{"id":218133,"date":"2025-05-23T09:05:21","date_gmt":"2025-05-23T09:05:21","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=218133"},"modified":"2025-05-23T09:05:23","modified_gmt":"2025-05-23T09:05:23","slug":"the-following-game-is-proposed-to-you","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/23\/the-following-game-is-proposed-to-you\/","title":{"rendered":"The following game is proposed to you."},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Certainty equivalent: U(C)=E[U(x)] -&gt; find C (certainty equivalent)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">4 Utility Theory. The following game is proposed to you. A fair coin will be tossed repeatedly until a tail appears (the first tail will end the game). The pot starts at \u00a31 and is doubled every time a head appears. You win whatever is in the pot when the game ends. Thus, you win \u00a31 if a tail appears on the first toss, \u00a32 if on the second, \u00a34 if on the third, \u00a38 if on the fourth, etc. In short, you win \u00a32-1 if the coin is tossed k times until the first tail appears. A reasonable way to analyse the value of this game is by calculating the certainty equivalent of the game&#8217;s random payoff. Calculate the expected utility and the certainty equivalent of the game in the following situations:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">a risk-neutral utility function: u(x) = x (hint: note that the probability of the first tail appearing in the k-th toss is )<br>b power utility function: u(x) = bx with b = -1;<br>c logarithmic utility function: u(x) = log2 (x) (hint: use the formula Dk-1 (k-1)\/2 = 1).<br>The three parts carry, respectively, 40%, 30%, and 30% of the marks.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>U(C) = E[U(X)]<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">where $X$ is the random payoff of the game.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Game description:<\/h2>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The game ends the first time a tail appears.<\/li>\n\n\n\n<li>Probability that the first tail appears on toss $k$ is:<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>P(k) = \\left(\\frac{1}{2}\\right)^k<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(because the first $k-1$ tosses are heads and the $k$-th toss is tail, probability $(1\/2)^{k-1} \\times (1\/2) = (1\/2)^k$).<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The payoff if the first tail appears on toss $k$ is:<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>X_k = 2^{k-1}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h1 class=\"wp-block-heading\">Part (a) Risk-neutral utility $u(x) = x$<\/h1>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate expected payoff (expected utility)<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>E[X] = \\sum_{k=1}^{\\infty} P(k) \\cdot X_k = \\sum_{k=1}^{\\infty} \\left(\\frac{1}{2}\\right)^k \\cdot 2^{k-1} = \\sum_{k=1}^{\\infty} \\frac{2^{k-1}}{2^k} = \\sum_{k=1}^\\infty \\frac{1}{2} = \\infty<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Since $\\frac{2^{k-1}}{2^k} = \\frac{1}{2}$, the infinite sum of $\\frac{1}{2}$ diverges to infinity.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Thus,<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>E[X] = \\infty<br>$$<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Certainty equivalent<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Since $u(x) = x$, $U(C) = C$. The certainty equivalent $C$ satisfies:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>C = E[X] = \\infty<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So the certainty equivalent is infinite under risk-neutral utility, meaning the player values the game infinitely high because the expected payoff diverges.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h1 class=\"wp-block-heading\">Part (b) Power utility $u(x) = x^b$ with $b = -1$<\/h1>\n\n\n\n<p class=\"wp-block-paragraph\">Here,<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>u(x) = x^{-1} = \\frac{1}{x}<br>$$<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Expected utility<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>E[U(X)] = \\sum_{k=1}^\\infty P(k) u(X_k) = \\sum_{k=1}^\\infty \\left(\\frac{1}{2}\\right)^k \\cdot \\frac{1}{2^{k-1}} = \\sum_{k=1}^\\infty \\frac{1}{2^k} \\cdot 2^{-(k-1)} = \\sum_{k=1}^\\infty \\frac{1}{2^k} \\cdot \\frac{1}{2^{k-1}} = \\sum_{k=1}^\\infty \\frac{1}{2^{2k-1}}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Simplify:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>E[U(X)] = \\sum_{k=1}^\\infty \\frac{1}{2^{2k-1}} = \\frac{1}{2} \\sum_{k=1}^\\infty \\left(\\frac{1}{4}\\right)^k = \\frac{1}{2} \\cdot \\frac{\\frac{1}{4}}{1 &#8211; \\frac{1}{4}} = \\frac{1}{2} \\cdot \\frac{\\frac{1}{4}}{\\frac{3}{4}} = \\frac{1}{2} \\cdot \\frac{1}{3} = \\frac{1}{6}<br>$$<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Find certainty equivalent $C$<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">We want $U(C) = E[U(X)]$, so<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>C^{-1} = \\frac{1}{6} \\implies C = 6<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h1 class=\"wp-block-heading\">Part (c) Logarithmic utility $u(x) = \\log_2(x)$<\/h1>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate expected utility<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>E[U(X)] = \\sum_{k=1}^\\infty P(k) \\cdot \\log_2(X_k) = \\sum_{k=1}^\\infty \\left(\\frac{1}{2}\\right)^k \\cdot \\log_2(2^{k-1}) = \\sum_{k=1}^\\infty \\left(\\frac{1}{2}\\right)^k (k-1)<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Use the hint that:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\sum_{k=1}^\\infty (k-1)\\left(\\frac{1}{2}\\right)^k = 1<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Therefore,<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>E[U(X)] = 1<br>$$<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Find certainty equivalent $C$<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>U(C) = \\log_2(C) = 1 \\implies C = 2^1 = 2<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h1 class=\"wp-block-heading\"><strong>Summary:<\/strong><\/h1>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th>Utility Function<\/th><th>Expected Utility $E[U(X)]$<\/th><th>Certainty Equivalent $C$<\/th><\/tr><\/thead><tbody><tr><td>Risk-neutral $u(x) = x$<\/td><td>$\\infty$<\/td><td>$\\infty$<\/td><\/tr><tr><td>Power $u(x) = x^{-1}$<\/td><td>$\\frac{1}{6}$<\/td><td>6<\/td><\/tr><tr><td>Logarithmic $u(x) = \\log_2(x)$<\/td><td>1<\/td><td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<h1 class=\"wp-block-heading\"><\/h1>\n\n\n\n<p class=\"wp-block-paragraph\">This problem explores the valuation of a St. Petersburg-type game under different utility functions, focusing on the certainty equivalent $C$. The game payoff doubles each time the coin lands heads before the first tail, leading to payoffs $X_k = 2^{k-1}$ with probabilities $P(k) = (1\/2)^k$.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Under a <strong>risk-neutral<\/strong> utility $u(x) = x$, the expected payoff is<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>E[X] = \\sum_{k=1}^\\infty \\left(\\frac{1}{2}\\right)^k 2^{k-1} = \\sum_{k=1}^\\infty \\frac{1}{2} = \\infty<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The expected value diverges because the doubling payoff outweighs the rapidly decreasing probability. Thus, the certainty equivalent, which satisfies $U(C) = E[U(X)]$, is infinite. This reflects the classical paradox where the expected value of the game is unbounded, and a risk-neutral player values the game infinitely.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">For the <strong>power utility<\/strong> function $u(x) = x^{-1}$, which represents risk aversion, the expected utility is finite:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>E[u(X)] = \\frac{1}{6}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Solving for $C$ yields $C = 6$. This finite certainty equivalent means the player values the game at a fixed amount (\u00a36) reflecting their aversion to the game&#8217;s large variance and infinite theoretical expectation.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Finally, with the <strong>logarithmic utility<\/strong> $u(x) = \\log_2(x)$, which models moderate risk aversion, the expected utility sums to 1. The certainty equivalent $C$ satisfies $\\log_2(C) = 1$, so $C=2$. This is lower than the power utility certainty equivalent, indicating more risk aversion: the player values the game less.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">In conclusion, the risk preference modeled by the utility function critically impacts the valuation of a stochastic payoff. Risk-neutral players focus on expected value and get infinite certainty equivalents, while risk-averse players discount the value, producing finite and more realistic certainty equivalents.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-154.png\" alt=\"\" class=\"wp-image-218134\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Certainty equivalent: U(C)=E[U(x)] -&gt; find C (certainty equivalent) 4 Utility Theory. The following game is proposed to you. A fair coin will be tossed repeatedly until a tail appears (the first tail will end the game). The pot starts at \u00a31 and is doubled every time a head appears. You win whatever is in the [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-218133","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218133","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=218133"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218133\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218133"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218133"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218133"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}