マクロ及びミクロ経済学
出題の趣旨・解答例

Micro and Macroeconomics
Solutions

1.
( )α ( )β
Y
K
H
=A
L
L
L
α β
y = Ak h( )α ( )β
A2 k2
h2
y2
=
y1
A k
h
( 1) 1 ( 1)
( )
( )
y2
A2
k2
h2
log
= log
+ α log+ β log
.
y1
A1
k1
h1
Because log(A2 /A1 ), log(k2 /k1 ), and log(h2 /h1 ) each increase by 1%, log(y2 /y1 )should increase by (1 + α + β)%.

2. (1) X’s profit maximization problem is maxy yD(y) − C(y), so the first-ordercondition isyD′ (y) + D(y) − C ′ (y) = 0
and the second-order condition isyD′′ (y) + 2D′ (y) − C ′′ (y) < 0.
In our specification, each becomes() 1a(1 − b) b+1y=c−ab(1 − b)y −b−1 − c < 0.
The SOC always holds because the left-hand side is negative. Hence theequilibrium quantity is given by the above.
(2) X’s profit maximization problem is maxy py − C(y), for which the FOC′′
is p − C ′ (y) = 0 and the SOC is −C (y) < 0. The SOC always holds.
Combining the FOC and the equilibrium condition p = D(y), we have1D(y) = C ′ (y), ory=1( a ) b+1.
c
Because b ∈ (0, 1), the monopolist produces less than the competitive
firm. As b → 0 (that is, when the demand elasticity is infinite), themonopolistic supply and the competitive supply coincide.

3. Because x = σz + µ, where z ∼ N (0, 1), we have
∫ ∞z2
1
E[u(x)] =−e−ρ(σz+µ) √ e− 2 dz
2π
∫−∞
∞
2 2
1
1
2 ρ σ
=
− √ e− 2 (z+ρσ) + 2 −ρµ dz
2π
−∞
∫ ∞
ρ2 σ 2112−ρµ2√ e− 2 (z+ρσ) dz
= −e
2π
−∞
ρσ 2

= −e−ρ(µ− 2 )ρσ 2= u(µ −),
2
where the second from the last equality is because √12π e− 2 (z+ρσ) is the density∫∞12function of N (−ρσ, 1) and hence −∞ √12π e− 2 (z+ρσ) = 1. Thus the certainty
122

equivalent is µ − ρσ2 .

4. (1) 3 C2 + 1 = 4. The first three correspond to pure-strategy equilibria in
which exactly two players choose to act and the public goal is achieved.
There is also an equilibrium in which no one acts.
(2) Let’s say one of the three players is Kate. Suppose the other two players
follow a mixed strategy in which they choose ‘Act’ with probability pand ‘Not act’ with probability 1 − p. Then, (i) Kate’s expected payoﬀ
from choosing ‘Act’ is −C(1 − p)2 + (1 − C)[2p(1 − p) + p2 ]. Also, (ii)her expected payoﬀ from choosing ‘Not act’ is p2 . (i) − (ii) = 0 gives
2p2 − 2p + C = 0,or√

1 − 2C2= 0.1, 0.9.

p =

2

1±

That is, in one symmetric equilibrium, everyone acts with probability
0.1. In the other symmetric equilibrium, everyone acts with probability
0.9.

5. (1) First, derive the reaction function for each firm. The profit maximization
problem for firm 1 is:max π1 = (a − q1 − q2 − c1 )q1
FOC gives a − q2 − c1 = 2q1 . Similarly, the reaction function for firm
2 is a − q1 − c2 = 2q2 . By solving these two equations, q1∗ = a+c23−2c1 ,q2∗ = a+c13−2c2 , and thus Q∗ = q1∗ + q2∗ = 2a−c31 −c2 . Equilibrium price isobtained by substituting Q∗ into the inverse demand function. Therefore,
P ∗ = a+c31 +c2 .

(2) Firm 1’s profit is π1∗ = (a+c29−2c1 ) . Firm 2’s profit is π2∗ = (a+c19−2c2 ) .
2
1 −c2 )
. Producer
Consumer surplus is CS ∗ = (a − P ∗ ) × Q∗ × 12 = (2a−c18(a+c2 −2c1 )2(a+c1 −2c2 )2∗∗∗surplus is P S = π1 + π2 =
+
.
9
9
2

2

6. (1) Making a conjecture that the constraint should hold with equality, theLagrangian is:
L = rK + wL + λ(ȳ − zK a L1−a ).
The optimal choice (K ∗ , L∗ , λ) is the solution to the following FOCs:r = λza(K ∗ /L∗ )a−1(1)

w = λz(1 − a)(K ∗ /L∗ )a(2)ȳ = z(K ∗ )a (L∗ )1−a .

(3)(2) Dividing (2) by (1) yields
w
K∗
a
·
.
=
L∗
1−a r

3

Use the constraint to find factor demands:ȳ = z(K ∗ )a (L∗ )1−a
( ∗ )a
K
= z
L∗
L∗
)a
(
w
a
= z
·
L∗ ,
1−a r
or

(
∗

−1

∗

−1

L = ȳz

1−a r
·
a
w

)a
.

Similarly, we have(
K = ȳz

a
w
·
1−a r

)1−a

Also, by (1),λ = r(za)−1 (K ∗ /L∗ )1−a
)1−a
(
w
a
−1
·
= r(za)1−a r( )a ( w )1−a
−1 r
= z
.
a
1−a
Thus we have obtained the solution (K ∗ , L∗ , λ).

(3) First, calculating K ∗ ×(1)+L∗ ×(2) gives the optimized objective:rK ∗ + wL∗ = λza(K ∗ )a (L∗ )1−a + λz(1 − a)(K ∗ )a (L∗ )1−a
= λaȳ + λ(1 − a)ȳ
= λȳ.
Second, notice that λ depends on many parameters but not on ȳ. These
two together imply that the cost function is linear in the amount of
output (i.e., the marginal cost is constant).

1
7. (1) b̄ is obtained by b̄v61 −p1 = b̄v62 −p2 ⇔ b̄ = 6 pv22 −p. Because b is distributed−v1
uniformly on [2, 8] and the number of consumers is 6, demand for firm
−p1
1 is q1 (p1 , p2 ) = b̄ − 2 = 6 pv22 −v− 2. Similarly, demand for firm 2 is1p2 −p1q2 (p1 , p2 ) = 8 − b̄ = 8 − 6 v2 −v1 .

4

(2) Firm 1’s maximization problem is:()
p2 − p1
max π1 (p1 , p2 ) = (p1 − cv1 ) 6−2
v2 − v1
1
1
FOC with respect to p1 gives p1 = p2 +cv− v2 −v. A similar problem2
62(v2 −v1 )p1 +cv2
to firm 2 gives p2 = 2 +. Solving these two equations gives
3
c(v2 +2v1 )2(v2 −v1 )∗equilibrium prices: p1 =+and p∗2 = c(2v23+v1 ) + 7(v29−v1 ) .
3
9
[
]
[
]2
2
Equilibrium profits are π1∗ = 2(v23−v1 ) c + 23 and π2∗ = 2(v23−v1 ) 37 − c .

(3) No. If firm 1 increases their quality v1 , the price competition becomes[]2
tougher and therefore its equilibrium profit (π1∗ = 2(v23−v1 ) c + 32 ) becomes smaller.

5

経済思想
出題の趣旨・解答例

問題Ⅰ.

経済思想に関する基本的な知識を問う問題である。カール・マルクス

は経済学のみならず、この世界の在り方そのものに大きな影響を与えてきた。
思想としては、『資本論』のほかに、『経済学・哲学草稿』などの初期の論稿が重要な文献となる。資本主義と社会主義（共産主義）の体制選択問題や、現代の資本主義社会をいかに改革するかという問題に、マルクスの思想をどのように活かすことができるのか、あるいはできないのか。本設問は、マルクスの理解と評価に関する基本的な知見を問うている。

問題Ⅱ.

現代の経済思想に関する基本的な質問である。センの学問的功績は、

社会的選択理論・厚生経済学・開発経済学・現代正義論など多岐に及ぶ。彼の制度論的なテーマは、貧困問題や開発問題における資源配分方法の仕組みの解
明と、社会選択プロセスの解明である。それを支える理念的なテーマは、福祉
理念や自由の理念への分析である。前者の観点から、センは近代経済学の利益追求システムを批判的に検討し、そのシステムへの依存を飢餓の原因と位置づ
けつつ、それを改革するための民主的手続きとして社会選択プロセスの解明を試みる。また後者の観点から、近代経済学のシステムの基礎となる合理性の仮定を批判的に検討し、個人は自己利益のみを追求するのではなく、自らの目的自体を検討する合理性と自由をもつべきであると唱えた。そこから、センは自由の指標として潜在能力アプローチを提唱した。本設問においては、このような多岐に及ぶセンの学問的功績に対して、センの理論的分析や規範的理念の特徴を理解しているかが問われる。

統計学
問題 I（解答例）

1. 他の条件を一定にした時, 飲食店でタバコが禁止されている地域では,喫煙量が平均して 2.825 本少なくなる.
2. t = −0.501/0.167 = 3.
3. 0.771 − 2 ∗ 0.009age < 0 を解いて, age > 42.8.
4. 不均一分散下では推定量は線形不偏推定量となり, 線形モデルの中で最も効率の良い推定量ではなくなる.
5. 下記のような検定が考えられる.
 Breusch-Pagan (BP) test(a) 残差を計算し ûi を得る.
(b) û2 = γ0 + γ1 lincome + γ2 lcigpric + γ3 educ + γ4 age + γ5 agesq +γ6 restaurn + ϵ を OLS 推定する.
(c) H0 : 全ての係数（切片を除く）が同時に 0 を F 検定/LM 検定.
 White test(a) 残差を計算し ûi を得る.
(b) û2 = γ0 + γ1 lincome + γ2 lcigpric + γ3 educ + γ4 age + γ5 agesq +γ6 restaurn+interactions +polynomials +ϵ を OLS 推定する.
(c) H0 : 全ての係数（切片を除く）が同時に 0 を F 検定/LM 検定.
 White test (special version)(a) 残差を計算し ûi を得る.
(b) 予測値 ŷi を得る.
(c) û2 = δ0 + δ1 ŷ + δ2 ŷ 2 + ϵ を OLS 推定する.
(d) H0 : 全ての係数（切片を除く）が同時に 0 を F 検定/LM 検定.
1
6. 各観測値に wi = √lincomeで重みづけを行ない,OLS 推定を行なう. こiのとき, 誤差項は均一分散を持つため, 推定量は最良線形不偏推定量(BLUE) の性質をもつ.

7. (1) 説明変数と誤差項の間に相関がある場合を指す.

(2) このとき OLS 推定量は一致性を失う.
(3) 操作変数法.educ と相関を持つが, 誤差項と相関を持たないような
操作変数 z を準備する.z によって説明される educ の外生的な変動部分を説明変数に利用して推定を行なう (2SLS).
1. When other conditions are held constant, in areas where smoking is
prohibited in restaurants, the average smoking quantity decreases by
2.825 cigarettes.
2. t = −0.501/0.167 = 3.
3. Solve 0.771 − 2 ∗ 0.009 ∗ age < 0 to get age > 42.8.
4. Under heterosckdasticity, the estimator becomes a linear unbiased estimator, no longer being the most eﬃcient estimator within the linear
model.
5. The following tests are considered: Breusch-Pagan (BP) test(a) Calculate the residuals to obtain ûi .
(b) Estimate û2 = γ0 + γ1 lincome + γ2 lcigpric + γ3 educ + γ4 age +γ5 agesq + γ6 restaurn + ϵ, using OLS.
(c) Test H0 : All coeﬃcients excluding an intercept are jointly
equal to 0, using an F-test/LM-test.
 White test(a) Calculate the residuals to obtain ûi .
(b) Estimate û2 = γ0 + γ1 lincome + γ2 lcigpric + γ3 educ + γ4 age +γ5 agesq + γ6 restaurn + interactions + polynomials + ϵ, using
OLS.
(c) Test H0 : All coeﬃcients excluding an intercept are jointly
equal to 0, using an F-test/LM-test.
 White test (special version)(a) Calculate the residuals to obtain ûi .
(b) Obtain predicted values ŷi .
(c) Estimate û2 = δ0 + δ1 ŷ + δ2 ŷ 2 + ϵ, using OLS.

(d) Test H0 : All coeﬃcients excluding an intercept are jointly
equal to 0, using an F-test/LM-test.
6. We perform OLS estimation with each observation weighted by wi =√ 1
. In this case, the error term has homoskedasticity, and the eslincomei
timator possesses the properties of the Best Linear Unbiased Estimator(BLUE).
7. (1) It refers to a situation where there is correlation between the explanatory variables and the error term.
(2) In such cases, the OLS estimators lose consistency.
(3) Instrumental Variable (IV) method can deal with the endogeneity.
We prepare an instrumental variable z that is correlated with educbut uncorrelated with the error term. Then, we use the exogenous
variation of educ explained by z as an explanatory variable in theestimation (2SLS).
問題 II（解答例）

1. E [Xi ] and V [Xi ] are,E [Xi ] = p, V [Xi ] = p(1 − p).
[ ][ ]2. E X̄ and V X̄ are,[ ] p(1 − p)
[ ]E X̄ = p, V X̄ =.
n
3. When the sample size is large, the asymptotic distribution of the random variable √ X̄−pis the standard normal distribution and thatp(1−p)/nof the random variable
degree-of-freedom.

n(X̄−p)p(1−p)

2is the chi-squared distribution with 1

4. From the deﬁnition of the constant c0.95 ,
]
[ (
)2
[(
(
]
n X̄ − pc0.95 ) 2
c0.95 )
≤ c0.95 = Pr 1 +p − 2 X̄ +p + X̄ 2 ≤ 0 .
0.95 = Prp(1 − p)n2nSince(c0.95 )2 (c0.95 ) 2 c0.95 X̄(1 − X̄) ( c0.95 )2
X̄ +
− 1+
X̄ =
+
> 0,
2n
n
n
2n
the quadratic equation always has two distinct real solutions. Therefore,√√( c0.95 )2( c0.95 )2 
c0.95 X̄(1−X̄)
c0.95 X̄(1−X̄)
c0.95
c0.95
+
+
X̄ + 2nX̄ + 2n
n
2n
n
2n

Pr 
≤p≤
c0.95 −
c0.95
c0.95 +
c0.95
1+ n
1+ n
1+ n
1+ n
Using the constant c0.95 , ﬁnd the 95% conﬁdence interval of the parameter p.
5. Note thatX̄ + c0.95
2n=
1 + c0.95n∑ni=1 Xi + c0.95 /2

n + c0.95.

In the case of the 95% conﬁdence interval, c0.95 ≈ 4 (When the 97.5%
point of the standard normal is given to be 1.96 (≈ 2), the 95%
point of the chi-squared distribution with 1 degree-of-freedom is c0.95 =(1.96)2 ≈ 22 ). Using this fact, the “four plus rule” estimate formula isderived as follows,∑nX̄ + c0.95
2ni=1 Xi + 2.
c0.95 ≈
1+ n
n+4
This estimation formula is used to avoid unstable estimation of thesample proportion when the estimate is very close to 0 or 1 or when
the sample size is relatively small.

経営学
出題の趣旨・解答例
問題Ⅰ
本設問の趣旨は，経営戦略論分野における基本知識の習得度を確認することである。S C P モデルとは，産業組織論にもとづく理論的フレームワークであり，業界構造（structure）によって企業行動（conduct）とパフォーマンス
(performance)を説明するモデルである。業界構造はその業界の競合企業の数や製品の均質度，参入と退出のコストなどによって測定される。業界構造の特性
は企業が取りうる行動の選択肢に影響を及ぼすが，その状況下で企業は競争優位を築くために価格調整による環境適応や製品差別化，談合などの戦略を追求する。パフォーマンスは企業レベルと社会レベル（たとえば生産と配分の効率性など）という 2 種類に分けられる。SCP モデルの意義として，外部環境の機会や脅威についての分析や事業領域の選択に際して有用であること，などが挙げられる。他方，その限界として，同一業界内の企業間の差異が看過されていること，業界構造と企業行動が動態的に相互に影響を及ぼし合う視点が欠如していることなどが挙げられる。本設問への答案により，（1）経営学，特に経営戦略論の基本概念に関する理解の正確性，(2)修士課程での学修に求められる論
理的思考の力量を評価することができる。
Question I.
The aim of this question is to evaluate applicants’ understanding of thebasic knowledge in the field of strategic management. The S-C-P (structureconduct-performance) model is a theoretical framework to understand therelationship among industry structure, corporate behavior, and performance.
Industry structure is measured by some factors such as the number ofcompeting firms in an industry, homogeneity of products in an industry, costof entry and exit in an industry. Under the constraint that characteristics of
the industry structure define the range of options that firms can take, they
pursue the strategies such as price taking, product differentiation, andcollusion in order to achieve competitive advantage. Performance can bedivided into two types: (1) firm level performance and (2) society levelperformance such as productive and allocative efficiency. This theoreticalmodel is useful in analyzing environmental opportunities and threats and inconsidering the options of business areas. However, this model is insufficient
to explain the reason why strategies or behavior among firms in a same
industry are different, and to understand the dynamic interaction between
industrial structure and corporate behavior. The way applicants answer thisquestion reveals both their understanding of the basic concept in businessstrategy, and their qualifications of logical and reasonable thinking skill.
問題Ⅱ
本設問は，経営組織論における機能別組織構造に関する知識の習得度を確認することを目的としている．解答に際し，①基本的特徴については，組織の基本機能（職能）別に活動がまとめられ部門が設けられている組織であること，②長所としては，各機能内において規模の経済性を高められることや，専門の知識・技能を深めることができる点，③短所としては，部門間のコンフリクトが起こりやすく，経営層に負担がかかるため，環境変化への対応の遅れや，イノベーションが生まれにくいことに触れつつ，適切に説明することが求められる．本設問への解答により，経営組織論の基本概念に関する理解の正確性，および修士課程での学修に求められる論理的な思考力を評価できる．
Question II.
This question aims to assess whether applicants have the basic
knowledge of the functional structure in organization theory. Applicantsneed to explain the following three points. First, the basic characteristics
are that departments are grouped by similar functions or work processes.
Second, the strengths are that the functional structure allows economies ofscale within functional departments and enables in-depth knowledge andskill development. Third, the weaknesses are that the functional structureleads to conflicts among departments, may cause decisions to pile on top
managers, slow response time to environmental changes, and results inless innovation.
The way applicants answer this question reveals both theirunderstanding of the key concept in organization theory and theirqualifications of logical and reasonable thinking skill.

会計学
出題の趣旨

問題Ⅰ．株式交付費や創立費などの繰延資産が，「過去の取引又は事象の結果として報告主体が支配している経済的資源」といった資産の定義に照らしてどのように位置づけられるかについての思考力を問うている．

問題Ⅱ． 本問は，戦略的コスト・マネジメントとしての原価企画（目標原価計算）について，標準原価計算と対比させながらその特徴を説明することを問うている．特に，以下の点に言及してもらいたい．

・製造段階でのコスト・コントロールと上流管理による原価低減
・フィードバック・コントロールとフィードフォワード・コントロール
・プロダクト・アウトとマーケット・イン
・プッシュ生産方式とプル生産方式

Question I.

An asset, which is defined by the IASB, is a present

economic resource controlled by the entity as a result of past
events. To answer this question reveals that applicants have askill in thinking about anything from the Conceptual Framework.
Question II.

This question asks the respondent to explain thecharacteristics of target costing as strategic cost management,
contrasting it with standard costing. The following
characteristics can be pointed out.
・cost control at the production stage and cost reduction throughupstream management
・feedback control and feedforward control
・product out and market in・push production and pull production

オペレーションズ・リサーチ

解説
問題 I．
１．
（１）
目的関数

min

w0 ,wA ,wB

{
制約式

2
2
+ 2σAB wA wB + σB2 wB
σA2 wArw0 + µA wA + µB wB = µ
w0 + wA + wB = 1（２）最適化問題を解くと，wA = 4/13, wB = 6/13, w0 = 3/13.
２．
（１）株価×割引率=95 円（２）取引手数料および貸株のレンタル料はかからないとする．

∗ 時刻 0 で，現物 A 株式 1 単位を 96 円で売り，同時に∗ 現金 95 円で割引国債を買う; そして∗ 1 年後に A 株式 1 株を 100 円で買う先渡し契約を結ぶ．
∗ 1 年後の株式代金は，割引債の償還金を充てる．
∗ これにより，時刻 0 では手元に 1 円が残る．

問題 II．
１．(IP) の目的関数と制約式における xi の係数をそれぞれ ai と bi で表す (i =1, 2, . . . , 8). 問題 (P) については，abii の降順に xi を見て，可能な限り大きい値を割り当てることで得られる解が最適である．すなわち，(x1 , x2 , x3 , x4 , x5 , x6 ,x7 , x8 ) = (0, 0, 1, 1, 0, 1, 0, 21 ) は最適解で，最適値は， 495である．
2２．(x1 , x2 , x3 , x4 , x5 , x6 , x7 , x8 ) = (1, 0, 1, 0, 1, 0, 0, 0) の目的関数値は 113 である．
113
(IP) の最適値を OP T で表す．１. より，OP T ≤ 495≥ 113である． OP
495 =
T
2
2

226
より，(x1 , x2 , x3 , x4 , x5 , x6 , x7 , x8 ) = (1, 0, 1, 0, 1, 0, 0, 0) の近似比は 226
以
495
495

上であるといえる．

1

問題 III．
下図における太線で示された辺からなる木は，s からの最短路木である．

2.6

3.2

6.0

3.1

1.0

3.7
2.3

s

2.3

2.1

1.2

4.8
3.6

3.1

1.4

1.5

2.5

2.1
3.7
1.7
3.2
2.4

1.3
4.4
2.2

4.5

1.8

English Version
Question I．
１．
（１）

objective functionmin

w0 ,wA ,wB{

22
σA2 wA+ 2σAB wA wB + σB2 wB

rw0 + µA wA + µB wB = µ
w0 + wA + wB = 1

constraint（２） Solving the optimization problem gives wA = 4/13, wB = 6/13, w0 =3/13.
２．
（１） Stock price multiplied by discount rate = 95 yen（２） Assume no transaction fees and there is no rental fee for stock lending.

∗ At time 0, sell 1 units of A-share for 96 yen by spot-trading, andat the same time
∗ buy a discount government bond with 95 cash; and∗ Enter a forward contract to buy 1 shares of A-share for 100 yen one
year later.
2∗ The redemption amount of the discount bond will be used for thestock payment after one year.
∗ This leaves 1 yen at time 0.

Question II．
１．For i = 1, 2, . . . , 8, let ai (resp., bi ) denote the coeﬃcient of xi in the objectivefunction (resp., the ﬁrst constraint) for (IP). A solution obtained by sorting xi
in a nonincreasing order of abii , and taking xi in this order to assign the highest
possible value to it is optimal to (P). Hence, (x1 , x2 , x3 , x4 , x5 , x6 , x7 , x8 ) =(0, 0, 1, 1, 0, 1, 0, 21 ) is an optimal solution to (P) and the optimal value for(P) is 495．
2
２．The objective function value for (x1 , x2 , x3 , x4 , x5 , x6 , x7 , x8 ) = (1, 0, 1, 0, 1, 0,0, 0) is 113. Denote the optimal value for (IP) by OP T . Note that OP T ≤
495
113
226
holds by Question 1. Since OP≥ 113
495 = 495 holds, it follows that the2T2

approximation ratio of the solution is at least 226.
495

Question III．
In the following ﬁgure, the tree whose edges are drawn in bold is a shortest pathtree from s.

2.6

3.2

6.0

3.1

3.7
2.3

s

2.3

1.2

4.8
3.6

3.1

1.4

2.5
1.8

3

1.0
2.1
1.5

2.1
3.7
1.7
3.2
2.4
4.5

1.3
4.4
2.2