemplar game shown in the picture more fun, it is now played with two kids R and B taking turns. R wins if the red vehicle passes through the exit, and B wins if he blocks so the red vehicle does not pass through the exit. The equilibrium outcome depends on whether R starts first or not.
    (d) To make the exemplar game shown in the picture more fun, it is played with two kids R and B taking turns. R wins if the red vehicle passes through the exit, and B wins if he blocks so the red vehicle does not pass through the exit. In equilibrium, R wins when B starts first. Hence there is a second mover advantage.
    (e) To make the exemplar game shown in the picture even more fun, it is played with three kids R1, R2 and B taking turns. R1 and R2 win if the red vehicle passes through the exit, and B wins if he blocks so the red vehicle does not pass through the exit. The equilibrium outcome, specifically on whether the red vehicle passes through the exit, is different from the equilibrium outcome in the question above where only a kid plays the exemplar game.
🖼️ 本題含圖表,以下為原卷對應頁面:
原卷第 2 頁原卷第 3 頁

登入後即可作答並保存紀錄。

這一題的完整詳解

這兩題是關於一個稱為 Rush Hour 的遊戲,結合了賽局理論與博弈論的觀念。遊戲目標是將紅色汽車移出棋盤。

題目 1

此題旨在考察對博弈論中一些基本概念的理解,特別是與遊戲的動態、策略以及賽局的可能結果相關的。

  • 核心概念: 賽局的性質(是否會無限進行)、後向歸納法(backward induction)、賽局均衡(equilibrium)。

  • 解題思路:

    • (a) This game can last forever. 雖然遊戲有目標(紅色汽車移出),但理論上,如果玩家不斷地進行無效移動,或者存在一些循環的策略,遊戲確實有可能無限期地進行下去。這在賽局理論中是可能的,尤其是在沒有明確終止條件或有無限多種移動順序時。
    • (b) The idea of backward induction suggests that during the process that the red vehicle passes through the exit, the yellow vehicle must have slid towards the left at least once. 後向歸納法是從終局往前推導。要讓紅色汽車(在 C1C2)通過出口(C6 附近),需要移動其他車輛。黃色汽車在 E4E5E6。為了讓紅色汽車移動,可能需要騰出 C3 或 C4 的空間。黃色汽車移動到 E4E5E6 的左邊(例如 E3E4E5)是為了給其他汽車(如橙色汽車 C5D5)讓路,以便橙色汽車移動,進而可能讓紅色汽車通過。因此,這個敘述有可能是正確的。
    • (c) In equilibrium, the red vehicle does not pass through the exit. 賽局的目標是讓紅色汽車通過出口。如果存在一個策略組合,能夠讓紅色汽車成功通過,那麼「不通過出口」就不會是均衡結果(除非存在某種阻止通過的機制,但題目描述的目標是通過)。然而,在某些情況下,由於棋盤的佈局和車輛的限制,可能確實無法通過,但這需要具體分析。
    • (d) The idea of backward induction suggests that the orange vehicle must move to A5B5. 橙色汽車在 C5D5。題目描述中,橙色汽車只能在 column 5 移動。A5B5 顯然不在 column 5。因此,這個敘述是錯誤的。
    • (e) In equilibrium, there could be a vehicle which never moves. 在賽局中,如果某個車輛的位置不影響其他關鍵車輛(特別是紅色汽車)的移動路徑,或者為了達到某個狀態而需要保持靜止,那麼它可能在整個遊戲過程中都保持不動。這是完全有可能的,尤其是在某些車輛的移動是為了給其他車輛讓路,而該車輛本身並非關鍵路徑上的阻礙時。

    基於以上分析,選項 (a) 和 (e) 似乎是比較可能正確的。 題目中提到「apply what you have learned from game theory」,這暗示了賽局理論中的一些普遍結論。

    • 進一步思考: 關於 (b),後向歸納法確實會考慮到為達成最終目標所必須的中間步驟,包括其他車輛的移動。黃色汽車(E4E5E6)要讓路給紅色汽車(C1C2)通過 C3、C4,橙色汽車(C5D5)需要移動,而橙色汽車的移動空間受限,可能需要黃色汽車向左移動來配合。
    • 關於 (c): 如果存在一種策略組合,能讓紅色汽車通過,那麼「不通過」就不是均衡。但如果遊戲設計上就存在「無法通過」的情況,那又另當別論。
    • 關於 (a) and (e): 遊戲理論中,無限循環(a)和靜止不動的參與者(e)都是可能存在的現象。

    綜合來看,(a) 和 (e) 是最符合賽局理論一般性結論的選項。

  • 答案: (a), (e)

題目 2

此題繼續探討 Rush Hour 遊戲,但引入了多個玩家(小孩)輪流進行的變體,並詢問賽局的均衡結果。

  • 核心概念: 多人賽局、輪流決策、先手優勢(first mover advantage)與後手優勢(second mover advantage)、均衡結果。

  • 解題思路:

    • (a) In a general game, increasing the number of vehicles makes the red vehicle pass through the exit less likely in equilibrium. 增加車輛意味著棋盤更擁擠,移動空間可能受限,需要更多步驟才能讓紅色汽車通過。因此,增加車輛數確實可能降低紅色汽車通過出口的機率。
🔒

後續完整解題步驟與【答案】

免費註冊,享三天全站完整詳解閱覽。

免費註冊

第 2 題10 分

  1. Continue from above.

(a) In a general game, increasing the number of vehicles makes the red vehicle pass through the exit less likely in equilibrium.
(b) In a general game, the red vehicle passes through the exit only when the number of vehicles is odd.
(c) To make the exemplar game shown in the picture more fun, it is now played with two kids R and B taking turns. R wins if the red vehicle passes through the exit, and B wins if he blocks so the red vehicle does not pass through the exit. The equilibrium outcome depends on whether R starts first or not.
(d) To make the exemplar game shown in the picture more fun, it is played with two kids R and B taking turns. R wins if the red vehicle passes through the exit, and B wins if he blocks so the red vehicle does not pass through the exit. In equilibrium, R wins when B starts first. Hence there is a second mover advantage.
(e) To make the exemplar game shown in the picture even more fun, it is played with three kids R1, R2 and B taking turns. R1 and R2 win if the red vehicle passes through the exit, and B wins if he blocks so the red vehicle does not pass through the exit. The equilibrium outcome, specifically on whether the red vehicle passes through the exit, is different from the equilibrium outcome in the question above where only a kid plays the exemplar game.

登入後即可作答並保存紀錄。

這一題的完整詳解

核心觀念

本題延續第 1 題的 Rush Hour 盤面,考的是輪流行動賽局的逆向思考:攻方(R)要讓紅車出去,需要「同時」達成好幾個條件;守方(B)只要每回合能把攻方上一手的進度復原,攻方就永遠湊不齊條件。判斷誰會贏,關鍵在於「攻方每輪能走幾手」對「守方每輪能撤銷幾手」。

解題方法

先把第 1 題的盤面整理清楚(列 A–F 由北而南、行 1–6 由西而東,出口在 C6 旁):

  • 紅車 C1C2,只能在 C 列水平移動;要出去必須 C3–C6 全空。
  • 藍車 A4B4C4(擋住 C4),只能在第 4 行上下移動。它上方已是盤緣,只能往下移到 D4E4F4,而這需要 E4、F4 同時空出。
  • 黃車 E4E5E6(佔 E4),只能在 E 列左右移動;白車 F4F5(佔 F4),只能在 F 列左右移動。
  • 橘車 C5D5(擋住 C5),只能在第 5 行上下移動,往上移到 A5B5 才會讓出 C5。

所以紅車出去之前,至少要先完成:黃車讓出 E4、白車讓出 F4(兩件事必須同時成立)→ 藍車下移 → 橘車上移 → 紅車駛出。

兩人輪流(R、B 各走一手)

R 每走一步「讓出 E4」或「讓出 F4」,B 下一手就把那輛車推回原位,E4 與 F4 永遠無法同時空出,藍車下不來,紅車永遠出不去。
若輪到 B 先動,B 也不會被迫破壞封鎖:B 隨時有無害的「閒著」可走,例如把橘車在 B5C5 與 C5D5 之間移動(C5 始終被佔住)、或把白車在 F3F4 與 F4F5 之間移動(F4 始終被佔住)。B 只要等 R 動了再復原即可。

🔒

後續完整解題步驟與【答案】

免費註冊,享三天全站完整詳解閱覽。

免費註冊

第 3 題

  1. An ancient figure named Kong Rong is famous for not picking the biggest pear. A modern figure, Rong Kong (RK hereafter), decides to follow suit. RK has a brother, BR. They have identical tastes: they both prefer more to less.

However, RK learns from Kong Rong but BR does not. So when RK picks first and BR picks second, RK deliberately chooses his second favorite to leave the first favorite to BR. On the other hand, if BR has picked before RK, RK certainly chooses his most favorite from the remaining.

(a) Facing a big, a medium and a small pear, if RK picks first and BR picks second, RK will choose the small pear.
(b) Facing a medium and a small pear, if BR picks first and RK picks second, RK will choose the small pear.
(c) Given enough observations, an outside observer who only observes RK's choice and the choice set when RK chooses will conclude RK's behavior satisfies the axiom of revealed preference.
(d) Given enough observations, an outside observer who only observes RK's choice and the choice set when RK chooses will conclude RK'S behavior violates the axiom of revealed preference.
(e) This year RK goes to extremes. No matter whether BR picks before or after him, RK deliberately chooses his second favorite when there are at least two different alternatives. Because RK always chooses his second favorite whenever possible, given enough observations, an outside observer who only observes RK's choice this year and the choice set when RK chooses will conclude RK's behavior satisfies the axiom of revealed preference.

🖼️ 本題含圖表,以下為原卷對應頁面:
原卷第 3 頁原卷第 4 頁

登入後即可作答並保存紀錄。

這一題的完整詳解

核心觀念

本題考查「顯示偏好」(revealed preference)與「顯示偏好公理」(Axiom of Revealed Preference, ARP)。

若個體在選擇集合 XX 中選擇 xx,且 y∈Xy\in X,則稱 xx 對 yy 具有直接顯示偏好:

x⪰Ryx \succeq^R y

ARP 的重點是:若 xx 被選出時 yy 同時可選,且 yy 被選出時 xx 也同時可選,則不能在 x≠yx\neq y 時同時成立。否則即形成互相顯示偏好的矛盾。

本題中,觀察者只看得到 RK 的「選擇」與「選擇集合」,看不到究竟是 RK 先選還是 BR 先選,因此必須直接依照不同選擇集合下的實際選擇判斷。


解題方法

依兩人的偏好完全相同,且都偏好「多於少」,梨子的偏好排序為:

大梨≻中梨≻小梨\text{大梨} \succ \text{中梨} \succ \text{小梨}

分別分析兩種情況:

  1. RK 先選:RK 故意選自己的第二偏好,將第一偏好留給 BR。
  2. BR 先選:BR 會先選自己最喜歡的梨子,RK 再從剩餘梨子中選最喜歡者。

接著將 RK 在不同選擇集合中的選擇,轉換成顯示偏好關係。


各選項分析

(a) 錯誤

面對大梨、中梨、小梨,若 RK 先選,RK 會故意選自己的第二偏好。

排序為:

大梨≻中梨≻小梨\text{大梨} \succ \text{中梨} \succ \text{小梨}

因此 RK 選擇中梨,並將大梨留給 BR,不會選擇小梨。

所以選項 (a) 錯誤。


(b) 正確

面對中梨與小梨,若 BR 先選,因為 BR 與 RK 偏好相同,BR 會先選中梨。

此時 RK 只能選剩下的小梨,因此:

BR選中梨⇒RK選小梨\text{BR選中梨}\quad\Rightarrow\quad \text{RK選小梨}

所以選項 (b) 正確。


(c) 錯誤

若 RK 先面對大梨、中梨、小梨,RK 會選中梨。因此,由選擇集合

{大梨,中梨,小梨}\{\text{大梨},\text{中梨},\text{小梨}\}

可知中梨被選出,且小梨同時可選,所以中梨對小梨具有顯示偏好:

🔒

後續完整解題步驟與【答案】

免費註冊,享三天全站完整詳解閱覽。

免費註冊

第 4 題

  1. We examine two events In Africa in this question, one on drought insurance and the other on human immunodeficiency viruses (HIV) testing.

Insurance: Crop farmers' fields are mostly rain-fed. Hence, insuring against drought is important. Traditional crop-insurance assesses farmers' losses due to a drought. And then farmers receive a payout. In recent years, a new kind of crop-insurance pays out automatically if a rainfall index, as estimated based on satellite monitoring of clouds, drops below a certain threshold.

HIV testing: HIV status is an important partner attribute where the prevalence is relatively high. However, it is hard to observe whether a potential partner is HIV-positive or -negative. Before, HIV testing is almost impossible because the cost is too high. A recent health intervention offers HIV test which is free and on-site, greatly reducing the cost of such test.

(a) This new type of insurance corrupts farmers' incentive to work hard because after they purchase the insurance, they will not suffer the loss of a drought.
(b) Because farmers cannot control the rain, there is less moral hazard of the new kind of insurance. That is, changes in farmers' behavior cannot make a payout more likely.
(c) The new insurance scheme could go wrong because monitoring clouds may not predict the subsequent rainfall perfectly.
(d) The HIV-negative people happily take the free test, to signal their health status. Hence, refusing to be tested sends a signal of the HIV-positive status.
(e) The HIV-positive group may be adversely affected by the intervention. Whether the welfare of all improves by the intervention depends on the proportion of the HIV-positive group.

🖼️ 本題含圖表,以下為原卷對應頁面:
原卷第 4 頁原卷第 5 頁

登入後即可作答並保存紀錄。

這一題的完整詳解

本題包含兩個獨立的子題,分別探討農業保險和 HIV 檢測的經濟學意涵,主要涉及資訊不對稱、道德風險、誘因設計以及訊號傳遞等概念。

子題一:農業保險

  • 核心概念: 道德風險(Moral Hazard)、誘因(Incentives)、保險設計。

  • 題目情境: 新型農業保險,根據衛星監測的降雨指數自動理賠,而非根據實際損失。

  • 解題思路:

    • (a) This new type of insurance corrupts farmers' incentive to work hard because after they purchase the insurance, they will not suffer the loss of a drought.
      • 新型保險基於降雨指數,而非實際損失。理論上,保險確實會降低農民承擔風險的意願,可能削弱其努力工作的誘因(例如,更努力地灌溉、採取防旱措施)。但「不承受損失」的說法過於絕對,因為保險賠付可能不足以彌補所有損失,且還有其他非金錢的損失(如時間、精力)。然而,誘因減弱是可能的。
    • (b) Because farmers cannot control the rain, there is less moral hazard of the new kind of insurance. That is, changes in farmers' behavior cannot make a payout more likely.
      • 這是關於道德風險的關鍵判斷。道德風險是指在契約關係中,一方的行為難以被另一方觀察或監測,導致該方可能採取對自己有利但對對方不利的行為。傳統保險中,農民的「努力程度」是影響實際損失的關鍵,而這種努力程度難以被保險公司完全監測,這就產生了道德風險。
      • 新型保險的理賠基於「降雨指數」,而不是農民的「實際損失」。農民無法控制下雨,也無法控制降雨指數。因此,農民的「努力程度」(例如,種植方式、灌溉投入)對保險理賠的影響非常小,甚至沒有。也就是說,農民的行為(努力或不努力)不太可能改變觸發理賠的條件(降雨指數低於閾值)。
      • 因此,與傳統保險相比,這種基於客觀指數的保險,其道德風險確實可能較低。
      • 這個敘述是正確的。
🔒

後續完整解題步驟與【答案】

免費註冊,享三天全站完整詳解閱覽。

免費註冊

第 5 題

  1. Conspicuous consumption is the spending of money on luxury goods to publicly display the wealth of the buyer. Terry is wealthy. His consumption on sports car has this element of conspicuous consumption. When the relative price of the car is higher, he likes the car better. He consumes sports car and slippers. Because slippers are worn at home, they do not have this display-of-wealth function at all.

(a) Terry's demand of sports car may be upward-sloping.
(b) The sports car must be an inferior good to Terry.
(c) If the price of the sports car goes up, the change in the demand of the sports car due to the substitution effect could be nonnegative.
(d) Taxing the sports car makes them more expensive. Therefore, the consumption becomes even more conspicuous. Hence, taxing could not make Terry worse off.
(e) Conspicuous consumption is more commonplace, unlike Giffen goods which are elusive and difficult to identify.

🖼️ 本題含圖表,以下為原卷對應頁面:
原卷第 5 頁

登入後即可作答並保存紀錄。

這一題的完整詳解

核心觀念

本題考察炫耀性消費(conspicuous consumption)與凡勃倫財(Veblen good)的價格效果。

一般商品中,價格上升通常造成需求量下降;但炫耀性商品具有「價格越高,越能彰顯財富與身分」的特性,因此價格上升可能提高消費者對該商品的偏好,形成正向的身分效果。

可將運動車效用表示為:

U=U(c,s;pc)U=U(c,s;p_c)

其中,cc 為運動車消費量、ss 為拖鞋消費量、pcp_c 為運動車價格,且炫耀性消費表示:

∂U∂pc>0\frac{\partial U}{\partial p_c}>0

也就是運動車價格越高,其炫耀財富的功能越強。

解題方法

判斷本題時,須區分:

  1. 價格上升對商品需求的總效果。
  2. 運動車與拖鞋之間的替代效果。
  3. 價格上升所產生的炫耀性或身分效果。
  4. 價格變動對實質所得與福利的影響。

炫耀性效果足夠強時,價格上升雖然提高購買成本,卻也提高運動車的身分價值,因此運動車需求曲線得以向右上方傾斜。

選項分析

(a) 正確

Terry 認為運動車的相對價格越高,運動車越能展現財富,因此價格上升會提高其炫耀性價值。

若炫耀性效果大於一般替代效果與所得效果,則價格上升時需求量反而增加:

∂qc∂pc>0\frac{\partial q_c}{\partial p_c}>0

所以 Terry 對運動車的需求曲線得以向上傾斜。

(b) 錯誤

需求曲線向上傾斜不代表運動車必然是劣等財。

劣等財的定義是所得增加時需求量下降:

∂qc∂m<0\frac{\partial q_c}{\partial m}<0
🔒

後續完整解題步驟與【答案】

免費註冊,享三天全站完整詳解閱覽。

免費註冊

第 6 題

  1. This graph shows a change in a consumer's budget line from line A to line B (from the solid line to the dashed line).

Good 2
Line B
Line A

Which could explain this change in the budget line?

(a) Income has decreased, p₁ is unchanged, and p₂ has fallen.
(b) Income has increased, p₁ has fallen, and p₂ is unchanged.
(c) Income has increased, p₁ is unchanged, and p₂ has increased.
(d) Income is unchanged, p₁ has increased, and p₂ has fallen.
(e) Income is unchanged, p₁ has increased, and p₂ is unchanged.

🖼️ 本題含圖表,以下為原卷對應頁面:
原卷第 6 頁

登入後即可作答並保存紀錄。

這一題的完整詳解

本題考查消費者預算線的變動與其可能原因。預算線的斜率和截距的變化可以反映價格和所得的變動。

核心概念: 預算線(Budget Line),價格變動,所得變動。

題目情境: 預算線從線 A (實線) 變動到線 B (虛線)。線 A 的斜率較陡,與 Good 1 軸的截距較大;線 B 的斜率較平緩,與 Good 1 軸的截距較小。

解題思路:

預算線的標準方程為 p1x1+p2x2=Mp_1 x_1 + p_2 x_2 = M,其中 p1p_1 是 Good 1 的價格,p2p_2 是 Good 2 的價格,x1x_1 是 Good 1 的消費量,x2x_2 是 Good 2 的消費量,MM 是所得。
將此方程改寫成 x2=Mp2−p1p2x1x_2 = \frac{M}{p_2} - \frac{p_1}{p_2} x_1。

  • Mp2\frac{M}{p_2} 是 Good 2 的垂直截距。
  • Mp1\frac{M}{p_1} 是 Good 1 的水平截距。
  • p1p2\frac{p_1}{p_2} 是預算線的斜率(絕對值)。

從圖中可以看出:

  1. 垂直截距(Good 2 的最大消費量): 從線 A 到線 B,垂直截距變小了。這意味著 Mp2\frac{M}{p_2} 變小了。
  2. 水平截距(Good 1 的最大消費量): 從線 A 到線 B,水平截距變大了。這意味著 Mp1\frac{M}{p_1} 變大了。
  3. 斜率: 從線 A 到線 B,預算線變得更平緩。線 A 的斜率絕對值 p1Ap2A\frac{p_1^A}{p_2^A} 大於線 B 的斜率絕對值 p1Bp2B\frac{p_1^B}{p_2^B}。這意味著 p1p2\frac{p_1}{p_2} 的比例變小了。

現在我們逐一分析選項:

  • (a) Income has decreased, p₁ is unchanged, and p₂ has fallen.

    • 所得 MM 下降:垂直截距 Mp2\frac{M}{p_2} 下降,水平截距 Mp1\frac{M}{p_1} 下降。
    • p1p_1 不變,p2p_2 下降:垂直截距 Mp2\frac{M}{p_2} 上升,水平截距 Mp1\frac{M}{p_1} 不變。斜率 p1p2\frac{p_1}{p_2} 上升。
    • 綜合來看,所得下降和 p2p_2 下降會產生相反的垂直截距變化。但題目中垂直截距確定下降,水平截距確定上升。這個選項無法同時解釋這兩個變化。
    • 因此,(a) 錯誤。
  • (b) Income has increased, p₁ has fallen, and p₂ is unchanged.

    • 所得 MM 上升:垂直截距 Mp2\frac{M}{p_2} 上升,水平截距 Mp1\frac{M}{p_1} 上升。
    • p1p_1 下降,p2p_2 不變:垂直截距 Mp2\frac{M}{p_2} 上升,水平截距 Mp1\frac{M}{p_1} 上升。斜率 p1p2\frac{p_1}{p_2} 下降。
    • 這裡,所得增加和 p1p_1 下降都會導致水平截距上升(符合圖示)。
    • 所得增加會導致垂直截距上升,而 p2p_2 不變,所以所得增加會使垂直截距上升。
    • 但是,圖示中垂直截距是下降的。
    • 因此,(b) 錯誤。
  • (c) Income has increased, p₁ is unchanged, and p₂ has increased.

    • 所得 MM 上升:垂直截距 Mp2\frac{M}{p_2} 上升,水平截距 Mp1\frac{M}{p_1} 上升。
    • p1p_1 不變,p2p_2 上升:垂直截距 Mp2\frac{M}{p_2} 下降,水平截距 Mp1\frac{M}{p_1} 不變。
🔒

後續完整解題步驟與【答案】

免費註冊,享三天全站完整詳解閱覽。

免費註冊

第 7 題

  1. Amy faces an intertemporal budget constraint: c1+1+π1+ic2=M1+1+π1+iM2c_1 + \frac{1+\pi}{1+i} c_2 = M_1 + \frac{1+\pi}{1+i} M_2, where c1c_1 is "consumption this period", c2c_2 is "consumption next period", M1M_1 is "income this period", M2M_2 is "income next period", π\pi is an "inflation rate", and ii is interest rate (which we assume to be the same for savers and borrowers). Amy has a well-behaved preference. Suppose the interest rate changes and Amy's intertemporal budget line changes from the solid line to the dashed line.

C2C_2
X
Y
Z
C1C_1

(a) Amy's optimal consumption bundle is always at point y.
(b) Suppose Amy's optimal consumption bundle was at point x before the interest rate changes, she definitely will consume more this period (i.e. c1c_1 increases) after the interest rate changes and Amy's intertemporal budget line changes from the solid line to the dashed line.
(c) Suppose Amy's optimal consumption bundle was at point x before the interest rate changes, she may consume less this period (i.e. c1c_1 decreases) after the interest rate changes and Amy's intertemporal budget line changes from the solid line to the dashed line. It is because the substitution effect may outweigh the income effect.
(d) Suppose Amy's optimal consumption bundle was at point z before the interest rate changes, she definitely will consume more this period (i.e. c1c_1 increases) after the interest rate changes and Amy's intertemporal budget line changes from the solid line to the dashed line.
(e) Suppose Amy's optimal consumption bundle was at point z before the interest rate changes, she may consume less this period (i.e. c1c_1 decreases) after the interest rate changes and Amy's intertemporal budget line changes from the solid line to the dashed line. It is because the substitution effect may outweight the income effect.

🖼️ 本題含圖表,以下為原卷對應頁面:
原卷第 7 頁

登入後即可作答並保存紀錄。

這一題的完整詳解

核心觀念

本題考「跨期消費選擇」中利率變動的替代效果與所得效果。

令

q=1+π1+iq=\frac{1+\pi}{1+i}

原預算限制式為

c1+qc2=M1+qM2.c_1+qc_2=M_1+qM_2.

整理得

c2=M2+1+i1+π(M1−c1).c_2=M_2+\frac{1+i}{1+\pi}(M_1-c_1).

因此,預算線斜率為

dc2dc1=−1+i1+π.\frac{dc_2}{dc_1}=-\frac{1+i}{1+\pi}.

圖中虛線比實線陡,表示利率 ii 上升。兩條預算線都通過

(M1,M2),(M_1,M_2),

也就是圖中的 YY 點;YY 是 Amy 的收入禀賦點,不代表一定是最適消費點。

利率上升時:

  1. c1c_1 的機會成本提高,替代效果使 c1c_1 減少。
  2. 所得效果取決於 Amy 原本是儲蓄者還是借款者。

由圖可知:

  • XX 位於 YY 的左上方:c1<M1c_1<M_1,Amy 是儲蓄者。
  • ZZ 位於 YY 的右下方:c1>M1c_1>M_1,Amy 是借款者。

解題方法

比較新舊預算線在同一個 c1c_1 下的 c2c_2:

c2new−c2old=(1+inew1+π−1+iold1+π)(M1−c1).c_2^{\text{new}}-c_2^{\text{old}} = \left( \frac{1+i_{\text{new}}}{1+\pi} - \frac{1+i_{\text{old}}}{1+\pi} \right)(M_1-c_1).

因為 inew>ioldi_{\text{new}}>i_{\text{old}}:

🔒

後續完整解題步驟與【答案】

免費註冊,享三天全站完整詳解閱覽。

免費註冊

第 8 題

  1. The Independence Axiom says, if a consumer prefers lottery p to lottery q, then, for any other lottery r and number 0<α<10 < \alpha < 1, they must prefer αp+(1−α)r\alpha p + (1 - \alpha)r to αq+(1−α)r\alpha q + (1 - \alpha)r.

Choose which situation(s) VIOLATE the independence axiom.

(a) A consumer is willingness to pay 800tobuyalottery,inwhichtherea800 to buy a lottery, in which there a 80%chanceofwinningchance of winning1,000.
(b) A consumer is not willingness to pay 800tobuyalottery,inwhichtherea800 to buy a lottery, in which there a 80%chanceofwinningchance of winning1,000.
(c) A consumer prefers lottery A to lottery B, in which lottery A is a 100100% chance of getting 800,andlotteryBisa800, and lottery B is a 80%chanceofgettingchance of getting1,000. This consumer also prefers lottery C to lottery D, in which lottery C is a 2525% chance of getting 800,andlotteryDisa800, and lottery D is a 20%chanceofgettingchance of getting1,000.
(d) A consumer prefers lottery A to lottery B, in which lottery A is a 100% chance of 1million,andlotteryBisa101 million, and lottery B is a 10% chance of 5 million, a 89% chance of 1millionanda11 million and a 1% chance of 0. This consumer also prefers lottery C to lottery D, in which lottery C is a 11% chance of 1million,891 million, 89% chance of 0, and lottery D is 10% chance of 5millionanda905 million and a 90% chance of 0.
(e) A consumer prefers lottery C to lottery D, in which lottery C is a 3333% chance of getting 2500andaa2500 and a a 66%chanceofgettingchance of getting2400, and lottery D is a 100100% chance of getting 2,400.ThisconsumeralsopreferslotteryAtolotteryB,inwhichlotteryAisa2,400. This consumer also prefers lottery A to lottery B, in which lottery A is a 33%chanceofgettingchance of getting2500, and lottery B is a 3434% chance of getting $2,400.

🖼️ 本題含圖表,以下為原卷對應頁面:
原卷第 8 頁

登入後即可作答並保存紀錄。

這一題的完整詳解

核心觀念

獨立性公理要求:若偏好彩票 pp 勝於 qq,加入相同的共同彩票 rr 後,偏好方向仍須一致:

p≻q⟹αp+(1−α)r≻αq+(1−α)r,0<α<1.p \succ q \quad\Longrightarrow\quad \alpha p+(1-\alpha)r \succ \alpha q+(1-\alpha)r, \qquad 0<\alpha<1.

判斷時要找出兩組選擇是否只是把同一對彩票加入相同的共同結果後形成;若偏好方向相反,才違反公理。

解題方法與選項分析

(a) 不違反。 願意花 $800 買這張彩票,只表示消費者接受了這項選擇。題目沒有再給一組加入共同彩票後、偏好方向相反的選擇,因此不能據此判定違反獨立性。

(b) 不違反。 不願花 $800 買彩票,也只是一項單獨的選擇,可能反映消費者偏好確定的 $800。它同樣沒有呈現兩組相關彩票的偏好反轉。

(c) 不違反。 彩票 C 是彩票 A 以 25%25\% 的機率進行,另以 75%75\% 的機率得到 $0;彩票 D 對彩票 B 也採用相同混合方式。因此,C、D 是在 A、B 上加入相同共同結果後形成。題目給出消費者偏好 A 勝於 B,也偏好 C 勝於 D,方向一致,符合獨立性。

🔒

後續完整解題步驟與【答案】

免費註冊,享三天全站完整詳解閱覽。

免費註冊

第 9 題

  1. Suppose there is an exchange economy with two individuals A and B and two periods "now" and "later". In each period there is only one good they can consume: chocolate. Let's define the amount of chocolate consumed by each consumer either now or later as (NA,LA)(N_A, L_A) and (NB,LB)(N_B, L_B). Consumers can't store chocolate from one period to another because it might get wasted. However, they can exchange chocolate within each period. For example, A might want to give 1 unit of chocolate to B now in exchange of 2 units of chocolates later. In such a case we can say that B is paying A an interest rater = 1.

(a) Suppose (NA,LA)=(NB,LB)=(5,5)(N_A, L_A) = (N_B, L_B) = (5,5), Consumer A gets pretty excited about chocolate the first time she sees it such that her preference can be represented by the following Cobb-Douglas utility function: UA=NALA1/2U_A = N_A L_A^{1/2}. Consumer B is more patient and she is relatively indifferent between consuming more chocolate now or later. Her preference can be represented then by: UB=NBLBU_B = N_B L_B. In this case, the equilibrium price is r=12r = \frac{1}{2}.
(b) Suppose (NAe,LAe)=(10,0)(N_A^e, L_A^e) = (10, 0), and (NBe,LBe)=(0,10)(N_B^e, L_B^e) = (0,10), Consumer A gets pretty excited about chocolate the first time she sees it such that her preference can be represented by the following Cobb-Douglas utility function: UA=NALA1/2U_A = N_A L_A^{1/2}. Consumer B is more patient and she is relatively indifferent between consuming more chocolate now or later. Her preference can be represented then by: UB=NBLBU_B = N_B L_B. In this case, the origin endowment point is pareto efficient.
(c) Suppose (NAe,LAe)=(NBe,LBe)=(5,5)(N_A^e, L_A^e) = (N_B^e, L_B^e) = (5,5), Consumer A only cares about the total amount of chocolate she consumes. Her preference can be represented by the following utility function: UA=NA+LAU_A = N_A + L_A. Consumer B gets pretty excited about chocolate every time she sees it such that her preference can be represented by the following Cobb-Douglas utility function: UB=NBLBU_B = N_B L_B. In this case, the equilibrium price is r=1r = 1.
(d) Suppose (NAe,LAe)=(NBe,LBe)=(5,5)(N_A^e, L_A^e) = (N_B^e, L_B^e) = (5,5), Consumer A's utility function is UA=min⁡(NA,LA)U_A = \min(N_A, L_A), and consumer B's utility function is UB=min⁡(NB,LB)U_B = \min(N_B, L_B). One of the pareto efficient allo-cation is (NA,LA)=(6,5)(N_A, L_A) = (6,5), and (NB,LB)=(4,5)(N_B, L_B) = (4,5).
(e) Suppose in this economy, there are 10 chocolates now and there are 10 chocolate later. Consumer A's utility function is UA=NA+LAU_A = N_A + L_A, and consumer B's utility function is UB=NB+LBU_B = N_B + L_B. The contract curve is all points in the edgeworth box.

🖼️ 本題含圖表,以下為原卷對應頁面:
原卷第 9 頁

登入後即可作答並保存紀錄。

這一題的完整詳解

核心觀念

本題考查兩期交換經濟、競爭均衡、帕累托效率與契約曲線。

令 NN 表示現在的巧克力,LL 表示未來的巧克力。題目規定現在的 11 單位可交換未來的 1+r1+r 單位,因此未來巧克力相對於現在巧克力的價格為

q=11+r.q=\frac{1}{1+r}.

若現在巧克力價格標準化為 11,個人的預算限制為

Ni+qLi=Nie+qLie=wi.N_i+qL_i=N_i^e+qL_i^e=w_i.

對 Cobb-Douglas 效用函數

Ui=NiαLiβ,U_i=N_i^\alpha L_i^\beta,

其需求為

Ni=αα+βwi,Li=βα+βwiq.N_i=\frac{\alpha}{\alpha+\beta}w_i,\qquad L_i=\frac{\beta}{\alpha+\beta}\frac{w_i}{q}.

競爭均衡還必須滿足市場清算:

NA+NB=10,LA+LB=10.N_A+N_B=10,\qquad L_A+L_B=10.

帕累托效率則是指不存在另一個可行配置,使至少一人嚴格變好且其他人不變差。


選項分析

(a)

依掃描圖,選項敘述為均衡利率

r=25.r=\frac25.

兩人的初始禀賦皆為 (5,5)(5,5),故各自的財富為

w=5+5q.w=5+5q.

消費者 A 的效用為

UA=NALA1/2,U_A=N_A L_A^{1/2},

因此

NA=23w,LA=13wq.N_A=\frac23w,\qquad L_A=\frac{1}{3}\frac{w}{q}.

消費者 B 的效用為

UB=NBLB,U_B=N_B L_B,

因此

NB=12w,LB=12wq.N_B=\frac12w,\qquad L_B=\frac{1}{2}\frac{w}{q}.

由現在期巧克力市場清算:

10=NA+NB=(23+12)w=76w,10=N_A+N_B =\left(\frac23+\frac12\right)w =\frac76w,

所以

w=607.w=\frac{60}{7}.

又因為

5+5q=607,5+5q=\frac{60}{7},

得到

q=57.q=\frac57.

利用 q=1/(1+r)q=1/(1+r):

11+r=57⟹1+r=75⟹r=25.\frac{1}{1+r}=\frac57 \quad\Longrightarrow\quad 1+r=\frac75 \quad\Longrightarrow\quad r=\frac25.

因此依掃描圖的實際內容,(a) 正確。

題目文字轉錄寫成 r=12r=\frac12,但掃描圖中可辨識的數值是 r=25r=\frac25;若依文字轉錄的 1/21/2 判斷,該選項會是錯誤。


(b)

初始配置為

(NAe,LAe)=(10,0),(NBe,LBe)=(0,10).(N_A^e,L_A^e)=(10,0),\qquad (N_B^e,L_B^e)=(0,10).

兩人的效用分別為

UA=10⋅01/2=0,UB=0⋅10=0.U_A=10\cdot 0^{1/2}=0, \qquad U_B=0\cdot 10=0.

考慮另一個可行配置:

(NA,LA)=(5,5),(NB,LB)=(5,5).(N_A,L_A)=(5,5),\qquad (N_B,L_B)=(5,5).

此時

UA=55>0,UB=25>0.U_A=5\sqrt5>0,\qquad U_B=25>0.

兩人都嚴格變好,因此原始禀賦配置不是帕累托有效。

所以 (b) 錯誤。


(c)

兩人的初始禀賦仍為 (5,5)(5,5),故

w=5+5q.w=5+5q.

消費者 B 的效用為

UB=NBLB,U_B=N_B L_B,

因此

NB=w2,LB=w2q.N_B=\frac{w}{2},\qquad L_B=\frac{w}{2q}.

消費者 A 的效用為

🔒

後續完整解題步驟與【答案】

免費註冊,享三天全站完整詳解閱覽。

免費註冊

第 10 題

  1. Six commuters need to commute from city W to city E every day. Each commuter drives his/her own car. There are two routes, one is the north route (road "W → N" and then road "N→ E"), and the other is the south road (road "W → S" and then road "S → E"). Each commuter wants to minimize their travel time, which is an increasing function of the number of cars on that road. If x is the number of cars on a road, then the travel time for each of the four roads, in minutes, are as follows.

road "WN": 1+3x1+3x
road "N E": 20+x20+ x
road "WS": 20+x20+ x
road "S→ E": 1+3x1+3x

(a) Suppose there is no regulation. Each commuter can freely choose one of the two routes. In this case, at the equilibrium, the number of cars through the north route and the number of cars through the south route must be the same, the estimated travel time from the north route is 33 minutes.
(b) Suppose a social planner is interested in minimizing the total travel time. This social planner is going to impose a regulation based on the optimal solution. In this case, the estimated travel time is still the same as the equilibrium solution.
(c) There is a new highway connecting N and S, that is to say, there is a new road, road "N → S". If x is the number of cars on this road, the travel time for this road is 7+x7+x. Suppose there is no regulation. Each commuter can freely choose one of the three routes, "W → N → E", "W → S → E", or "W → N → S → E". At the new equilibrium, no one will use the new highway.
(d) There is a new highway connecting N and S, that is to say, there is a new road, road "N → S". If x is the number of cars on this road, the travel time for this road is 7+x7+x. Suppose there is no regulation. Each commuter can freely choose one of the three routes, "W → N → E", "W → S → E", or "W → N → S → E". At the new equilibrium, the estimated travel time is shorter than the estimated travel time at the old equilibrium when the high way, road "NS", is not built.
(e) There is a new highway connecting N and S, that is to say, there is a new road, road "N → S". If x is the number of cars on this road, the travel time for this road is 7+x7+x. Suppose there is no regulation. Each commuter can freely choose one of the three routes, "W → N → E", "W → S → E", or "W → N → S → E". Suppose a social planner is interested in minimizing the total travel time. This social planner is going to impose a regulation based on the new optimal solution (after the highway is built). In this case, the estimated travel time is the same as the new equilibrium solution when new highway is built.

🖼️ 本題含圖表,以下為原卷對應頁面:
原卷第 10 頁

登入後即可作答並保存紀錄。

這一題的完整詳解

核心觀念

本題考查:

  • Wardrop 均衡/納許均衡:使用中的路線,其總旅行時間不能透過單方面改走其他路線而降低。
  • 外部性:每位通勤者選擇路線時,會增加其他人所面對的道路壅塞成本。
  • 社會最適配置:使所有通勤者的總旅行時間最小。
  • Braess 悖論:新增道路後,個人理性選擇可能使均衡旅行時間反而上升。

一、尚未興建新道路

令走北線的人數為 nn,走南線的人數為 ss,則

n+s=6n+s=6

北線旅行時間為

TN(n)=(1+3n)+(20+n)=21+4nT_N(n)=(1+3n)+(20+n)=21+4n

南線旅行時間為

TS(s)=(20+s)+(1+3s)=21+4sT_S(s)=(20+s)+(1+3s)=21+4s

均衡時兩條路線都有人使用,因此

21+4n=21+4s21+4n=21+4s

配合 n+s=6n+s=6,得到

n=s=3n=s=3

每位通勤者的旅行時間為

TN(3)=TS(3)=21+4(3)=33T_N(3)=T_S(3)=21+4(3)=33

因此,原均衡配置為北線 33 人、南線 33 人,每人的旅行時間為 3333 分鐘。


二、社會最適配置

總旅行時間為

C(n,s)=n(21+4n)+s(21+4s)C(n,s)=n(21+4n)+s(21+4s)

由於 n+s=6n+s=6,可寫為

C(n,s)=126+4n2+4s2C(n,s)=126+4n^2+4s^2

在總人數固定時,n2+s2n^2+s^2 在兩者相等時最小,因此社會最適配置也是

n=s=3n=s=3

總旅行時間為

C(3,3)=3(33)+3(33)=198C(3,3)=3(33)+3(33)=198

平均每位通勤者的旅行時間為

1986=33\frac{198}{6}=33

所以原本的非管制均衡恰好也是社會最適配置。


三、興建 N→SN\to S 新道路後的均衡

令:

  • aa:走 W→N→EW\to N\to E 的人數
  • bb:走 W→S→EW\to S\to E 的人數
  • cc:走 W→N→S→EW\to N\to S\to E 的人數

則

a+b+c=6a+b+c=6

各道路流量為:

  • WNWN:a+ca+c
  • NENE:aa
  • WSWS:bb
  • SESE:b+cb+c
  • NSNS:cc

三條路線的旅行時間如下。

北線:

TN=1+3(a+c)+20+a=21+4a+3cT_N=1+3(a+c)+20+a =21+4a+3c

南線:

TS=20+b+1+3(b+c)=21+4b+3cT_S=20+b+1+3(b+c) =21+4b+3c

使用新道路的路線:

TH=1+3(a+c)+(7+c)+1+3(b+c)T_H=1+3(a+c)+(7+c)+1+3(b+c)

因為 a+b=6−ca+b=6-c,所以

TH=27+4cT_H=27+4c

在新均衡中,三條路線各有使用者時,令三者旅行時間相等:

21+4a+3c=21+4b+3c=27+4c21+4a+3c=21+4b+3c=27+4c

因此

4a−c=6,4b−c=64a-c=6,\qquad 4b-c=6
🔒

後續完整解題步驟與【答案】

免費註冊,享三天全站完整詳解閱覽。

免費註冊

其他考古題