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2015年2月25日 星期三

練乙錚:駁「未富先驕」說.論「陸客來港」稅 /「擠塞性界外效應」(congestion externality) - 經濟學 Economics (高斯定律 Coase Theorem)

http://forum.hkej.com/node/110367

https://sparkpost.wordpress.com/2014/02/20/%E7%B7%B4%E4%B9%99%E9%8C%9A%EF%BC%9A%E9%A7%81%E3%80%8C%E6%9C%AA%E5%AF%8C%E5%85%88%E9%A9%95%E3%80%8D%E8%AA%AA%EF%BC%8E%E8%AB%96%E3%80%8C%E9%99%B8%E5%AE%A2%E4%BE%86%E6%B8%AF%E3%80%8D%E7%A8%85/

「擠塞性界外效應」(congestion externality) - 經濟學 Economics (高斯定律 Coase Theorem)

練乙錚:駁「未富先驕」說.論「陸客來港」稅

原刊於 2014年2月20日 信報

梁派融合論者去年在公眾壓力之下提出「港人港奶」、「港人港地」等港字頭口號之後,本土派「切實執行」,結果不僅傷害了自由行至此地的大陸人民的感情,還引致北京官媒《環球網》上出現針對港人的「微言」。本來,後者並非是什麼值得大驚小怪的一回事,但對特區政府高層來說,恐怕卻是非同小可,再加上西環也開了腔,遂有三局一司一首高調出場聲色俱厲撻伐本土派,並在未有充分證據證明上周末廣東道事件中的「驅蝗」示威者有不法舉動之時,便聲稱要把示威者繩之以法。

不過,根據當日視頻,「驅蝗」、「迎蝗」兩派其實互有口角推撞,若是犯法要控告,則兩派應同時控告,特區政府何厚此而薄彼?至於口角推撞之餘的「驅蝗」行動和意識,頂多只能算是一些人的態度問題,而非干犯了什麼法律,官員或可來一番春風化雨,進行「深入細緻的思想工作」,但絕對談不上可繩之以法。因此,三局一司一首的言辭,說理不足而粗暴有餘,除了官字兩個口,實與當日在街上推撞的兩派市民的表現差不了多遠;作為領導人而未能趁機在市民面前樹立良好風範、轉移社會風氣,殊為可惜。


擠塞性界外效應

「驅蝗」意識的興起,源於公共空間出現因陸客過多而引致的「擠塞性界外效應」(congestion externality)。經濟學對這類問題有很透徹的分析,解決的辦法基本上有三個:一是把公共空間私有化,任何人要使用該空間便須向空間物權的擁有者付出市場決定的價格(這個辦法實際上顯然行不通,但觀政府只突出個別商戶周末在廣東道出現「驅蝗」行動時蒙受損失,便知高官潛意識裏早把該處公共空間視為商戶的私產);二是鼓勵空間的使用受益者與受損者定出彼此可接受的補償協議(這個辦法對事事率先維護商界利益的特區政府而言,無異於與虎謀皮);三是由政府負責徵收足夠高的空間使用稅,以減低擠塞性界外效應及賠償利益受損者的損失。最後這個辦法,經濟學教科書裏講得最多,筆者作一簡介。

辦法三首先要解決的問題就是,「向誰打稅?」。具體而言,只能有兩個選擇:陸客和商戶。商戶因陸客獲得額外利潤,所繳利得稅的稅率基本上是固定的,但陸客愈多,引起的公共空間擠塞不便卻是以超線性方式增加的,故現存的利得稅制顯然不足以解決問題;政府必須另外打稅。本來,由於市場機制一般有把稅擔自然攤派給市場買賣雙方的作用,政府把稅直接打在那一方身上,理論上並不重要(這是ECON101裏的一個必教課題,道理精妙而不深奧,但要解釋,得費一番唇舌,故於此不贅)。不過,考慮到(一)商戶也有做本地人生意的,而那部分的利潤不應再打公共空間使用稅(本地消費者已經付出稅金擔負公共空間的建設與維修);(二)很難估計商戶的利潤有多少是從陸客身上賺得;因此,最可行的辦法就是直接向陸客打稅了

不能漠視市民訴求

可惜,這個解決問題的正路辦法還未及讓公眾討論,梁振英便已經把倡議者打成「未富先驕」。這除了顯示他條件反射、政治上偏袒陸客而不惜犧牲港人空間使用者的利益之外,還暴露了他對經濟一竅不通。一般人不通沒關係,但管理一個「經濟城市」的領導人不學無術不懂經濟坐在那裏胡搞,大家看了好氣還是好笑?為香港利益起見,筆者提議政務司司長效法北京領導人那種好學模樣,找一位政府經濟師給這個特首補幾個小時的經濟課。

有人會說,遊客到發達國家旅遊購物,人家還要退稅,香港怎麼能對陸客來港打稅?那是未有考慮到空間、特別是公共空間在香港的矜貴。發達國的地方一般比較大,稅率比香港高,公共設施充足,大眾消費地點比香港多,行人道擠塞的情況不如香港;就算你到「小日本」東京的銀座或秋葉原,街道也相當廣闊,起碼到今天也容得下那些大陸客,東京人不必像香港人那樣,天天一出門便得面對要不要「斬腳趾避蝗蟲」的問題。

既應向陸客打稅,怎樣打?那裏打?向陸客徵收陸路入境稅是一個辦法,但此法的缺點是計人頭不計購物多少、逗留多久,因此不夠精準。還有其他辦法。例如:(一)徵收離境稅,不僅計人頭,還計購物多少(要申報、要保留單據準備接受隨機抽查,虛報查出要重罰)、計停留了多久(即日來回客以小時計)。(二)大陸結算的信用卡付款時徵收陸客消費稅;港人用陸卡在港購物繳稅後可每年一次過退稅;陸客帶入和滙入的私人「小額」款項打進境稅。

如此打稅,並不會過分影響陸客來港購物意欲,因為陸客如此大量購物,牟利多於自用,因此其需求彈性很低,打比較高的稅對購買量的影響也不會很大。梁振英顯然因為也不懂這個經濟道理,才會罵倡議打稅的人趕客、「未富先驕」。不過,港人也因此要有心理準備,即打了稅,公共空間依然會有很多陸客擠迫,不過整體而言可獲得相當高的抵償就是了。如果這樣也不能令陸客數目減至港人可接受的水平,那就必須提高稅率,或同時使用其他辦法如源頭減人、入境數量限制等;但後者又會有其他問題,特別是政治問題。筆者認為,還是只靠打足夠高的稅比較好。

這樣打稅,除了融合論者會反對之外,當然會影響若干業界利益,包括業界僱員利益,所以會有各種反對聲音和理由(如「擾民」等)。不過,旅遊業佔香港GDP的4.5%【註1】,陸客的貢獻是這個的一半不到,而且還可能因為陸客數量過高質量太低而令不少其他地區的遊客不願多來香港【註2】;而另一方面,陸客過多引起的問題卻是幾乎全港市民都能感受到的(除了那些生活在高級得陸客不會光臨的地方的富人)。因此,政府不能只顧少數人的觀點而漠視大多數市民的訴求。

為了阻止港人稀有的公共生活空間受陸客入侵而不斷收窄,向陸客徵收入境消費稅是經典辦法,公平公正,絕非仇視大陸人,而且都比由本土派以示威方式趕走陸客「高貴」;因此,特區政府特別是梁振英,明白了箇中道理之後,要反過來大力向中央政府解說清楚。如果大陸政府大陸人還是牛皮燈籠,依然認為港人此舉是「傷害大陸人感情」的話,那就對不起、沒辦法了。到時要看特區政府認為港人利益優先還是陸人感情重要;本土派更可視政府如何取捨再給顏色,當會得到更多港人撐腰支持。

「富而不驕」又如何

理論上,陸客來港引起的「擠塞性界外效應」以及用打稅作為解決辦法,與香港馬路車輛過多引起擠迫及空氣污染須徵收新車落地稅與燃油稅來緩衝,是完全同構的事情。如果有哪個特首今天出來說:「港人港府不要『未富先驕』徵收新車落地稅與燃油稅,以免減少經濟活動、影響GDP!」那就必然犯眾怒。

大家再看看加拿大,最近凍結了其國家移民政策中的投資移民類別,令每年為數幾萬身懷巨款急不及待的大陸富豪申請人望門興嘆(大陸人佔加國該類別移民申請人總數的八九成)。加國聯邦政府的理由其實也和香港本土派「驅蝗」的理由差不多:陸客太多了。難道人家也是「未富先驕」嗎?論人均GDP,香港和加拿大差不多;按購買力平價算,港人收入比加拿大人還高【註3】。但人家為了自身的整體利益,也開始對該類移民數量予以限制。有人提出「未富先驕」論嘲笑港人,顯然因為識見有限。

「未富先驕」論當然不值一哂,「富而不驕」又如何?香港似乎不少這種「修養」特別好的人士。像在前不久的一個飯局裏,筆者坐在一個本地「音樂世家」的富二代旁邊。該戶人家在香港早已發迹,近年家族生意已經做到上海等大陸大城市;這位富二代躊躇滿志,席間卻大罵一些港人不懂面北之道、不屑和大陸人打關係,「抵窮」。可以想見,這種人對着北官絕對溫良恭謹,一點不驕,否則怎會愈來愈富?人各有志,富人為了賺更多的錢對着掌權者哈腰,旁人毋庸說三道四,但如果認為所有人都應該如此、不然就是犯賤,那麼,那種「不驕」就帶有一種特殊的哲學味道,拿來治港,港人恐怕很快都變成「狗娘養的」。

作者為《信報》特約評論員

註1:見香港貿發局最新數據香港經貿概況。

註2 :可參考《852post》18日文章〈實施自由行大陸客升500% 溝淡外國旅客香港邁向「國內化」〉

註3 :見維基百科資料。



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2013年10月5日 星期六

1991年諾貝爾經濟學獎得主高斯 (Ronald Coase) 2013年9月2日於芝加哥逝世 (Economics)

1991年諾貝爾經濟學獎 Nobel Prize 得主高斯 (Ronald Coase) 2013年9月2日於芝加哥逝世


諾獎經濟巨人高斯逝世

2013-09-04 05:00:13

【明報專訊】1991年諾貝爾經濟學獎得主、交易成本理論提出者高斯(Ronald Coase)9月2日於芝加哥逝世,享年102歲。高斯被視為「芝加哥學派」代表人物,是新制度經濟學的鼻祖,產權理論奠基人,其理論對中國的經濟改革影響深遠。
肺炎奪命 享年102
芝加哥大學法律學院昨日在官方網站公布高斯死訊。高斯的助手、浙江大學科斯經濟研究中心國際主任王寧向內地財新網透露,前些日子高斯正興奮地準備今年10月造訪中國。但4周之前,他出現呼吸短促,診斷為肺炎,其後病情反覆,延至本周一離世。
「老先生此前已表示過,不願意過度治療,希望能與(去年10月去世的)太太早日相聚。」王寧稱,醫院遵從高斯的意願,從8月28日起放棄實質性治療。按高斯遺囑,其遺體將捐獻作科研,不安排告別儀式。
高斯1910年生於倫敦市郊的威爾斯登(Willesden),父母皆為郵局工人。他於1932年獲倫敦政經學院(LSE)經濟學學士學位,1951年獲倫敦大學博士學位,其後移民美國,曾任教於紐約州立大學水牛城分校與弗吉尼亞大學,1964年後主要任教於芝加哥大學,1991年因對經濟組織產生原理的闡述及推動法學、經濟史和組織理論的發展,獲諾貝爾經濟學獎。
提交易成本 創「高斯定律」
高斯的主要洞見在於以「交易成本」解釋企業的性質和規模。他於1937年發表的首篇重要論文——《企業的性質》(The Nature of the Firm ),便詮釋了這個觀點。他以通用汽車為例,指出廠方買入別家製的輪胎而不自行生產,是因省下的成本要比自家生產好處大得多。這類對「製造」或「購買」的決策,在半個世紀後成為商學院日常研習個案。
到了1960年,高斯發表《社會成本問題》(The Problem of Social Cost )論文,衍生了著名的「高斯定律」(Coase Theorem)。該理論認為,在市場交易成本為零的前設下,完善的產權界定可解决「外部效應」(externalities)問題。以噴黑煙的工廠污染為例,如果受影響居民擁有「不被污染權」,則工廠要排黑煙便要從受影響者購買「污染配額」,最終一個願買一個願賣,毋須政府干預,所謂「外部效應」就可解決。
(綜合報道)


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憾未訪中國 批一孩政策可致滅亡

2013-09-04 05:00:13

【明報專訊】高斯很關注中國的經濟發展,中港經濟學家如張維迎、盛洪、周其仁及張五常等亦深受其理論影響。高斯生前最後一本著作就是與王寧合著的《變革中國:市場經濟的中國之路》(How China Became Capitalist),惟他說過一生莫大遺憾是未到過中國,原計劃10月訪華的他如今願望永遠無法實現。
1990年代初期,高斯產權理論在中國很流行,甚至影響到國企產權制度改革。部分人認為低估賤賣或賠錢出賣國企是 「明晰產權」的最好方式,甚至有說「國有資產全部流失之日,就是市場經濟完全確立之時」。
高斯曾批評計劃生育「開錯藥方」,他表示,若中國維持一孩政策,最終可能走向滅亡,因為家庭單位是國民經濟所依仗的基礎。
高斯與曾在加哥大學待了兩年的張五常交情深厚。高斯稱,1980年代初,他極力建議張五常到香港大學任教,因港大是當時研究中國最新經濟改革最好的地方。
(綜合報道)

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2013-09-04   蘋果日報   兩岸國際

提出交易成本 界定產權重要性
經濟學巨擘 高斯逝世


【Ronald Harry Coase 1910─2013】
1991年諾貝爾經濟學獎得主、芝加哥學派代表人物高斯,前日(周一)在美國芝加哥聖約瑟醫院逝世,享年102歲。高斯提出交易成本理論及「高斯定律」(Coase Theorem),強調只要釐清產權,毋須政府干預,市場就可解決問題,對產權理論、法津經濟學及新制度經濟學有極大貢獻,被譽為「當代最具影響力經濟學家之一」。
高斯生前在美國芝加哥大學執教多年,芝大前日發表聲明公佈他的死訊,但未透露死因。這位芝加哥經濟學派代表人物,跟一般經濟學家不同,他早已認定數學「不是我杯茶」,故此研究着重於法律規定和企業,當中兩篇重要著作,奠定他在經濟學界的地位。

《公司的本質》 避開市場成本

高斯1929年入讀倫敦政經學院,受到新任教的南非開普敦大學商業教授普蘭特(Arnold Plant)影響至深,「普蘭特向我介紹了史密斯的『無形之手』,讓我意識到一個具競爭力的經濟體系是如何由價格制度協調」。

1931年他拿獎學金赴美國學習,研究美國工業的截然不同結構。他在福特汽車公司和其他公司進行實地研究,讓他對交易成本有深刻體會,回國後除了在大學教書,還着手將他在美國見聞的體會整理,終於在1937年發表了《公司的本質》(The Nature of the Firm),以交易成本,包括時間、費用和經常開支,解釋了公司性質和規模。他認為,機構的成立是為避開「市場成本」。

哪些生產工序自行生產、哪些外判的相對成本,決定了一間公司的規模和性質。以美國通用汽車為例,寧願外購輪胎,都不自行生產,因為在外判競投中省掉的成本,遠較自行生產的好處大。

高斯是首位以這理論解釋現代企業規模的人,而直至今時今日,「生產」與「購買」的決定,仍然是大學商學院定時進行的個案研究。
在1960年,高斯再發表《社會耗費問題》(The Problem of Social Cost),進一步探討交易成本,主張完善的產權界定可解決經濟活動產生的外在問題,即是經濟學上所稱的「界外效應」(Externality)。

這即著名的「高斯定律」,核心思想是產權交易和交易費用──如果交易費用為零,產權清晰而且能自由交換,即使由法律規定的權利分配不當時,市場會通過自由交換得以修正。
高斯1997年受訪時,就以簡單一句解釋:「人們會利用資源,務求生產出價值最高的東西。我覺得這是顯而易見的,根本不需要『高斯定律』。」

經濟學家張五常曾提出一個例子說明。如果地主有兩間並列的房子,其中一家養了一隻狗,日夜吠叫,騷擾鄰居,吠叫聲就是「界外效應」。地主可向養狗一家多收租金,作為「購買」養狗的權利;少收另一戶租金,作為承受狗吠聲的補償。計算及交易狗吠聲的費用雖然高,而且過程中沒有直接交易,但狗主已承擔了他所製造出來的「界外效應」,毋須政府干預。

1991年獲頒諾貝爾獎瑞典

瑞典皇家科學院的諾貝爾經濟學獎評選委員會,在1991年頒獎給高斯時,讚揚「發現和釐清了交易成本和產權,對公司架構和經濟運作的重要性」,「指出傳統基礎微觀經濟理論的不完整,因為它只包括生產和運輸成本,忽略了制訂和執行合約及管理公司的成本」。
美國經濟學家威廉森(Oliver Williamson)和溫特(Sidney Winter)在1993年著書談及高斯時,形容他是「當代最具影響力經濟學家之一」。

英國《金融時報》/彭博網


高斯生平

29/12/1910 生於倫敦,小時不良於行,雙腿用金屬支架,入讀殘障兒童學校
1923 12歲以獎學金入讀基爾伯恩中學(Kilburn Grammar School)
1927 大學入學試及格,中學畢業前已修畢倫敦大學首年課程
1929-31 入讀倫敦政經學院(LSE)商業學系,最後一年獲獎學金赴美學習一年
1932-34 任教於Dundee School of Economics and Commerce
1934-35 任教於利物浦大學
1935-51 任教於LSE
1937 發表《公司的本質》;與Marion Ruth Hartung結婚,兩人一直無所出
1948 以Rockefeller Fellowship獎金赴美9個月
1951 移居美國,任教於紐約州立大學水牛城分校(SUNY Buffalo)
1959 轉往維珍尼亞大學任教
1960 發表《社會耗費問題》
1964 加入芝加哥大學法學院,任《法律與經濟》期刊編輯至1982年
1982至今 成為芝大榮休教授
1991 獲諾貝爾經濟學獎
02/09/2013 卒於芝加哥,享年102歲

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遺憾從未踏足中國


高斯以產權理論傳世,對中國由共產主義發展市場經濟的制度變革充滿興趣,亦深刻影響中國經濟改革。他去年以101歲高齡出版的最後著作,就是《變革中國──市場經濟的中國之路》(How China Became Capitalist),又曾對合著的助手王寧說過,他的一生最大遺憾,就是有生之年沒去過中國。
2010年12月29日,即高斯100歲壽辰,中國十多家學術機構的學者在北京、上海和芝加哥三地舉行一場學術研討會為高斯賀壽,其間高斯在接受王寧的訪問中說,他對中國的濃厚興趣,可能是小時看《馬可孛羅遊記》時就有,後來中國推行改革開放,他就力促張五常去港大任教,因為他認為香港是研究中國經改的最好地方。

高斯定律影響中國經改

在張五常介紹下,高斯的產權理論在中國產生重大影響,新華網形容「在中國絕大部份經濟學人,都或多或少受到高斯的影響和啟發,而他更是在中國擁有大量忠實擁躉」,包括茅于軾、張維迎、盛洪、周其仁等內地最優秀的經濟學家。
高斯曾說,「經濟學的未來在中國」,理由一是中國人口多、人才多,理由二是認為中國市場經濟轉型的研究,將對經濟學發展有巨大貢獻。高斯2008年在中國改革開放30周年之際,用諾貝爾獎金在芝大舉辦大型中國經改研討會,後來亦成立「高斯中國學會」,推動中國經改研究。
中國農業銀行首席經濟學家向松祚表示,高斯說過「中國改革是世界上最偉大的奇蹟」,因為成功解決了全世界1/5人口的富裕問題,亦為其他國家經濟發展提供制度上借鑑。
向松祚說,高斯論證了產權明確保障,是市場經濟運作的最核心先決條件影響,中國的學者從高斯的理論分析出發展經濟的制度障礙,中國市場化改革的核心,也是越來越注重產權保護,「中國過去三十多年的改革,應該是高斯定律很好的實證應用」,今天改革深化,也是繼續完善產權制度,包括改革國有企業壟斷,對民營資本的產權保障。
他又透露,張五常和高斯在中國的好友,積極安排高斯今年10月到中國考察,可惜高斯等不及,留下遺憾。
新浪網/和訊網/新華網


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經濟學巨人高斯102歲離世

星島日報 – 2013年9月4日星期三上午5:48

(綜合報道)(星島日報報道)美國芝加哥大學宣布,在經濟學史上地位重要的殿堂級大師高斯(Ronald Coase),周二於芝加哥逝世,享年一百○二歲。高斯是新制度經濟學鼻祖、產權理論創始人,與本港經濟學家張五常淵源極深,其學說對中國經濟改革影響深遠,遺作正是新近出版的《變革中國-市場經濟的中國之路》。
  英國出生的高斯被視為芝加哥學派巨人。他的代表作是兩篇論文,其一是一九三七年的《公司的性質》(Nature of the Firm),講述交易成本的概念,該文奠定了現代企業理論的基礎,主要探討兩個問題:企業為甚麼會存在?企業的規模由甚麼因素決定?論文的核心內容便是以市場成本論與組織成本論來回答這兩個問題,寫這篇革命性文章時他只有二十多歲。
  另一篇代表作是一九六○年撰寫的《社會成本問題》(The Problem of Social Cost),講述政府干預市場的能力遠較經濟學家預期為低,論文成為後來著名的「高斯定理」(Coase Theorem)基礎;該理論指,清楚界定的財產權可以克服非市場力量。
  另外,他也提出著名的「高斯猜想」(Coase Conjecture),即是像售賣汽車、雪櫃等耐用品的壟斷商,在一段時間後也會開始與其他壟斷商競爭,因此最後仍必須以較低價格出售商品。
  雖然高斯提出的不少理論引發更多的學術討論,但他卻是因為《公司的性質》和《社會成本問題》兩篇舊文,令他在一九九一年得到諾貝爾經濟學獎。他對產權理論、法律經濟學與新制度經濟學都有極大貢獻。
  高斯一九一○年生於英國倫敦郊區,父母均為郵局職員。他取得獎學金後入讀名校倫敦經濟學院,在校內接觸了經濟學大師史密斯(Adam Smith)的「無形之手」理論,後來再取得獎學金赴美研究美國工業的架構。他一九五一年獲倫敦大學博士學位,婚後搬到美國,六四年後主要任教於芝加哥大學。
  高斯的最後一本著作是與中國弟子王寧合著的《變革中國-市場經濟的中國之路》,當時他已一百○一歲。



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2012年6月27日 星期三

Nash equilibrium, Game theory, Cournot competition - Wikipedia

Game theory - Wikipedia
http://en.wikipedia.org/wiki/Game_theory


Cournot competition - Wikipedia
http://en.wikipedia.org/wiki/Cournot_equilibrium



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Nash equilibrium - Wikipedia
http://en.wikipedia.org/wiki/Nash_equilibrium

Nash equilibrium

From Wikipedia, the free encyclopedia

In game theory, the Nash equilibrium (named after John Forbes Nash, who proposed it) is a solution concept of a game involving two or more players, in which each player is assumed to know the equilibrium strategies of the other players, and no player has anything to gain by changing only his own strategy unilaterally.[1]:14 If each player has chosen a strategy and no player can benefit by changing his or her strategy while the other players keep theirs unchanged, then the current set of strategy choices and the corresponding payoffs constitute a Nash equilibrium.
Stated simply, Amy and Phil are in Nash equilibrium if Amy is making the best decision she can, taking into account Phil's decision, and Phil is making the best decision he can, taking into account Amy's decision. Likewise, a group of players are in Nash equilibrium if each one is making the best decision that he or she can, taking into account the decisions of the others.

Contents

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Applications

Game theorists use the Nash equilibrium concept to analyze the outcome of the strategic interaction of several decision makers. In other words, it provides a way of predicting what will happen if several people or several institutions are making decisions at the same time, and if the outcome depends on the decisions of the others. The simple insight underlying John Nash's idea is that we cannot predict the result of the choices of multiple decision makers if we analyze those decisions in isolation. Instead, we must ask what each player would do, taking into account the decision-making of the others.
Nash equilibrium has been used to analyze hostile situations like war and arms races[2] (see Prisoner's dilemma), and also how conflict may be mitigated by repeated interaction (see Tit-for-tat). It has also been used to study to what extent people with different preferences can cooperate (see Battle of the sexes), and whether they will take risks to achieve a cooperative outcome (see Stag hunt). It has been used to study the adoption of technical standards[citation needed], and also the occurrence of bank runs and currency crises (see Coordination game). Other applications include traffic flow (see Wardrop's principle), how to organize auctions (see auction theory), the outcome of efforts exerted by multiple parties in the education process,[3] and even penalty kicks in soccer (see Matching pennies).[4]

History

A version of the Nash equilibrium concept was first used by Antoine Augustin Cournot in his theory of oligopoly (1838). In Cournot's theory, firms choose how much output to produce to maximize their own profit. However, the best output for one firm depends on the outputs of others. A Cournot equilibrium occurs when each firm's output maximizes its profits given the output of the other firms, which is a pure strategy Nash Equilibrium.
The modern game-theoretic concept of Nash Equilibrium is instead defined in terms of mixed strategies, where players choose a probability distribution over possible actions. The concept of the mixed strategy Nash Equilibrium was introduced by John von Neumann and Oskar Morgenstern in their 1944 book The Theory of Games and Economic Behavior. However, their analysis was restricted to the special case of zero-sum games. They showed that a mixed-strategy Nash Equilibrium will exist for any zero-sum game with a finite set of actions. The contribution of John Forbes Nash in his 1951 article Non-Cooperative Games was to define a mixed strategy Nash Equilibrium for any game with a finite set of actions and prove that at least one (mixed strategy) Nash Equilibrium must exist in such a game.
Since the development of the Nash equilibrium concept, game theorists have discovered that it makes misleading predictions (or fails to make a unique prediction) in certain circumstances. Therefore they have proposed many related solution concepts (also called 'refinements' of Nash equilibrium) designed to overcome perceived flaws in the Nash concept. One particularly important issue is that some Nash equilibria may be based on threats that are not 'credible'. Therefore, in 1965 Reinhard Selten proposed subgame perfect equilibrium as a refinement that eliminates equilibria which depend on non-credible threats. Other extensions of the Nash equilibrium concept have addressed what happens if a game is repeated, or what happens if a game is played in the absence of perfect information. However, subsequent refinements and extensions of the Nash equilibrium concept share the main insight on which Nash's concept rests: all equilibrium concepts analyze what choices will be made when each player takes into account the decision-making of others.

Definitions

Informal definition

Informally, a set of strategies is a Nash equilibrium if no player can do better by unilaterally changing his or her strategy. To see what this means, imagine that each player is told the strategies of the others. Suppose then that each player asks himself or herself: "Knowing the strategies of the other players, and treating the strategies of the other players as set in stone, can I benefit by changing my strategy?"
If any player would answer "Yes", then that set of strategies is not a Nash equilibrium. But if every player prefers not to switch (or is indifferent between switching and not) then the set of strategies is a Nash equilibrium. Thus, each strategy in a Nash equilibrium is a best response to all other strategies in that equilibrium.[5]
The Nash equilibrium may sometimes appear non-rational in a third-person perspective. This is because it may happen that a Nash equilibrium is not Pareto optimal.
The Nash equilibrium may also have non-rational consequences in sequential games because players may "threaten" each other with non-rational moves. For such games the subgame perfect Nash equilibrium may be more meaningful as a tool of analysis.

Formal definition

Let (S, f) be a game with n players, where Si is the strategy set for player i, S=S1 × S2 ... × Sn is the set of strategy profiles and f=(f1(x), ..., fn(x)) is the payoff function for x \in S. Let xi be a strategy profile of player i and x-i be a strategy profile of all players except for player i. When each player i \in {1, ..., n} chooses strategy xi resulting in strategy profile x = (x1, ..., xn) then player i obtains payoff fi(x). Note that the payoff depends on the strategy profile chosen, i.e., on the strategy chosen by player i as well as the strategies chosen by all the other players. A strategy profile x* \in S is a Nash equilibrium (NE) if no unilateral deviation in strategy by any single player is profitable for that player, that is
\forall i,x_i\in S_i, x_i \neq x^*_{i} :  f_i(x^*_{i}, x^*_{-i}) \geq f_i(x_{i},x^*_{-i}).
A game can have either a pure-strategy or a mixed Nash Equilibrium, (in the latter a pure strategy is chosen stochastically with a fixed frequency). Nash proved that if we allow mixed strategies, then every game with a finite number of players in which each player can choose from finitely many pure strategies has at least one Nash equilibrium.
When the inequality above holds strictly (with > instead of \geq) for all players and all feasible alternative strategies, then the equilibrium is classified as a strict Nash equilibrium. If instead, for some player, there is exact equality between x^*_i and some other strategy in the set S, then the equilibrium is classified as a weak Nash equilibrium.

Examples

Coordination game


A sample coordination game showing relative payoff for player1 / player2 with each combination
Player 2 adopts strategy APlayer 2 adopts strategy B
Player 1 adopts strategy A4 / 41 / 3
Player 1 adopts strategy B3 / 12 / 2

The coordination game is a classic (symmetric) two player, two strategy game, with an example payoff matrix shown to the right. The players should thus coordinate, both adopting strategy A, to receive the highest payoff; i.e., 4. If both players chose strategy B though, there is still a Nash equilibrium. Although each player is awarded less than optimal payoff, neither player has incentive to change strategy due to a reduction in the immediate payoff (from 2 to 1).
A famous example of this type of game was called the Stag hunt; in the game two players may choose to hunt a stag or a rabbit, the former providing more meat (4 utility units) than the latter (1 utility unit). The caveat is that the stag must be cooperatively hunted, so if one player attempts to hunt the stag, while the other hunts the rabbit, he will fail in hunting (0 utility units), whereas if they both hunt it they will split the payload (2, 2). The game hence exhibits two equilibria at (stag, stag) and (rabbit, rabbit) and hence the players' optimal strategy depend on their expectation on what the other player may do. If one hunter trusts that the other will hunt the stag, he should hunt the stag; however if he suspects that the other will hunt the rabbit, he should hunt the rabbit. This game was used as an analogy for social cooperation, since much of the benefit that people gain in society depends upon people cooperating and implicitly trusting one another to act in a manner corresponding with cooperation.
Another example of a coordination game is the setting where two technologies are available to two firms with compatible products, and they have to elect a strategy to become the market standard. If both firms agree on the chosen technology, high sales are expected for both firms. If the firms do not agree on the standard technology, few sales result. Both strategies are Nash equilibria of the game.
Driving on a road, and having to choose either to drive on the left or to drive on the right of the road, is also a coordination game. For example, with payoffs 100 meaning no crash and 0 meaning a crash, the coordination game can be defined with the following payoff matrix:

The driving game
Drive on the LeftDrive on the Right
Drive on the Left100, 1000, 0
Drive on the Right0, 0100, 100

In this case there are two pure strategy Nash equilibria, when both choose to either drive on the left or on the right. If we admit mixed strategies (where a pure strategy is chosen at random, subject to some fixed probability), then there are three Nash equilibria for the same case: two we have seen from the pure-strategy form, where the probabilities are (0%,100%) for player one, (0%, 100%) for player two; and (100%, 0%) for player one, (100%, 0%) for player two respectively. We add another where the probabilities for each player is (50%, 50%).

Prisoner's dilemma

(note differences in the orientation of the payoff matrix)

Example PD payoff matrix
CooperateDefect
Cooperate3, 30, 5
Defect5, 01, 1

The Prisoner's Dilemma has the same payoff matrix as depicted for the Coordination Game, but now C > A > D > B. Because C > A and D > B, each player improves his situation by switching from strategy #1 to strategy #2, no matter what the other player decides. The Prisoner's Dilemma thus has a single Nash Equilibrium: both players choosing strategy #2 ("defect"). What has long made this an interesting case to study is the fact that D < A (i.e., "both defect" is globally inferior to "both cooperate"). That is, both players would be better off if they both chose the other strategy, but doing so is not an equilibrium.

Network traffic


Sample network graph. Values on edges are the travel time experienced by a 'car' travelling down that edge. x is the number of cars travelling via that edge.
An application of Nash equilibria is in determining the expected flow of traffic in a network. Consider the graph on the right. If we assume that there are x "cars" traveling from A to D, what is the expected distribution of traffic in the network?
This situation can be modeled as a "game" where every traveler has a choice of 3 strategies, where each strategy is a route from A to D (either ABD, ABCD, or ACD). The "payoff" of each strategy is the travel time of each route. In the graph on the right, a car travelling via ABD experiences travel time of (1+x/100)+2, where x is the number of cars traveling on edge AB. Thus, payoffs for any given strategy depend on the choices of the other players, as is usual. However, the goal in this case is to minimize travel time, not maximize it. Equilibrium will occur when the time on all paths is exactly the same. When that happens, no single driver has any incentive to switch routes, since it can only add to his/her travel time. For the graph on the right, if, for example, 100 cars are travelling from A to D, then equilibrium will occur when 25 drivers travel via ABD, 50 via ABCD, and 25 via ACD. Every driver now has a total travel time of 3.75.
Notice that this distribution is not, actually, socially optimal. If the 100 cars agreed that 50 travel via ABD and the other 50 through ACD, then travel time for any single car would actually be 3.5 which is less than 3.75. This is also the Nash equilibrium if the path between B and C is removed, which means that adding an additional possible route can decrease the efficiency of the system, a phenomenon known as Braess's Paradox.

Competition game


A competition game
Player 2 chooses '0'Player 2 chooses '1'Player 2 chooses '2'Player 2 chooses '3'
Player 1 chooses '0'0, 02, -22, -22, -2
Player 1 chooses '1'-2, 21, 13, -13, -1
Player 1 chooses '2'-2, 2-1, 32, 24, 0
Player 1 chooses '3'-2, 2-1, 30, 43, 3

This can be illustrated by a two-player game in which both players simultaneously choose an integer from 0 to 3 and they both win the smaller of the two numbers in points. In addition, if one player chooses a larger number than the other, then he/she has to give up two points to the other.
This game has a unique pure-strategy Nash equilibrium: both players choosing 0 (highlighted in light red). Any other choice of strategies can be improved if one of the players lowers his number to one less than the other player's number. In the table to the right, for example, when starting at the green square it is in player 1's interest to move to the purple square by choosing a smaller number, and it is in player 2's interest to move to the blue square by choosing a smaller number. If the game is modified so that the two players win the named amount if they both choose the same number, and otherwise win nothing, then there are 4 Nash equilibria (0,0...1,1...2,2...and 3,3).

Nash equilibria in a payoff matrix

There is an easy numerical way to identify Nash equilibria on a payoff matrix. It is especially helpful in two-person games where players have more than two strategies. In this case formal analysis may become too long. This rule does not apply to the case where mixed (stochastic) strategies are of interest. The rule goes as follows: if the first payoff number, in the duplet of the cell, is the maximum of the column of the cell and if the second number is the maximum of the row of the cell - then the cell represents a Nash equilibrium.

A Payoff Matrix - Nash Equilibria in bold
Option AOption BOption C
Option A0, 025, 405, 10
Option B40, 250, 05, 15
Option C10, 515, 510, 10

We can apply this rule to a 3×3 matrix:
Using the rule, we can very quickly (much faster than with formal analysis) see that the Nash Equilibria cells are (B,A), (A,B), and (C,C). Indeed, for cell (B,A) 40 is the maximum of the first column and 25 is the maximum of the second row. For (A,B) 25 is the maximum of the second column and 40 is the maximum of the first row. Same for cell (C,C). For other cells, either one or both of the duplet members are not the maximum of the corresponding rows and columns.
This said, the actual mechanics of finding equilibrium cells is obvious: find the maximum of a column and check if the second member of the pair is the maximum of the row. If these conditions are met, the cell represents a Nash Equilibrium. Check all columns this way to find all NE cells. An N×N matrix may have between 0 and N×N pure strategy Nash equilibria.

Stability

The concept of stability, useful in the analysis of many kinds of equilibria, can also be applied to Nash equilibria.
A Nash equilibrium for a mixed strategy game is stable if a small change (specifically, an infinitesimal change) in probabilities for one player leads to a situation where two conditions hold:
  1. the player who did not change has no better strategy in the new circumstance
  2. the player who did change is now playing with a strictly worse strategy.
If these cases are both met, then a player with the small change in his mixed-strategy will return immediately to the Nash equilibrium. The equilibrium is said to be stable. If condition one does not hold then the equilibrium is unstable. If only condition one holds then there are likely to be an infinite number of optimal strategies for the player who changed. John Nash showed that the latter situation could not arise in a range of well-defined games.
In the "driving game" example above there are both stable and unstable equilibria. The equilibria involving mixed-strategies with 100% probabilities are stable. If either player changes his probabilities slightly, they will be both at a disadvantage, and his opponent will have no reason to change his strategy in turn. The (50%,50%) equilibrium is unstable. If either player changes his probabilities, then the other player immediately has a better strategy at either (0%, 100%) or (100%, 0%).
Stability is crucial in practical applications of Nash equilibria, since the mixed-strategy of each player is not perfectly known, but has to be inferred from statistical distribution of his actions in the game. In this case unstable equilibria are very unlikely to arise in practice, since any minute change in the proportions of each strategy seen will lead to a change in strategy and the breakdown of the equilibrium.
The Nash equilibrium defines stability only in terms of unilateral deviations. In cooperative games such a concept is not convincing enough. Strong Nash equilibrium allows for deviations by every conceivable coalition.[6] Formally, a Strong Nash equilibrium is a Nash equilibrium in which no coalition, taking the actions of its complements as given, can cooperatively deviate in a way that benefits all of its members.[7] However, the Strong Nash concept is sometimes perceived as too "strong" in that the environment allows for unlimited private communication. In fact, Strong Nash equilibrium has to be Pareto efficient. As a result of these requirements, Strong Nash is too rare to be useful in many branches of game theory. However, in games such as elections with many more players than possible outcomes, it can be more common than a stable equilibrium.
A refined Nash equilibrium known as coalition-proof Nash equilibrium (CPNE)[6] occurs when players cannot do better even if they are allowed to communicate and make "self-enforcing" agreement to deviate. Every correlated strategy supported by iterated strict dominance and on the Pareto frontier is a CPNE.[8] Further, it is possible for a game to have a Nash equilibrium that is resilient against coalitions less than a specified size, k. CPNE is related to the theory of the core.
Finally in the eighties, building with great depth on such ideas Mertens-stable equilibria were introduced as a solution concept. Mertens stable equilibria satisfy both forward induction and backward induction. In a Game theory context stable equilibria now usually refer to Mertens stable equilibria.

Occurrence

If a game has a unique Nash equilibrium and is played among players under certain conditions, then the NE strategy set will be adopted. Sufficient conditions to guarantee that the Nash equilibrium is played are:
  1. The players all will do their utmost to maximize their expected payoff as described by the game.
  2. The players are flawless in execution.
  3. The players have sufficient intelligence to deduce the solution.
  4. The players know the planned equilibrium strategy of all of the other players.
  5. The players believe that a deviation in their own strategy will not cause deviations by any other players.
  6. There is common knowledge that all players meet these conditions, including this one. So, not only must each player know the other players meet the conditions, but also they must know that they all know that they meet them, and know that they know that they know that they meet them, and so on.

Where the conditions are not met

Examples of game theory problems in which these conditions are not met:
  1. The first condition is not met if the game does not correctly describe the quantities a player wishes to maximize. In this case there is no particular reason for that player to adopt an equilibrium strategy. For instance, the prisoner’s dilemma is not a dilemma if either player is happy to be jailed indefinitely.
  2. Intentional or accidental imperfection in execution. For example, a computer capable of flawless logical play facing a second flawless computer will result in equilibrium. Introduction of imperfection will lead to its disruption either through loss to the player who makes the mistake, or through negation of the common knowledge criterion leading to possible victory for the player. (An example would be a player suddenly putting the car into reverse in the game of chicken, ensuring a no-loss no-win scenario).
  3. In many cases, the third condition is not met because, even though the equilibrium must exist, it is unknown due to the complexity of the game, for instance in Chinese chess.[9] Or, if known, it may not be known to all players, as when playing tic-tac-toe with a small child who desperately wants to win (meeting the other criteria).
  4. The criterion of common knowledge may not be met even if all players do, in fact, meet all the other criteria. Players wrongly distrusting each other's rationality may adopt counter-strategies to expected irrational play on their opponents’ behalf. This is a major consideration in “Chicken” or an arms race, for example.

Where the conditions are met

Due to the limited conditions in which NE can actually be observed, they are rarely treated as a guide to day-to-day behaviour, or observed in practice in human negotiations. However, as a theoretical concept in economics and evolutionary biology, the NE has explanatory power. The payoff in economics is utility (or sometimes money), and in evolutionary biology gene transmission, both are the fundamental bottom line of survival. Researchers who apply games theory in these fields claim that strategies failing to maximize these for whatever reason will be competed out of the market or environment, which are ascribed the ability to test all strategies. This conclusion is drawn from the "stability" theory above. In these situations the assumption that the strategy observed is actually a NE has often been borne out by research[citation needed].[verification needed]

NE and non-credible threats


Extensive and Normal form illustrations that show the difference between SPNE and other NE. The blue equilibrium is not subgame perfect because player two makes a non-credible threat at 2(2) to be unkind (U).
The Nash equilibrium is a superset of the subgame perfect Nash equilibrium. The subgame perfect equilibrium in addition to the Nash Equilibrium requires that the strategy also is a Nash equilibrium in every subgame of that game. This eliminates all non-credible threats, that is, strategies that contain non-rational moves in order to make the counter-player change his strategy.
The image to the right shows a simple sequential game that illustrates the issue with subgame imperfect Nash equilibria. In this game player one chooses left(L) or right(R), which is followed by player two being called upon to be kind (K) or unkind (U) to player one, However, player two only stands to gain from being unkind if player one goes left. If player one goes right the rational player two would de facto be kind to him in that subgame. However, The non-credible threat of being unkind at 2(2) is still part of the blue (L, (U,U)) Nash equilibrium. Therefore, if rational behavior can be expected by both parties the subgame perfect Nash equilibrium may be a more meaningful solution concept when such dynamic inconsistencies arise.

Proof of existence

Proof using the Kakutani fixed point theorem

Nash's original proof (in his thesis) used Brouwer's fixed point theorem (e.g., see below for a variant). We give a simpler proof via the Kakutani fixed point theorem, following Nash's 1950 paper (he credits David Gale with the observation that such a simplification is possible).
To prove the existence of a Nash Equilibrium, let r_i(\sigma_{-i}) be the best response of player i to the strategies of all other players.
 r_i = \arg\max_{\sigma_i} u_i (\sigma_i,\sigma_{-i})
Here, \sigma \in \Sigma, where \Sigma = \Sigma_i \times \Sigma_{-i}, is a mixed strategy profile in the set of all mixed strategies and  u_i is the payoff function for player i. Define a set-valued function r = (r_i(\sigma_{-i}),r_{-i}(\sigma_{i})) such that r: \Sigma \rightarrow 2^\Sigma . The existence of a Nash Equilibrium is equivalent to  r having a fixed point.
Kakutani's fixed point theorem guarantees the existence of a fixed point if the following four conditions are satisfied.
  1.  \Sigma is compact, convex, and nonempty.
  2. r(\sigma) is nonempty.
  3. r(\sigma) is convex.
  4. r(\sigma) is upper hemicontinuous
Condition 1. is satisfied from the fact that \Sigma  is a simplex and thus compact. Convexity follows from players' ability to mix strategies.  \Sigma is nonempty as long as players have strategies.
Condition 2. is satisfied because players maximize expected payoffs which is continuous function over a compact set. The Weierstrass Extreme Value Theorem guarantees that there is always a maximum value.
Condition 3. is satisfied as a result of mixed strategies. Suppose  \sigma_i, \sigma'_i \in r(\sigma_{-i}) , then  \lambda \sigma_i + (1-\lambda) \sigma'_i \in r(\sigma_{-i}) . i.e. if two strategies maximize payoffs, then a mix between the two strategies will yield the same payoff.
Condition 4. is satisfied by way of Berge's maximum theorem. Because  u_i is continuous and compact,  r(\sigma_i) is upper hemicontinuous.
Therefore, there exists a fixed point in  r and a Nash Equilibrium.[10]
When Nash made this point to John von Neumann in 1949, von Neumann famously dismissed it with the words, "That's trivial, you know. That's just a fixed point theorem." (See Nasar, 1998, p. 94.)

Alternate proof using the Brouwer fixed-point theorem

We have a game G=(N,A,u) where N is the number of players and A = A_1 \times \ldots \times A_N is the action set for the players. All of the action sets A_i are finite. Let \Delta = \Delta_1 \times \ldots \times \Delta_N denote the set of mixed strategies for the players. The finiteness of the A_is ensures the compactness of \Delta.
We can now define the gain functions. For a mixed strategy \sigma \in \Delta, we let the gain for player i on action a \in A_i be
\text{Gain}_i(\sigma,a) = \max \{0, u_i(a, \sigma_{-i}) - u_i(\sigma_{i}, \sigma_{-i})\}.\
The gain function represents the benefit a player gets by unilaterally changing his strategy. We now define g = (g_1,\ldots,g_N) where
g_i(\sigma)(a) = \sigma_i(a) + \text{Gain}_i(\sigma,a)\
for \sigma \in \Delta, a \in A_i. We see that
\sum_{a \in A_i} g_i(\sigma)(a) = \sum_{a \in A_i} \sigma_i(a) + \text{Gain}_i(\sigma,a)
= 1 + \sum_{a \in A_i} \text{Gain}_i(\sigma,a) > 0.\
We now use g to define f: \Delta \rightarrow \Delta as follows. Let

f_i(\sigma)(a) = \frac{g_i(\sigma)(a)}{\sum_{b \in A_i} g_i(\sigma)(b)}
for a \in A_i. It is easy to see that each f_i is a valid mixed strategy in \Delta_i. It is also easy to check that each f_i is a continuous function of \sigma, and hence f is a continuous function. Now \Delta is the cross product of a finite number of compact convex sets, and so we get that \Delta is also compact and convex. Therefore we may apply the Brouwer fixed point theorem to f. So f has a fixed point in \Delta, call it \sigma^*.
I claim that \sigma^* is a Nash Equilibrium in G. For this purpose, it suffices to show that

\forall 1 \leq i \leq N, ~ \forall a \in A_i, ~ \text{Gain}_i(\sigma^*,a) = 0 \text{.}
This simply states the each player gains no benefit by unilaterally changing his strategy which is exactly the necessary condition for being a Nash Equilibrium.
Now assume that the gains are not all zero. Therefore, \exists i, 1 \leq i \leq N, and a \in A_i such that \text{Gain}_i(\sigma^*, a) > 0. Note then that

\sum_{a \in A_i} g_i(\sigma^*, a) = 1 + \sum_{a \in A_i} Gain_i(\sigma^*,a) > 1.
So let C = \sum_{a \in A_i} g_i(\sigma^*, a). Also we shall denote \text{Gain}(i,\cdot) as the gain vector indexed by actions in A_i. Since f(\sigma^*) = \sigma^* we clearly have that f_i(\sigma^*) = \sigma^*_i. Therefore we see that

\sigma^*_i = \frac{g_i(\sigma^*)}{\sum_{a \in A_i} g_i(\sigma^*)(a)}
\Rightarrow
\sigma^*_i = \frac{\sigma^*_i + \text{Gain}_i(\sigma^*,\cdot)}{C}
\Rightarrow
C\sigma^*_i = \sigma^*_i + \text{Gain}_i(\sigma^*,\cdot)

\left(C-1\right)\sigma^*_i = \text{Gain}_i(\sigma^*,\cdot)
\Rightarrow
\sigma^*_i = \left(\frac{1}{C-1}\right)\text{Gain}_i(\sigma^*,\cdot).
Since C > 1 we have that \sigma^*_i is some positive scaling of the vector \text{Gain}_i(\sigma^*,\cdot). Now I claim that

\sigma^*_i(a)(u_i(a_i, \sigma^*_{-i}) - u_i(\sigma^*_i, \sigma^*_{-i}))
=
\sigma^*_i(a)\text{Gain}_i(\sigma^*, a)
\forall a \in A_i. To see this, we first note that if \text{Gain}_i(\sigma^*, a) > 0 then this is true by definition of the gain function. Now assume that \text{Gain}_i(\sigma^*, a) = 0. By our previous statements we have that

\sigma^*_i(a) = \left(\frac{1}{C-1}\right)\text{Gain}_i(\sigma^*, a) = 0
and so the left term is zero, giving us that the entire expression is 0 as needed.
So we finally have that

0 = u_i(\sigma^*_i, \sigma^*_{-i}) - u_i(\sigma^*_i, \sigma^*_{-i})


  =  \left(\sum_{a \in A_i} \sigma^*_i(a)u_i(a_i, \sigma^*_{-i})\right) - u_i(\sigma^*_i, \sigma^*_{-i})


  =  \sum_{a \in A_i} \sigma^*_i(a) (u_i(a_i, \sigma^*_{-i}) - u_i(\sigma^*_i, \sigma^*_{-i}))


  = \sum_{a \in A_i} \sigma^*_i(a) \text{Gain}_i(\sigma^*, a) \quad \text{ by the previous statements }


  = \sum_{a \in A_i} \left( C -1 \right) \sigma^*_i(a)^2 > 0
where the last inequality follows since \sigma^*_i is a non-zero vector. But this is a clear contradiction, so all the gains must indeed be zero. Therefore \sigma^* is a Nash Equilibrium for G as needed.

Computing Nash equilibria

If a player A has a dominant strategy s_A then there exists a Nash equilibrium in which A plays s_A. In the case of two players A and B, there exists a Nash equilibrium in which A plays s_A and B plays a best response to s_A. If s_A is a strictly dominant strategy, A plays s_A in all Nash equilibria. If both A and B have strictly dominant strategies, there exists a unique Nash equilibrium in which each plays his strictly dominant strategy.
In games with mixed strategy Nash equilibria, the probability of a player choosing any particular strategy can be computed by assigning a variable to each strategy that represents a fixed probability for choosing that strategy. In order for a player to be willing to randomize, his expected payoff for each strategy should be the same. In addition, the sum of the probabilities for each strategy of a particular player should be 1. This creates a system of equations from which the probabilities of choosing each strategy can be derived.[5]

Examples


Matching pennies
Player B plays HPlayer B plays T
Player A plays H−1, +1+1, −1
Player A plays T+1, −1−1, +1

In the matching pennies game, player A loses a point to B if A and B play the same strategy and wins a point from B if they play different strategies. To compute the mixed strategy Nash equilibrium, assign A the probability p of playing H and (1−p) of playing T, and assign B the probability q of playing H and (1−q) of playing T.
E[payoff for A playing H] = (−1)q + (+1)(1−q) = 1−2q
E[payoff for A playing T] = (+1)q + (−1)(1−q) = 2q−1
E[payoff for A playing H] = E[payoff for A playing T] ⇒ 1−2q = 2q−1 ⇒ q = 1/2
E[payoff for B playing H] = (+1)p + (−1)(1−p) = 2p−1
E[payoff for B playing T] = (−1)p + (+1)(1−p) = 1−2p
E[payoff for B playing H] = E[payoff for B playing T] ⇒ 2p−1 = 1−2p ⇒ p = 1/2
Thus a mixed strategy Nash equilibrium in this game is for each player to randomly choose H or T with equal probability.

See also




[edit] Notes

  1. ^ Osborne, Martin J., and Ariel Rubinstein. A Course in Game Theory. Cambridge, MA: MIT, 1994. Print.
  2. ^ Schelling, Thomas, The Strategy of Conflict, copyright 1960, 1980, Harvard University Press, ISBN 0-674-84031-3.
  3. ^ G. De Fraja, T. Oliveira, L. Zanchi (2010), 'Must Try Harder: Evaluating the Role of Effort in Educational Attainment'. "The Review of Economis and Statistics" 92:3. pp. 577-597
  4. ^ P. Chiappori, S. Levitt, and T. Groseclose (2002), 'Testing Mixed-Strategy Equilibria When Players Are Heterogeneous: The Case of Penalty Kicks in Soccer'. American Economic Review 92, pp. 1138-51.
  5. ^ a b von Ahn, Luis. "Preliminaries of Game Theory". http://www.scienceoftheweb.org/15-396/lectures_f11/lecture09.pdf. Retrieved 2008-11-07. 
  6. ^ a b B. D. Bernheim, B. Peleg, M. D. Whinston (1987), "Coalition-Proof Equilibria I. Concepts", Journal of Economic Theory 42 (1): 1–12, DOI:10.1016/0022-0531(87)90099-8. 
  7. ^ R. Aumann, (1959), Acceptable points in general cooperative n-person games in "Contributions to the Theory of Games IV", Princeton Univ. Press, Princeton, N.J.. 
  8. ^ D. Moreno, J. Wooders (1996), "Coalition-Proof Equilibrium", Games and Economic Behavior 17 (1): 80–112, DOI:10.1006/game.1996.0095. 
  9. ^ Nash proved that a perfect NE exists for this type of finite extensive form game[citation needed] – it can be represented as a strategy complying with his original conditions for a game with a NE. Such games may not have unique NE, but at least one of the many equilibrium strategies would be played by hypothetical players having perfect knowledge of all 10150 game trees[citation needed].
  10. ^ Fudenburg, Drew, and Jean Tirole. Game Theory: Massachusetts Institute of Technology, 1991.

[edit] References

[edit] Game theory textbooks

[edit] Original Nash papers

[edit] Other references

  • Mehlmann, A. The Game's Afoot! Game Theory in Myth and Paradox, American Mathematical Society (2000).
  • Nasar, Sylvia (1998), "A Beautiful Mind", Simon and Schuster, Inc.

[edit] External links



This page was last modified on 23 June 2012 at 17:55.



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