﻿ 带有反比佣金率和保留价的拍卖模型研究
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 北京化工大学学报(自然科学版)  2017, Vol. 44 Issue (6): 106-110  DOI: 10.13543/j.bhxbzr.2017.06.017 0

### 引用本文

YANG WeiXing. Study of an auction model with both a reserve price and an inverse commission rate[J]. Journal of Beijing University of Chemical Technology (Natural Science), 2017, 44(6): 106-110. DOI: 10.13543/j.bhxbzr.2017.06.017.

### 文章历史

Study of an auction model with both a reserve price and an inverse commission rate
YANG WeiXing
Faculty of Science, Beijing University of Chemical Technology, Beijing 100029, China
Abstract: In order to reflect the actual situation, this paper reports a study of a sealed-bid auction model with both a reserve price and an inverse commission rate. Firstly, we find the equilibrium bidding strategy in both first-price and second-price sealed-bid auctions. Secondly, we obtain the expected revenue of the bidder, seller and auction house, then these results are compared using two models-one with a fixed commission rate and the other without a commission rate. Thirdly, we study the influence of the reserve price and the coefficient in the inverse commission rate on the equilibrium bidding strategy, and on the expected revenues of the bidder, seller and auction house. The results indicate that the expected revenue of the seller has a positive correlation with both r and k, whereas the expected revenue of the bidder has a positive correlation with k but a negative correlation with r.
Key words: sealed-bid auction    inverse commission rate    reserve price    expected revenue

1 假设和记号

2 投标者的投标策略 2.1 一级价格密封式拍卖

 $\pi \left( {v,A} \right) = \left[ {v - \left( {1 + \frac{1}{{kA}}} \right)A} \right]{F^{n - 1}}\left( {b_1^{ - 1}\left( A \right)} \right)$ (1)

 $\frac{{\partial \pi \left( {v,A} \right)}}{{\partial A}}\left| {_{A = {b_1}\left( v \right)}} \right. = 0$ (2)

 $\begin{array}{l} \frac{{\partial \pi \left( {v,A} \right)}}{{\partial v}}\left| {_{A = {b_1}\left( v \right)}} \right. = \frac{{\partial \pi \left( {v,A} \right)}}{{\partial v}}\left| {_{A = {b_1}\left( v \right)}} \right. + \frac{{\partial \pi \left( {v,A} \right)}}{{\partial A}}\\ \frac{{\partial A}}{{\partial v}}\left| {_{A = {b_1}\left( v \right)}} \right. = \frac{{\partial \pi \left( {v,A} \right)}}{{\partial v}}\left| {_{A = {b_1}\left( v \right)}} \right. = {F^{n - 1}}\left( {b_1^{ - 1}\left( A \right)} \right)\left| {_{A = {b_1}\left( v \right)}} \right. = \\ {F^{n - 1}}\left( v \right) \end{array}$ (3)

 $\pi \left( {v,{b_1}\left( v \right)} \right) - \pi \left( {{v_ * },{b_1}\left( {{v_ * }} \right)} \right) = \int_{{v_ * }}^v {{F^{n - 1}}\left( x \right){\rm{d}}x} ,v \ge {v_ * }$

 $\pi \left( {{v_ * },{b_1}\left( {{v_ * }} \right)} \right) = 0,\pi \left( {v,{b_1}\left( v \right)} \right) = \left[ {v - \left( {1 + \frac{1}{{k{b_1}\left( v \right)}}} \right){b_1}\left( v \right)} \right]{F^{n - 1}}\left( v \right)$

B(v)=$\left( 1+\frac{1}{k{{b}_{1}}\left( v \right)} \right){{b}_{1}}\left( v \right)$，则由上面两式可解出

 $B\left( v \right) = v - \frac{{\int_{{v_ * }}^v {G\left( y \right){\rm{d}}y} }}{{G\left( v \right)}}$ (4)

 $\left( {1 + \frac{1}{{k{b_1}\left( v \right)}}} \right){b_1}\left( v \right) = B\left( v \right) = v - \frac{{\int_{{v_ * }}^v {G\left( y \right){\rm{d}}y} }}{{G\left( v \right)}} = \frac{1}{{G\left( v \right)}}\int_{{v_ * }}^v {yg\left( y \right){\rm{d}}y}$ (5)

 ${b_1}\left( v \right) = B\left( v \right) - \frac{1}{k} = v - \frac{{\int_{{v_ * }}^v {G\left( y \right){\rm{d}}y} }}{{G\left( v \right)}} - \frac{1}{k}$ (6)

 ${b_1}\left( v \right) = B\left( v \right) - \frac{1}{k} = v - \frac{{\int_{{v_ * }}^v {G\left( y \right){\rm{d}}y} }}{{G\left( v \right)}} - \frac{1}{k},v \ge {v_ * }$

 $B\left( v \right) = v - \frac{{\int_{{v_ * }}^v {G\left( y \right){\rm{d}}y} }}{{G\left( v \right)}}$
 $G\left( v \right) = {F^{n - 1}}\left( v \right)$
 ${v_ * } = r + \frac{1}{k}$

 $\begin{array}{l} \frac{{\partial {b_1}\left( v \right)}}{{\partial r}} = \frac{{G\left( {{v_ * }} \right)}}{{G\left( v \right)}} > 0\\ \frac{{\partial {b_1}\left( v \right)}}{{\partial k}} = - \frac{1}{{{k^2}}}\frac{{G\left( {{v_ * }} \right)}}{{G\left( v \right)}} + \frac{1}{{{k^2}}} = \frac{1}{{{k^2}}}\left[ {1 - \frac{{G\left( {{v_ * }} \right)}}{{G\left( v \right)}}} \right] > 0 \end{array}$ (7)

2.2 二级价格密封式拍卖

 $\pi = \int_{\underline v }^v {\left[ {v - \left( {1 + \frac{1}{{k{b_2}\left( y \right)}}} \right){b_2}\left( y \right)} \right]{\rm{d}}{F^{n - 1}}\left( y \right)}$ (8)

 $\left[ {v - \left( {1 + \frac{1}{{k{b_2}\left( y \right)}}} \right){b_2}\left( y \right)} \right]{\left[ {{F^{n - 1}}\left( v \right)} \right]^\prime } = 0$ (9)

 ${b_2}\left( v \right) = v - \frac{1}{k},v \ge {v_ * }$ (10)

3 收益比较 3.1 卖者的收益比较

 $ER_1^{\rm{S}} = E\left[ {{b_1}\left( {{v_{\left[ {1,n} \right]}}} \right)} \right] = \int_{\underline v }^{\bar v} {{b_1}\left( v \right){f_{1,n}}\left( v \right){\rm{d}}v} = \int_{\underline v }^{\bar v} {\left[ {B\left( v \right) - \frac{1}{k}} \right]n{F^{n - 1}}\left( v \right)f\left( v \right){\rm{d}}v}$ (11)

 $ER_2^{\rm{S}} = E\left[ {{b_2}\left( {{v_{\left[ {2,n} \right]}}} \right)} \right] = \int_{\underline v }^{\bar v} {{b_2}\left( v \right){f_{2,n}}\left( v \right){\rm{d}}v} = \int_{\underline v }^{\bar v} {\left[ {v - \frac{1}{k}} \right]{f_{2,n}}\left( v \right){\rm{d}}v}$ (12)

 $\begin{array}{l} {f_{2,n}}\left( v \right) = n\left( {n - 1} \right){F^{n - 2}}\left( v \right)\left( {1 - F\left( v \right)} \right)f\left( v \right) = \\ {n^2}{F^{n - 2}}\left( v \right)f\left( v \right) - {n^2}{F^{n - 1}}\left( v \right)f\left( v \right) - n{F^{n - 2}}\left( v \right)f\left( v \right) + \\ n{F^{n - 1}}\left( v \right)f\left( v \right) = n{F^{n - 2}}\left( v \right)f\left( v \right)\left[ {n\left( {1 - F\left( v \right)} \right) - 1} \right] + \\ n{F^{n - 1}}\left( v \right)f\left( v \right) \end{array}$ (13)

n比较大时，有n(1－F(v))>1，从而得出f2, n(v)>nFn－1(v)f(v)=f1, n(v)，进一步得出ER1S < ER2S

 $\left\{ \begin{array}{l} \frac{{\partial ER_1^{\rm{S}}}}{{\partial k}} = \int_{\underline v }^{\bar v} {\frac{{\partial {b_1}\left( v \right)}}{{\partial k}}{f_{1,n}}\left( v \right){\rm{d}}v} > 0\\ \frac{{\partial ER_2^{\rm{S}}}}{{\partial k}} = \int_{\underline v }^{\bar v} {\frac{{\partial {b_2}\left( v \right)}}{{\partial k}}{f_{2,n}}\left( v \right){\rm{d}}v} > 0\\ \frac{{\partial ER_1^{\rm{S}}}}{{\partial r}} = \int_{\underline v }^{\bar v} {\frac{{\partial {b_1}\left( v \right)}}{{\partial k}}{f_{1,n}}\left( v \right){\rm{d}}v} > 0 \end{array} \right.$ (14)

3.2 拍卖行的收益比较

 $\begin{array}{l} \;\;\;\;\;\;ER_1^{\rm{H}} = E\left[ {\frac{1}{{k{b_1}\left( v \right)}}{b_1}\left( {{v_{\left[ {1,n} \right]}}} \right)} \right] = \int_{\underline v }^{\bar v} {\frac{1}{{k{b_1}\left( v \right)}}{b_1}\left( v \right)} \\ {f_{1,n}}\left( v \right){\rm{d}}v = \int_{\underline v }^{\bar v} {\frac{1}{k}{f_{1,n}}\left( v \right){\rm{d}}v} = \frac{1}{k} \end{array}$ (15)

 $\begin{array}{l} \;\;\;\;\;\;ER_2^{\rm{H}} = E\left[ {\frac{1}{{k{b_2}\left( v \right)}}{b_2}\left( {{v_{\left[ {2,n} \right]}}} \right)} \right] = \int_{\underline v }^{\bar v} {\frac{1}{{k{b_2}\left( v \right)}}{b_2}\left( v \right)} \\ {f_{2,n}}\left( v \right){\rm{d}}v = \int_{\underline v }^{\bar v} {\frac{1}{k}{f_{2,n}}\left( v \right){\rm{d}}v} = \frac{1}{k} \end{array}$ (16)

 $ER_1^{\rm{S}} + ER_1^{\rm{H}} = \int_{\underline v }^{\bar v} {B\left( v \right){f_{1,n}}\left( v \right){\rm{d}}v}$ (17)

 $ER_2^{\rm{S}} + ER_2^{\rm{H}} = \int_{\underline v }^{\bar v} {v{f_{2,n}}\left( v \right){\rm{d}}v}$ (18)

3.3 投标者的收益比较

 $\begin{array}{l} {\pi _1} = \left[ {v - \left( {1 + \frac{1}{{k{b_1}\left( v \right)}}} \right){b_1}\left( v \right)} \right]{F^{n - 1}}\left( v \right) = \\ \left[ {v - {b_1}\left( v \right) - \frac{1}{k}} \right]{F^{n - 1}}\left( v \right) = \left[ {v - \left( {v - \frac{{\int_{{v_ * }}^v {G\left( y \right){\rm{d}}y} }}{{G\left( v \right)}} - } \right.} \right.\\ \left. {\left. {\frac{1}{k}} \right) - \frac{1}{k}} \right]{F^{n - 1}}\left( v \right) = \frac{{\int_{{v_ * }}^v {G\left( y \right){\rm{d}}y} }}{{G\left( v \right)}}{F^{n - 1}}\left( v \right) = \int_{{v_ * }}^v {{F^{n - 1}}\left( y \right){\rm{d}}y} \end{array}$ (19)

 $\begin{array}{l} {\pi _2} = \int_{{v_ * }}^v {\left[ {v - \left( {1 + \frac{1}{{k{b_2}\left( v \right)}}} \right){b_2}\left( v \right)} \right]{\rm{d}}{F^{n - 1}}\left( y \right)} = \\ \int_{{v_ * }}^v {\left[ {v - {b_2}\left( v \right) - \frac{1}{k}} \right]{\rm{d}}{F^{n - 1}}\left( y \right) = } \int_{{v_ * }}^v {\left[ {v - \left( {y - \frac{1}{k}} \right) - } \right.} \\ \left. {\frac{1}{k}} \right]{\rm{d}}{F^{n - 1}}\left( y \right) = \int_{{v_ * }}^v {\left( {v - y} \right){\rm{d}}{F^{n - 1}}\left( y \right)} = \int_{{v_ * }}^v {{F^{n - 1}}\left( y \right)} \\ {\rm{d}}y + {F^{n - 1}}\left( {{v_ * }} \right)\left( {{v_ * } - v} \right) \le {\pi _1} \end{array}$ (20)

 $\frac{{\partial {\pi _2}}}{{\partial {v_ * }}} = \left( {n - 1} \right){F^{n - 2}}\left( {{v_ * }} \right)f\left( {{v_ * }} \right)\left( {{v_ * } - v} \right) \le 0$

4 结论

(1) 假设佣金率与成交价是反比关系, 本文通过研究一个同时带有保留价和反比佣金率的拍卖模型，给出了一般分布下的一级价格密封式拍卖和二级价格密封式拍卖的均衡投标策略，并且发现二者均与k成正比，与保留价r也成正比。

(2) 通过计算卖方、拍卖行、投标者三方的预期收益，发现在本文的模型中，收益等价定理不再成立；但是当投标人数较多时，卖者采取第二价格拍卖比较有利；卖者的预期收益与k及保留价r均成正相关。

(3) 投标者在一级价格密封式拍卖中获得的预期收益大于二级价格密封式拍卖，二者均和k成正比，和r成反比。

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