{"id":5462,"date":"2016-09-21T19:44:38","date_gmt":"2016-09-22T00:44:38","guid":{"rendered":"http:\/\/www.ssc.wisc.edu\/~jfrees\/?page_id=5462"},"modified":"2019-01-06T13:09:18","modified_gmt":"2019-01-06T19:09:18","slug":"loss-data-analytics-problems","status":"publish","type":"page","link":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/loss-data-analytics\/loss-data-analytics-problems\/","title":{"rendered":"Frequency Guided Tutorials"},"content":{"rendered":"\r\n<!--  Show Hide Stuff -->\r\n<script language=\"javascript\">function toggleStrat(id1,id2) {\r\n   var ele = document.getElementById(id1);\r\n   var text = document.getElementById(id2);\r\n   if (ele.style.display == \"block\") {\r\n      ele.style.display = \"none\";\r\n      text.innerHTML = \"Show Strategy\";\r\n   } else {\r\n      ele.style.display = \"block\";\r\n      text.innerHTML = \"Hide Strategy\";\r\n   }\r\n}       \r\n<\/script> \r\n\n<p>Here are a set of exercises that guide the viewer through some of the theoretical foundations of <em>Loss Data Analytics<\/em>. Each tutorial is based on one or more questions from the professional actuarial examinations &#8211; typically the Society of Actuaries Exam C.<\/p>\n<p><strong>Tutorial Structure.<\/strong> Each guided tutorial has a strategy set that describes the context. When you hit the &#8220;Start quiz&#8221; button, you begin the tutorial that is comprised of a series of mini-questions designed to lead you to the target question. At each stage, hints are provided as well as feedback on the correct solution of each mini-question.<\/p>\n<p><strong>Your Assignment.<\/strong> In reviewing these exercises, ideally the viewer will: <\/p>\n<ul>\n<li>Work the problem posed referring only to basic theory<\/li>\n<li>Even if you get the answer correct, review the strategy for this type of problem by clicking (revealing) the <font color=\"red\">Strategy for &#8230;<\/font>  header<\/li>\n<li>If you feel comfortable with the strategy and got the problem correct, then you may choose to move on. However, you might also decide to follow the step-by-step process for solving the problem by clicking on the &#8220;Start Quiz&#8221; button. It is not really a quiz \u2014 it is a guided tutorial.<\/li>\n<\/ul>\n<p><a id=\"displayText135\" href=\"javascript:toggleStrat('toggleText135','displayText135');\"><strong>Strategy for Solving Law of Iterated Expectations and<br \/>\nLaw of Total Variation Problems<\/strong><\/a><\/p>\n<div id=\"toggleText135\" style=\"display: none\">\n<hr \/>\n<p>In some situations, we only observe a single outcome but can conceptualize an outcome as resulting from a two (or more) stage process. These are called <strong>two-stage<\/strong>, or &#8220;<strong>hierarchical<\/strong>,&#8221; type situations. Some special cases include:<\/p>\n<ol>\n<li> problems where the parameters of the distribution are random variables,<\/li>\n<li> mixture problems, where stage 1 represents the type of subpopulation and stage 2 represents a random variable with a distribution that depends on population type <\/li>\n<li> an aggregate distribution, where stage 1 represents the number of events and stage two represents the amount per event.<\/li>\n<\/ol>\n<p>In these situations, the law of iterated expectations can be useful. The law of total variation is a special case that is particularly helpful for variance calculations.<\/p>\n<p>To apply these rules, <\/p>\n<ul>\n<li> Identify the random variable that is being conditioned upon, typically a stage 1 outcome (that is not observed).<\/li>\n<li> Conditional on the stage 1 outcome, calculate summary measures such as a mean, variance, and the like. <\/li>\n<li> There are several results of the step (ii), one for each stage 1 outcome. Then, combine these results using the iterated expectations or total variation rules.<\/li>\n<\/ul>\n<hr \/>\n<\/div>\n<p><div class=\"scbb-content-box scbb-content-box-gray\"><br \/>\n<strong>Iterated Expectations 135<\/strong><br \/>\nThe number of workplace injuries, \\(N\\), occurring in a factory on any given day is Poisson distributed with mean \\( \\lambda \\). The parameter \\( \\lambda \\) is a random variable that is determined by the level of activity in the factory, and is uniformly distributed on the interval [0, 3].\n<p>(a) Calculate E(<em>N<\/em>).<br \/>\n(b) Calculate Var(<em>N<\/em>).<br \/>\n[WpProQuiz 55]<br \/>\n<\/p><\/div><br \/>\n\r\n\r\n<hr \/>\r\n\r\n<\/p>\n<p><a id=\"displayText166\" href=\"javascript:toggleStrat('toggleText166','displayText166');\"><strong>Strategy for Solving &#40; (a,b,0) &#41; and &#40; (a,b,1) &#41; Problems<\/strong><\/a><\/p>\n<div id=\"toggleText166\" style=\"display: none\">\n<hr \/>\n<p>For  \\( (a,b,0) \\) problems:<\/p>\n<ul>\n<li> Examine the recursive expression and identify the distribution as \\( (a,b,0) \\) or \\( (a,b,1) \\)<\/li>\n<li> Based on the sign of \\( a \\), identify the distribution as either binomial ( \\(a \\lt 0\\) ), Poisson ( \\(a=0\\) ), or negative binomial ( \\(a \\gt 0\\) ). <\/li>\n<li> Using the values of \\( a \\) and \\( b \\), identify the parameters of the distribution<\/li>\n<li> With the entire parametric distribution, you can compute any quantity desired, such as the starting value \\(p_0\\), expectation, and so forth.<\/li>\n<li> For some problems, you may need to interchange these steps. For example, in question 166, you can use the starting value to help determine the parameters of the distribution<\/li>\n<\/ul>\n<p>For  \\( (a,b,1) \\) problems, the strategy is similar. Here, one needs extra information (typically, the starting value \\(p_0^M\\)) in order to determine the proportionality constant<br \/>\n$$ c=\\frac{1-p_0^M}{1-p_0^0}$$<br \/>\nYou can then use the relationship \\( p_0^M = c p_0^0 \\) to determine the modified distribution. <\/p>\n<hr \/>\n<\/div>\n<p><div class=\"scbb-content-box scbb-content-box-gray\"><br \/>\n<strong>\\( (a,b,0) \\) and \\( (a,b,1) \\) 166<\/strong><br \/>\nA discrete probability distribution has the following properties:<br \/>\n$$p_k = d\\left( 1+\\frac{1}{k}\\right) p_{k-1}  \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (1)$$ for \\( k= 1, 2, &#8230; \\).<br \/>\n(a) Assume that \\( p_0 = 0.5 \\). Calculate \\(d\\).<br \/>\n(b) Under the assumptions of part (a), calculate \\(p_3 \\).<br \/>\n(c) Assume that equation (1) holds for only \\( k= 2, 3,  &#8230; \\). Using the value of  \\(d\\) determined in part (a) and a modified assumed value \\( p_0^M = 0.3 \\), calculate a modified \\(p_3^M \\).\n<hr \/>\n<p>[WpProQuiz 54]<\/p><\/div><br \/>\n\r\n\r\n<hr \/>\r\n\r\n<br \/>\n<a id=\"displayText156\" href=\"javascript:toggleStrat('toggleText156','displayText156');\"><strong>Strategy for Obtaining Maximum Likelihood Estimators for Frequency Problems<\/strong><\/a><\/p>\n<div id=\"toggleText156\" style=\"display: none\">\n<hr \/>\n<ul>\n<li> Retrieve the general expression for the log-likelihood<\/li>\n<li> Using the problem information, identify the count variables<\/li>\n<li>Using the problem information, identify the probabilities<\/li>\n<li>Differentiate the log-likelihood<\/li>\n<li>Set the log-likelihood equal to zero and solve for the parameter estimate<\/li>\n<\/ul>\n<hr \/>\n<\/div>\n<p>\r\n\r\n\r\n\r\n<br \/>\n<div class=\"scbb-content-box scbb-content-box-gray\"><br \/>\n<strong>Poisson Likelihood 156<\/strong><br \/>\nYou are given:<br \/>\n(i) The number of claims follows a Poisson distribution with mean \\( \\lambda \\).<br \/>\n(ii) Observations other than 0 and 1 have been deleted from the data.<br \/>\n(iii) The data contain an equal number of observations of 0 and 1.<br \/>\nCalculate the maximum likelihood estimate of \\( \\lambda \\).\n<hr \/>\n<p>[WpProQuiz 53]<\/p><\/div><\/p>\n<hr \/>\n<p><a id=\"displayText90\" href=\"javascript:toggleStrat('toggleText90','displayText90');\"><strong>Strategy for Solving Mixture Problems<\/strong><\/a><\/p>\n<div id=\"toggleText90\" style=\"display: none\">\n<hr \/>\n<p>In mixture problems, we only observe a single outcome but can conceptualize an outcome as resulting from a two (or more) stage process. These are called <strong>two-stage<\/strong>, or &#8220;<strong>hierarchical<\/strong>,&#8221; type situations. Specifically, stage 1 represents the type of subpopulation and stage 2 represents a random variable with a distribution that depends on population type.<\/p>\n<p>In special instances such as Poisson-gamma, we know that a gamma mixture of Poisson variables has a negative binomial distribution. For other problems, one applies general rules from probability such as the law of iterated expectations, by using conditional probabilities through the hierarchical nature of the problem. To apply these rules, <\/p>\n<ul>\n<li> Identify the random variable that is being conditioned upon, typically a stage 1 outcome (that is not observed).<\/li>\n<li> Conditional on the stage 1 outcome, calculate summary measures such as a mean, variance, and the like. <\/li>\n<li> There are several results of the step (ii), one for each stage 1 outcome. Then, combine these results using, for example, the iterated expectations.<\/li>\n<\/ul>\n<hr \/>\n<\/div>\n<p><div class=\"scbb-content-box scbb-content-box-gray\"><br \/>\n<strong>Mixtures 90<\/strong><br \/>\nActuaries have modeled auto windshield claim frequencies. They have concluded that the number of windshield claims filed per year per driver follows the Poisson distribution with parameter \\( \\lambda \\), where \\( \\lambda \\) follows the gamma distribution with mean 3 and variance 3.<br \/>\nCalculate the probability that a driver selected at random will file no more than 1 windshield claim next year.<br \/>\n[WpProQuiz 56]<br \/>\n<\/div><br \/>\n\r\n\r\n<hr \/>\r\n\r\n<br \/>\n<a id=\"displayText71\" href=\"javascript:toggleStrat('toggleText71','displayText71');\"><strong>Strategy for Model Comparison\/Selection Problems<\/strong><\/a><\/p>\n<div id=\"toggleText71\" style=\"display: none\">\n<hr \/>\n<p>For model comparison\/selection problems, consider the following outline of steps:<\/p>\n<ol>\n<li> Identify the probabilities of outcomes under a model. For some problems, you will need to compute a maximum likelihood (or other type of) estimator to get these probabilities<\/li>\n<li> From the outcome probabilities, determine how many counts one expects for each outcome under the model <\/li>\n<li> Compare the actual and expected outcomes by computing a chi-square goodness of fit statistics<\/li>\n<li> For some problems, you are asked to compare the goodness of fit statistic to a chi-square reference distribution. For this distribution, the parameter is known as a degree of freedom, say, <em>df<\/em>. Use <em>df<\/em> = the number of cells minus 1 minus the number of estimated parameters. (If there are no estimated parameters, just use <em>df<\/em> = the number of cells minus 1 )<\/li>\n<\/ol>\n<hr \/>\n<\/div>\n<p>\r\n\r\n\r\n\r\n<br \/>\n<div class=\"scbb-content-box scbb-content-box-gray\"><br \/>\n<strong>Model Comparison 71<\/strong><br \/>\nYou are investigating insurance fraud that manifests itself through claimants who file claims with respect to auto accidents with which they were not involved. Your evidence consists of a distribution of the observed number of claimants per accident and a standard distribution for accidents on which fraud is known to be absent. The two distributions are summarized below:\n<p>$$<br \/>\n{\\scriptsize<br \/>\n\\begin{matrix}<br \/>\n\\begin{array}{ccc}<br \/>\n    \\hline<br \/>\n    \\text{Number of Claimants} &#038; \\text{Standard}  &#038; \\text{Observed Number} \\\\<br \/>\n   \\text{ per Accident} &#038; \\text{Probability}  &#038;  \\text{of Accidents}  \\\\<br \/>\n    \\hline<br \/>\n    1     &#038; 0.25  &#038; 235 \\\\<br \/>\n    2     &#038; 0.35  &#038; 335 \\\\<br \/>\n    3     &#038; 0.24  &#038; 250 \\\\<br \/>\n    4     &#038; 0.11  &#038; 111 \\\\<br \/>\n    5     &#038; 0.04  &#038; 47 \\\\<br \/>\n    6+    &#038; 0.01  &#038; 22 \\\\<br \/>\n    \\hline<br \/>\n    Total &#038; 1.00  &#038; 1000 \\\\<br \/>\n    \\hline \\\\<br \/>\n  \\end{array}<br \/>\n\\end{matrix}<br \/>\n}<br \/>\n$$<br \/>\n(a) Determine the result of a chi-square goodness of fit statistic.<br \/>\n(b) Determine the result of a chi-square test of the null hypothesis that there is no fraud in the observed accidents.<\/p>\n<hr \/>\n<p>[WpProQuiz 57]<\/p><\/div><\/p>\n<p><div class=\"alignleft\"><a href=\"https:\/\/users.ssc.wisc.edu\/~ewfrees\/loss-data-analytics\/introduction-to-loss-data-analytics\/1-5-further-reading-and-references\/\" title=\"1.5 Further Reading and References\">&#9668 Previous page<\/a><\/div><div class=\"alignright\"><a href=\"https:\/\/users.ssc.wisc.edu\/~ewfrees\/loss-data-analytics\/chapter-3-modeling-loss-severity\/\" title=\"Chapter 3. Modeling Loss Severity\">Next page &#9658<\/a><\/div><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Here are a set of exercises that guide the viewer through some of the theoretical foundations of Loss Data Analytics. Each tutorial is based on one or more questions from the professional actuarial examinations &#8211; &hellip;<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":5824,"menu_order":1,"comment_status":"closed","ping_status":"closed","template":"","meta":{"jetpack_post_was_ever_published":false},"jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/P8cLPd-1q6","acf":[],"_links":{"self":[{"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/5462"}],"collection":[{"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/comments?post=5462"}],"version-history":[{"count":61,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/5462\/revisions"}],"predecessor-version":[{"id":6044,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/5462\/revisions\/6044"}],"up":[{"embeddable":true,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/5824"}],"wp:attachment":[{"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/media?parent=5462"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}