In the article by Hashizume and Nariu (2020), the following errors were published.
(On page 847) The Correspondence section should be Ryo Hashizume, Faculty of Art and Design (Correspondence Education), Kyoto University of the Arts: 2–116 Uryuyama, Kitashirakawa, Sakyo‐ku, Kyoto, Japan. Email: rhashizum@gmail.com.
(On page 850) Equations 1 and 2 should be
(1)
(2)
(On page 852) The maximization problem of firm
i is expressed by
(On page 853) Equations 9, 10‐1, 10‐2, 11‐1, 11‐2, 12‐1, 12‐2, 12‐3, and 12‐4-9, 10‐1, 10‐2, 11‐1, 11‐2, 12‐1, 12‐2, 12‐3, and 12‐4 should be
(9)
(10-1)
(10-2)
(11-1)
(11-2)
(12-1)
(12-2)
(12-3)
(12-4)
(On page 857) Proposition 2 should be as follows:
where
∂f(
e,
n)/
∂n <
∂f(
e,
n)/
∂e ≦ 0.
Proof.From (8‐1)–(8‐2) and (10‐1)–(10‐2), it follows that πPP/πQQ = (1 − b2)(qPP)2/(qQQ)2. Let di = d, i = 1, 2, and then we have
Hence, we obtain the result: πPP/πQQ ⋛ 1 1 – b2 ⋛ [f(e, n)]2. By simple calculation, we get ■
(On page 859) Equations 14‐1 and 14‐2-14‐1 and 14‐2 should be
(14-1)
(14-2)
(On page 862, 863) Under Section A2 | Proof of Lemma 1 should be
(i) From (8‐1), (10‐1), and (12‐1)–(12‐2), we get
(ii) From (8‐1) and (10‐1), it follows that
(iii) From (8‐1), (10‐1), and (12‐1), we have
(iv) From (8‐1), (10‐1), and (12‐1)–(12‐2), we get
which lead to
by (A1)–(A3).
(On page 863, 864) Under Section A3 | Proof of Proposition 1 should be
(i) From (8‐1) and (10‐1), we have
where
A(
n) is defined in the proof of Lemma 1‐(ii). First, let us check the positivity of the coefficient of
di. Multiplying the coefficient by (1 −
n), we obtain
where the inequality follows from (1 −
n)
2 > (
en –
b)
2 by (A1) and (2 −
n)(2 –
n –
b2) − (1 −
n)(3 –
n –
b2) = 1 –
b2 > 0. Hence, the coefficient of
di is positive. Noting that
A(
n) ≧ 0 for any
n ∈ [0, 1),
holds if e
n ≦
b. Similarly, we have
holds if e
n ≦
b.
Now, we consider the case where
en >
b. For the more efficient firm's price, we have
On the other hand, for price of the less efficient firm, we get
Therefore, we get the desired result.
(ii) Suppose that
en ≧
b. From (8‐2) and (10‐2), it follows that
. Hence, we only need to show
. By (8‐1) and (10‐1), we get
where the inequality follows from (
DQ/
DP) ≧ [(2 −
n)/(2 −
n –
b2)]
2 and [(2 −
n –
b2)
di + (
en –
b)
dj]/ [(2 −
n)
di + (
en –
b)
dj]> (2 –
n –
b2)/(2 −
n), by
en ≧
b. Here, we have
where the inequality is due to
n ≧
b. As a result, we obtain
.
(iii) From (ii), the statement is true if en ≧ b. Hence, it suffices to show that when en < b holds. From (8‐2) and (10‐2), we get
■
(On page 864) Under Section A4 | Proof of Corollary 1 should be
Let
e = 0. From Proposition 2, we have
Hence, the necessary and sufficient condition for πPP > πQQ is given by < n. ■
(On pages 864, 865) Under Section A5 | Proof of Proposition 3 should be
From (12‐1)–(12‐4), it follows that
From (A3), (1 −
n)
dj> (
b –
en)
di holds, and we have
On the other hand, by applying a similar argument for the less efficient firm, we have
These have been corrected in the online version of the paper. We regret any inconvenience caused by these errors.
在Hashizume和Nariu(2020)的文章中,发布了以下错误。
(在第847页)通讯部分应为京都艺术大学艺术与设计学院(函授教育)桥久亮(Ryo Hashizume):日本京都左京区北岚河Uryuyama邮编2–116。电子邮件:rhashizum@gmail.com。
(在第850页)公式1和2应该是
(1)
(2)
(在第857页)提议2应该如下:
其中
∂F(
ê,
Ñ)/
∂n <
∂F(
ê,
Ñ)/
?E ≦0。
证明。从(8-1) - (8-2)和(10-1) - (10-2),它遵循π PP / π QQ =(1 - b 2)(q PP)2 /(q QQ)2。令d i = d,i = 1,2,然后有
因此,我们得到的结果:π PP / π QQ ⋛1 1 - b 2 ⋛[ ˚F(ë,Ñ)] 2。通过简单的计算,我们得到■
(在第859页)公式14-1和14-2-14-1和14-2应为
(14-1)
(14-2)
(在第862、863页上)在A2节中| 引理1的证明应为
(i)从(8-1),(10-1)和(12-1)–(12-2)中,我们得到
(iii)从(8-1),(10-1)和(12-1),我们有
(iv)从(8-1),(10-1)和(12-1)–(12-2)中,我们得到
由(A1)–(A3)导致。
(第863、864页)在A3节中| 命题1的证明应为
(i)从(8-1)和(10-1),我们有
其中引理1-(ii)的证明中定义了
A(
n)。首先,让我们检查
d i系数的正性。将系数乘以(1-
n),我们得到
其中不等式由(1-
n)
2 >(
en -
b)
2乘以(A1)和(2-
n)(2 –
n –
b 2)−(1 −
n)(3 –
n –
b 2) = 1 –
b 2 >0。因此,
d i的系数为正。注意到
甲(
Ñ)≧0对于任何
Ñ ∈[0,1),
保持如果E
Ñ ≦
b。同样,我们
认为如果E
ñ ≦
b。
现在,我们考虑
en >
b的情况。对于更有效的公司价格,我们有
因此,我们得到了预期的结果。
(ⅱ)假设
烯≧
b。从(8-2)和(10-2),得出
。因此,我们只需要显示
。通过(8-1)和(10-1),我们得到
其中不等式源自(
D Q /
D P)≥[(2-
n)/(2-
n –
b 2)]
2和[(2-
n –
b 2)
d i +(
en –
b)
d j ] / [(2 -
ñ)
ð我+(
烯-
b)
d Ĵ ]>(2 -
ñ -
b 2)/(2 -
ñ),通过
连接≧
b。在这里,我们有
其中不等式是由于
Ñ ≧
b。结果,我们得到
。
(iii)由(II),该声明是真实的,如果连接≧ b。因此,足以证明当en <b成立时。从(8‐2)和(10‐2),我们得到
■
(在第864页)在A4节下|保留所有权利。推论1的证明应为
因此,对于必要和充分条件π PP > π QQ由下式给出< Ñ。■
(在第864、865页上)在A5节下| 命题3的证明应为
从(A3),(1 −
n)
d j >(
b –
en)
d i成立,我们有
另一方面,通过对效率较低的公司应用类似的论点,我们可以
这些已在论文的在线版本中得到纠正。对于这些错误给您带来的不便,我们深表歉意。