{"id":6260,"date":"2017-04-02T21:12:06","date_gmt":"2017-04-03T02:12:06","guid":{"rendered":"http:\/\/www.ssc.wisc.edu\/~jfrees\/?page_id=6260"},"modified":"2019-01-06T13:09:18","modified_gmt":"2019-01-06T19:09:18","slug":"aggregate-loss-guided-tutorials","status":"publish","type":"page","link":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/loss-data-analytics\/aggregate-loss-guided-tutorials\/","title":{"rendered":"Aggregate Loss 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<hr \/>\n<p><a id=\"displayText1\" href=\"javascript:toggleStrat('toggleText1','displayText1');\"><strong>Collective Risk Model: General considerations<\/strong><\/a><\/p>\n<div id=\"toggleText1\" style=\"display: none\">\n<hr \/>\n<p>Consider the <em>collective risk model<\/em>, which has the representation;<\/p>\n<p>$$S=X_1+&#8230;+X_N$$<\/p>\n<p>Where \\(S\\) the aggregate loss of random number \\(N\\) of individual claims \\(X_1,&#8230;,X_N\\). Here, we assume that conditional on \\(N=n\\) , \\(X_1,&#8230;,X_n\\) are \\(i.i.d\\) random variables. Also, the frequency  \\(N\\) and severity \\(Y\\) are independent<\/p>\n<p>Using the law of iterated expectations, the mean of  \\(S\\) is given by;<\/p>\n<p>$$\\begin{array}{ll}<br \/>\nE(S)<br \/>\n&#038;=E[E(S|N)]=E[NE(X)]=E(N)E(X) \\\\<br \/>\n&#038;=\\mu E(N)\\quad \\quad \\end{array}$$ <\/p>\n<p>Where \\(\\mu=E(X_i)\\).<\/p>\n<p>Using the law of total variation, the variance of \\(S\\) is given by;<br \/>\n$$\\begin{array}{ll}<br \/>\nVar(S)<br \/>\n&#038;=E[Var(S|N)]+Var[E(S|N)] \\\\<br \/>\n&#038;=E[N Var(X)]+Var[N E(X)] \\\\<br \/>\n&#038;=E(N)Var(X)+Var(N)E(X)^2 \\\\<br \/>\n&#038;=\\sigma^2E(N)+\\mu^2Var(N) \\quad \\quad \\end{array}$$ <\/p>\n<p>Where \\(Var(X_i)=\\sigma^2\\). <\/p>\n<p>Special case: Poisson distributed frequency. \\(N\\sim Poisson(\\lambda)\\)<br \/>\n$$\\begin{array}{ll}<br \/>\nE(N)&#038;=Var(N)=\\lambda \\\\<br \/>\nVar(S)&#038;=\\lambda(\\sigma^2+\\mu^2)=\\lambda E(X^2)<br \/>\n\\end{array}$$ <\/p>\n<hr \/>\n<\/div>\n<hr \/>\n<p><a id=\"displayText95\" href=\"javascript:toggleStrat('toggleText95','displayText95');\"><strong>Strategy for Aggregate Claim Distribution Problems<\/strong><\/a><\/p>\n<div id=\"toggleText95\" style=\"display: none\">\n<hr \/>\n<p>The <strong>pdf(pmf)<\/strong> of \\(S\\) is given by;<\/p>\n<p>$$f_S(s)=\\sum_{n=0}^{\\infty}\\Pr(N=n)\\cdot f_X^{*n}(s) \\quad \\quad (1)$$<\/p>\n<p>This means that to calculate \\(f_S(s)\\), we must sum up over all possibilities of \\(n\\) claims summing up to \\(s\\). The n-fold convolution of \\(f_X\\) is \\(f_X^{*n}\\) and \\(f_X^{*0}(0)=1\\). To obtain the <strong>cdf<\/strong> of \\(S\\), its given by;<\/p>\n<p>$$F_S(s)=\\sum_{i=0}^{s}f_S(i)\\quad \\quad (2)$$<\/p>\n<p>To apply these rules, <\/p>\n<ul>\n<li> From the question, identify the distribution of the claim frequency  \\(N\\) and severity \\(X\\).<\/li>\n<li> If the question requires you to calculate the<strong> pdf(pmf)<\/strong> of \\(S\\) ,\\(f_S(s)\\), its is given by equation (1) above. <\/li>\n<li> If the question requires you to calculate the <strong>cdf<\/strong>, \\(F_S(s)\\) it is given by equation (2).<\/li>\n<\/ul>\n<hr \/>\n<\/div>\n<div class=\"scbb-content-box scbb-content-box-gray\"><br \/>\n<strong>Aggregate Claim Distribution SOA #113<\/strong><br \/>\nThe number of claims , \\(N\\), made on an insurance portfolio follows the following distribution:<br \/>\n$$<br \/>\n{\\scriptsize<br \/>\n\\begin{matrix}<br \/>\n\\begin{array}{ccc}<br \/>\n    \\hline<br \/>\n    n &#038; \\Pr(N=n) \\\\<br \/>\n    \\hline<br \/>\n    0     &#038; 0.7 \\\\<br \/>\n    2     &#038; 0.2 \\\\<br \/>\n    3     &#038; 0.1 \\\\<br \/>\n    \\hline \\\\<br \/>\n  \\end{array}<br \/>\n\\end{matrix}<br \/>\n}<br \/>\n$$ If a claim occurs, the benefit is 0 or 10 with probability 0.8 and 0.2, respectively. The number of claims and the benefit for each claim are independent. \n<p>(a) Calculate the expected value of \\(S\\), \\(E(S)\\), and variance of \\(S\\) ,\\(Var(S)\\)<br \/>\n(b) Calculate the probability that aggregate benefits will exceed expected benefits by more than 2 standard deviations.<br \/>\n[WpProQuiz 82]<br \/>\n<\/p><\/div>\n\r\n\r\n<hr \/>\r\n\r\n\n<div class=\"scbb-content-box scbb-content-box-gray\"><br \/>\n<strong>Aggregate Claim Distribution SOA #95<\/strong><br \/>\nThe number of claims in a period has a geometric distribution with mean 4.The amount of each claim \\(X\\) follows \\(\\Pr(X=x)=0.25\\), \\(x=1,2,3,4\\). The number of claims and the claim amounts are independent. \\(S\\) is the aggregate claim amount in the period. \n<p>Calculate \\(F_S(3)\\)<\/p>\n<p>[WpProQuiz 81]<br \/>\n<\/p><\/div>\n<p><a id=\"displayText118\" href=\"javascript:toggleStrat('toggleText118','displayText118');\"><strong>Strategy for Aggregate Loss Distribution Approximation Problems<\/strong><\/a><\/p>\n<div id=\"toggleText118\" style=\"display: none\">\n<hr \/>\n<p>When the number of observed losses \\(n\\),is large, the observed aggregate losses are approximated normally distributed. To use normal approximation for aggregate loss distributions, \\(F_S(s)\\), we use the formula below:<br \/>\n$$F_S(s)=\\Pr(S \\leq s)=\\Phi\\left(\\frac{s-\\mu_s}{\\sigma_s}\\right) \\quad \\quad (1)$$<\/p>\n<p>Where;<br \/>\n$$\\mu_s=E(S) \\quad \\text{and} \\quad \\sigma_s=\\sqrt{Var(S)}$$<\/p>\n<p>When \\(n\\) is small then the distribution of the aggregate losses is positively skewed and hence its better to approximate aggregate losses with a distribution that is positively skewed (e.g LogNormal distribution ). To use LogNormal approximations:<br \/>\n$$F_S(s)=\\Pr(S \\leq s)=\\Pr(\\ln S \\leq \\ln s)=\\Phi\\left(\\frac{\\ln s-\\mu_s}{\\sigma_s}\\right)   \\quad \\quad (2)$$<\/p>\n<p>Where \\(u_s\\) and \\(\\sigma_s\\) are found using method of moments : <\/p>\n<p>$$<br \/>\n\\left\\{\\begin{array}{ll}<br \/>\n\\exp\\{\\mu_s+\\sigma_s^2\/2\\}=E(S)\\\\<br \/>\n\\exp\\{2\\mu_s+2\\sigma_s^2\\}=E(S^2)=E(S)^2+Var(S) \\end{array}\\right. $$<\/p>\n<p>To apply these rules, <\/p>\n<ul>\n<li> From the question, identify the distribution of the claim frequency  \\(N\\) and severity \\(X\\).<\/li>\n<li> Calculate  \\(E(S)\\) and \\(Var(S)\\) using the appropriate formulae <\/li>\n<li> Using equation (1) compute the aggregate loss distribution function \\(F_S(s)\\).<\/li>\n<li> When \\(n\\) is small,use equation (2) for a LogNormal approximation for the aggregate loss distribution function \\(F_S(s)\\) .<\/li>\n<\/ul>\n<hr \/>\n<\/div>\n<div class=\"scbb-content-box scbb-content-box-gray\"><br \/>\n<strong>Normal Approximation SOA #118<\/strong><br \/>\nFor an individual over 65:<br \/>\n(i)  The number of pharmacy claims is a Poisson random variable with mean 25.<br \/>\n(ii) The amount of each pharmacy claim is uniformly distributed between 5 and 95.<br \/>\n(iii) The amounts of the claims and the number of claims are mutually independent. \n<p>Determine the probability that aggregate claims for this individual will exceed 2000 using the normal approximation. <\/p>\n<p>[WpProQuiz 83]<br \/>\n<\/p><\/div>\n\r\n\r\n<hr \/>\r\n\r\n\n<p><a id=\"displayText99\" href=\"javascript:toggleStrat('toggleText99','displayText99');\"><strong>Strategy for Stop Loss Premium Problems<\/strong><\/a><\/p>\n<div id=\"toggleText99\" style=\"display: none\">\n<hr \/>\n<p>The stop loss premium is the expected aggregate losses <em>above<\/em> of aggregate deductible. Here, aggregate loss means the sum of individual claims payments after individual claim modifications such as policy limits and deductibles but before aggregate modifications. An aggregate deductible is a deductible applied to aggregate losses rather than individual losses. <\/p>\n<p>Assuming the frequency  \\(N\\) and severity \\(X\\) are independent, the mean of  aggregate losses,\\(S\\) , is given by;<br \/>\n$$E(S)=E(N)E(X) \\quad \\quad (1)$$<\/p>\n<p>With an aggregate deductible \\(d\\), the stop loss premium is the mean aggregate loss \\(E(S)\\) minus the expected value of the aggregate loss limited to \\(d\\) , \\(E(S\\wedge d)\\). The stop loss premium is given by :<br \/>\n$$E[(S-d)_+]=E(S)-E(S\\wedge d) \\quad \\quad (2)$$<\/p>\n<p>For discrete aggregate loss distributions:<br \/>\n$$E(S\\wedge d)=\\sum_{s\\lt d} s f_S(s)  + d (1-P(S\\lt d)) $$<\/p>\n<p>For continuous aggregate loss distributions:<br \/>\n$$E(S\\wedge d)=\\int_0^d s f_S(s) ds + \\int_d^{\\infty} d f_S(s) ds \\\\<br \/>\n= \\int^d_0 (1-F(s)) ds . $$<\/p>\n<p>To apply these rules, <\/p>\n<ul>\n<li> From the question, identify the distribution of the claim frequency  \\(N\\) and severity \\(X\\).<\/li>\n<li> Calculate the mean aggregate loss \\(E(S)\\) and the expected value of the aggregate loss limited to \\(d\\), \\(E(S\\wedge d)\\), using the appropriate formulae <\/li>\n<li> Using equation (2) compute the stop loss premium \\(E[(S-d)_+]\\) .<\/li>\n<\/ul>\n<hr \/>\n<\/div>\n<div class=\"scbb-content-box scbb-content-box-gray\"><br \/>\n<strong>Stop loss premium SOA #99<\/strong><br \/>\nFor a certain company, losses follow a Poisson frequency distribution with mean 2 per year, and the amount of a loss is 1, 2, or 3, each with probability 1\/3. Loss amounts are independent of the number of losses, and of each other.<br \/>\nAn insurance policy covers all losses in a year, subject to an annual aggregate deductible of 2. \n<p>Calculate the expected claim payments for this insurance policy. <\/p>\n<p>[WpProQuiz 84]<br \/>\n<\/p><\/div>\n\r\n\r\n<hr \/>\r\n\r\n\n<p><a id=\"displayText212\" href=\"javascript:toggleStrat('toggleText212','displayText212');\"><strong>Strategy Coverage Modifications Problems<\/strong><\/a><\/p>\n<div id=\"toggleText212\" style=\"display: none\">\n<hr \/>\n<p>Consider an insurance contract with deductible <em>d<\/em>. This causes the number of claims for amounts greater than zero to change. Let \\(N^L\\) be the number of losses and given \\(X_i\\), the ground-up losses for \\(i=1,&#8230;,n^L\\), denote the number of claims by \\(N^P\\) . We assume , given  \\(N^L=n^L\\) , \\(X_i\\) are \\(i.i.d\\) with common distribution \\(X\\). Then; <\/p>\n<p>$$N^P=I(X_1 \\gt d)+&#8230;+I(X_{N^L}\\gt d)$$<\/p>\n<p>It can be shown that for distributions of the \\((a,b,0)\\) class \\(N_L\\) and \\(N_P\\) follow the same distribution but with different parameters. Let \\(v=\\Pr(X \\gt d)\\), then we have: <\/p>\n<p>$$<br \/>\n{\\scriptsize<br \/>\n\\begin{matrix}<br \/>\n\\begin{array}{lll}<br \/>\n    \\hline<br \/>\n    \\text{Distribution} &#038; N^L &#038; N^P \\\\<br \/>\n    \\hline<br \/>\n     \\text{Poisson} &#038; \\lambda &#038; v\\lambda\\\\<br \/>\n    \\text{Binomial} &#038; n, p &#038; n, vp\\\\<br \/>\n     \\text{Negative binomial} &#038; r,\\beta &#038; r, v\\beta \\\\<br \/>\n    \\hline \\\\<br \/>\n  \\end{array}<br \/>\n\\end{matrix}<br \/>\n}<br \/>\n$$<\/p>\n<p>Also, recall that \\(X^L\\) , the random variable for the payment per loss with deductible \\(d\\) is given by :<br \/>\n$$X^L=<br \/>\n\\left\\{\\begin{array}{ll}<br \/>\n0 &#038;  X_i \\lt d \\\\<br \/>\nX_i-d &#038; X_i \\geq d \\end{array} \\right. $$<\/p>\n<p>And \\(X^P\\),the random variable for the per payment on an insurance with deductible \\(d\\) is the payment per loss variable conditioned on \\(X>d\\) or  <\/p>\n<p>$$X^P=<br \/>\n\\left\\{\\begin{array}{ll}<br \/>\nundefined &#038;  X_i \\lt d \\\\<br \/>\nX_i-d &#038; X_i \\geq d \\end{array} \\right. $$<\/p>\n<p>In modeling aggregate losses with per loss deductibles, one way is to model the number of losses \\(N^L\\) in which no modification is needed for the frequncy model and use the payment per loss variable ,\\(X^L\\). Another way is to use the modified frequency model \\(N^P\\) with the per payment random variable \\(X^P\\). Then, we can express the aggregate loss below: <\/p>\n<p>$$\\begin{array}{ll}<br \/>\nS<br \/>\n&#038;=X_1^L+&#8230;+X_{N^L}^L\\\\<br \/>\n&#038;=X_1^P+&#8230;+X_{N^P}^P<br \/>\n\\end{array}$$<\/p>\n<p>If \\(X\\) follows uniform, exponential or Pareto then using the per payment approach makes calculation easier.<\/p>\n<p>To apply these rules, <\/p>\n<ul>\n<li> From the question, identify coverage modification, the distribution of the claim frequency  \\(N\\) and severity \\(X\\).<\/li>\n<li> Model the aggregate claims using the per-loss apporach  or using the per-payment approach <\/li>\n<\/ul>\n<hr \/>\n<\/div>\n<div class=\"scbb-content-box scbb-content-box-gray\"><br \/>\n<strong>Coverage Modifications SOA #212<\/strong><br \/>\nFor an insurance:<br \/>\n(i) The number of losses per year has a Poisson distribution with \\(\\lambda= 10 \\)<br \/>\n(ii) Loss amounts are uniformly distributed on (0, 10).<br \/>\n(iii) Loss amounts and the number of losses are mutually independent.<br \/>\n(iv) There is an ordinary deductible of 4 per loss. \n<p>(a) Calculate the expected value of aggregate payments in a year.<br \/>\n(b) Calculate the variance of aggregate payments in a year.<br \/>\n[WpProQuiz 85]<br \/>\n<\/p><\/div>\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":12,"featured_media":0,"parent":5824,"menu_order":5,"comment_status":"closed","ping_status":"closed","template":"","meta":{"jetpack_post_was_ever_published":false},"jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/P8cLPd-1CY","acf":[],"_links":{"self":[{"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/6260"}],"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\/12"}],"replies":[{"embeddable":true,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/comments?post=6260"}],"version-history":[{"count":80,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/6260\/revisions"}],"predecessor-version":[{"id":6343,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/6260\/revisions\/6343"}],"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=6260"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}