{"id":6157,"date":"2017-02-25T18:08:53","date_gmt":"2017-02-26T00:08:53","guid":{"rendered":"http:\/\/www.ssc.wisc.edu\/~jfrees\/?page_id=6157"},"modified":"2019-01-06T13:09:18","modified_gmt":"2019-01-06T19:09:18","slug":"loss-data-analyticscredibility-guided-tutorials","status":"publish","type":"page","link":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/loss-data-analytics\/loss-data-analyticscredibility-guided-tutorials\/","title":{"rendered":"Credibility 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=\"displayText2\" href=\"javascript:toggleStrat('toggleText2','displayText2');\"><strong>Strategy for Limited Fluctuation Credibility Problems<\/strong><\/a><\/p>\n<div id=\"toggleText2\" style=\"display: none\">\n<hr \/>\n<p>Let \\(X_1,X_2&#8230;X_n\\) be <em>iid<\/em>. Actuaries use the phrase <em>full credibility<\/em> to mean that we can use sample mean \\(\\bar{X}\\) for pricing. To achieve this standard, it is common to require the sample size <em>n<\/em> to be large enough so that \\(\\bar{X}\\) is within 5% of the mean (\\(\\mu\\)) at least \\(90\\%\\) of the time. Using \\(r=0.05\\) and \\(p=0.9\\), then we require <em>n<\/em> large enough so that<br \/>\n$$\\Pr\\left((1-r)\\mu\\leq\\overline{X}\\leq(1+r)\\mu\\right)\\geq p .$$ To find the smallest value of <em>n<\/em>, replace the inequality by equality and use the central limit to solve for <em>n<\/em> as follows:<br \/>\n$$\\begin{array}{ll}<br \/>\np<br \/>\n&#038;=\\Pr\\left(\\frac{(1-r)\\mu-\\mu}{\\sigma\/\\sqrt{n}}\\leq \\frac{\\overline{X}-\\mu}{\\sigma\/\\sqrt{n}}\\leq \\frac{(1+r)\\mu-\\mu}{\\sigma\/\\sqrt{n}}\\right) \\\\<br \/>\n&#038;=\\Pr\\left(\\frac{-r\\mu}{\\sigma\/\\sqrt{n}}\\leq \\frac{\\overline{X}-\\mu}{\\sigma\/\\sqrt{n}}\\leq \\frac{r\\mu}{\\sigma\/\\sqrt{n}}\\right) \\\\<br \/>\n&#038; =\\Pr\\left(-y_p\\leq N(0,1) \\leq y_p\\right) .\\end{array}$$ With this, \\( y_p={\\Phi}^{-1}\\left((1+p)\/2\\right) \\) is a quantile from the standard normal distribution. For example, if <em>p<\/em> = 0.90, then  \\( y_{0.9}={\\Phi}^{-1}\\left(0.95\\right) = 1.645.\\) From the solution, you see that $y_p=\\frac{r\\mu}{\\sigma\/\\sqrt{n}}.$ Solving for <em>n<\/em> yields<br \/>\n$$ n_f \\geq \\left(\\frac{y_p \\sigma}{r\\mu}\\right)^2= \\left(\\frac{y_p}{r}\\right)^2 CV^2  \\quad \\quad &#8230; (1)$$<br \/>\nHere, \\(n_f\\) is the exposure needed for full credibility  and \\(CV = \\sigma\/\\mu\\) is the coefficient of variation. <\/p>\n<p><strong>Note:<\/strong> If you are establishing the standard for full credibility for <strong>pure premium<\/strong> or <strong>aggregate losses<\/strong>, the formulae for the expected number of claims needed for full credibility is :<br \/>\n$$E(N)\\times n_f    \\quad \\quad &#8230; (2)$$ Where \\(E(N)\\) is the expected number of claims and \\(n_f\\) is the number of exposures required for full credibility calculated from equation (1) above. <\/p>\n<p>When there is inadequate experience for full credibility , we calculate the credibility factor \\(Z\\) by<br \/>\n$$Z=<br \/>\n\\left\\{\\begin{array}{ll}<br \/>\n\\sqrt{\\frac{n}{n_f}} &#038; \\text{if} \\quad n\\leq n_f \\\\<br \/>\n1  &#038; \\text{otherwise}\\end{array} \\right. &#8230; (3)$$<br \/>\nWe use \\(Z\\) to calculate the credibility premium \\(P_c\\) given by;<br \/>\n$$P_c=Z\\overline{X} + (1-Z)M  \\quad \\quad &#8230; (4)$$ Here, \\(M\\) is the <em>manual premium<\/em> which is  generally part of the problem givens. <\/p>\n<p>To apply these rules, <\/p>\n<ul>\n<li> Using the question, identify the distribution of the claims, \\(p\\) and \\(r\\).<\/li>\n<li> Using equation (1) above calculate the standard for full credibility in terms of exposure. <\/li>\n<li> If the question requires you to establish the standard for full credibility for claim sizes(severity) in terms of exposure units, the exposure unit is a claim, so the the standard expressed in exposure units is the same as the standard for number of claims and it is given by equation (1). But if you are required to establish the standard for full credibility for aggregate losses in terms of expected number of claims, its is given by equation (2) above. <\/li>\n<li> If the question requires you to calculate the credibility factor, it is given by equation (3).<\/li>\n<li> If the question requires you to calculate the credibility premium, it is given by equation (4).<\/li>\n<\/ul>\n<hr \/>\n<\/div>\n<p><div class=\"scbb-content-box scbb-content-box-gray\"><br \/>\n<strong>Credibility-Poisson Frequency SOA #2<\/strong><br \/>\nYou are given:<br \/>\n(i) The number of claims has a Poisson distribution.<br \/>\n(ii) Claim sizes have a Pareto distribution with parameters \\(\\theta=0.5\\) and \\(\\alpha=6\\).<br \/>\n(iii) The number of claims and claim sizes are independent.<br \/>\n(iv) The number of claims in the first year was 1200.<br \/>\n(v) The aggregate loss in the first year was 6.75 million.<br \/>\n(vi) The manual premium for the first year was 5.00 million.<br \/>\n(vii) The exposure in the second year is identical to the exposure in the first year.<br \/>\n(vii) The observed pure premium should be within \\(2\\%\\) of the expected pure premium \\(90\\%\\) of the time. \n<p>(a) Calculate the expected number of claims needed for full credibility.<br \/>\n(b) Determine the limited fluctuation credibility net premium(in millions) for the second year. <\/p>\n<p>[WpProQuiz 73]<br \/>\n<\/p><\/div><br \/>\n\r\n\r\n<hr \/>\r\n\r\n<\/p>\n<div class=\"scbb-content-box scbb-content-box-gray\"><br \/>\n<strong>Credibility-Non Poisson Frequency SOA #65<\/strong><br \/>\nYou are given the following information about a general liability book of business comprised of 2500 insureds:<br \/>\n(i)  \\(X_i=\\sum_{j=1}^{N_i}Y_{ij}\\) is a random variable representing the annual loss of the \\(i\\)th insured.<br \/>\n(ii) \\(N_1,N_2,\u2026,N_{2500}\\) are independent and identically distributed random variables following a negative binomial distribution with parameters \\(r=2\\) and \\(\\beta=0.2\\).<br \/>\n(iii) \\(Y_{i1},Y_{i2},\u2026,Y_{iN_i}\\), are independent and identically distributed random variables following a Pareto distribution with \\(\\alpha=3\\) and \\(\\theta=1000\\).<br \/>\n(iv) The full credibility standard is to be within \\(5\\%\\) of the expected aggregate losses \\(90\\%\\) of the time.<br \/>\nUsing limited fluctuation credibility theory, calculate the partial credibility of the annual loss experience for this book of business. \n<hr \/>\n<p>[WpProQuiz 75]<br \/>\n<\/p><\/div>\n\r\n\r\n<hr \/>\r\n\r\n\n<p><a id=\"displayText219\" href=\"javascript:toggleStrat('toggleText219','displayText219');\"><strong>Strategy for B\u00fchlmann Credibility<\/strong><\/a><\/p>\n<div id=\"toggleText219\" style=\"display: none\">\n<hr \/>\n<p>The B\u00fchlmann credibility is sometimes known as <em>greatest accuracy<\/em> credibility. As before, \\(X_1,X_2, \\ldots, X_n\\) are from a policyholder. Let \\(\\theta\\) be a latent random variable associated with a policyholder and \\(\\{X_i|\\theta\\}\\) are <em>iid<\/em>, Then:<br \/>\n$$\\begin{array}{lll}<br \/>\n\\mu(\\theta) &#038; =E(X_i| \\theta) &#038;=\\text{&#8220;hypothetical mean&#8221;} \\\\<br \/>\nv(\\theta) &#038; =Var(X_i| \\theta) &#038;=\\text{&#8220;process variance&#8221;} \\end{array}$$<br \/>\nThe credibility premium is :<br \/>\n$$P_c=Z\\overline{X} + (1-Z)M  \\quad \\quad (1)$$As before, \\(M\\) is the manual premium which is often given to be \\(\\mu=E(X)\\). In this case, the credibility factor \\(Z\\) is<br \/>\n$$\\begin{array}{ll}<br \/>\nZ &#038; =\\frac{n}{n+k} &#038; &#038;  \\quad \\quad (2)\\\\<br \/>\n\\end{array}$$ where <em>k<\/em> is the ratio of &#8220;expectation of the process variance&#8221;  to the &#8220;variance of the hypothetical mean&#8221;<br \/>\n$$ k = \\frac{\\mathrm{E~} v(\\theta)}{\\mathrm{Var~}\\mu(\\theta)}<br \/>\n= \\frac{\\mathrm{E}\\left(\\mathrm{Var}(X_i| \\theta)\\right)}{\\mathrm{Var}\\left(\\mathrm{E}(X_i| \\theta)\\right)} \\quad \\quad (3)<br \/>\n$$ As before, <em>n<\/em> is the number of observations. <\/p>\n<p>To apply these rules, <\/p>\n<ul>\n<li> Use the problem to identify the conditional <em>model distribution<\/em> of each observation given \\(\\theta\\), \\(f_{X|\\theta}(x|\\theta)\\) and the <em>prior distribution<\/em> of \\(\\theta\\), \\(\\pi(\\theta)\\).<\/li>\n<li> Using the conditional <em>model distribution<\/em> \\(f_{X|\\theta}(x|\\theta)\\), find the hypothetical mean \\(E(X|\\theta)\\) and process variance \\(Var(X|\\theta)\\). <\/li>\n<li> Using the results above and the prior distribution \\(\\pi(\\theta)\\), Calculate \\(k\\) using equation (3) above.  <\/li>\n<li> To get credibility factor \\(Z\\) use equation (2) above.<\/li>\n<li> If the question requires you to calculate the credibility premium, it is given by equation (1).<\/li>\n<\/ul>\n<hr \/>\n<\/div>\n<p><div class=\"scbb-content-box scbb-content-box-gray\"><br \/>\n<strong>B\u00fchlmann Credibility SOA #219<\/strong>\n<p>For a portfolio of policies, you are given:<br \/>\n(i)   The annual claim amount on a policy has probability density function:<br \/>\n$$f(x|\\theta)=\\frac{2x}{{\\theta}^2}, \\quad 0\\lt x \\lt \\theta $$<br \/>\n(ii)  The prior distribution of \\(\\theta\\) has density function:<br \/>\n$$\\pi(\\theta)=4{\\theta}^3, \\quad \\quad 0\\lt\\theta\\lt 1 $$<br \/>\n(iii)  A randomly selected policy had claim amount 0.1 in Year 1. <\/p>\n<p>Calculate the B\u00fchlmann credibility estimate of the claim amount for the selected policy in Year 2 <\/p>\n<p>[WpProQuiz 76]<br \/>\n<\/p><\/div><br \/>\n\r\n\r\n<hr \/>\r\n\r\n<\/p>\n<div class=\"scbb-content-box scbb-content-box-gray\"><br \/>\n<strong>B\u00fchlmann Credibility- Aggregate losses SOA #8<\/strong><br \/>\nYou are given:<br \/>\n(i) Claim counts follow a Poisson distribution with mean \\(\\theta\\).\n<p>(ii) Claim sizes follow an exponential distribution with mean \\(10\\theta\\).<\/p>\n<p>(iii) Claim counts and claim sizes are independent, given \\(\\theta\\).<\/p>\n<p>(iv) The prior distribution has probability density function:<br \/>\n$$\\pi(\\theta)=\\frac{5}{{\\theta}^6}, \\quad\\quad \\theta\\gt 1$$<\/p>\n<p>Calculate B\u00fchlmann\u2019s \\(k\\) for aggregate losses.<\/p>\n<hr \/>\n<p>[WpProQuiz 77]<br \/>\n<\/p><\/div>\n\r\n\r\n<hr \/>\r\n\r\n\n<p><a id=\"displayText21\" href=\"javascript:toggleStrat('toggleText21','displayText21');\"><strong>Strategy for B\u00fchlmann-Straub Credibility<\/strong><\/a><\/p>\n<div id=\"toggleText21\" style=\"display: none\">\n<hr \/>\n<p>The B\u00fchlmann model assumes one exposure in every period but the B\u00fchlmann-Straub model generalizes B\u00fchlmann model to a case where there are $m_j$ exposures in period $j$. Then for the B\u00fchlmann-Straub model: <\/p>\n<p>$$<br \/>\nVar(X_j| \\theta)= \\frac{Var(X| \\theta)}{m_j} ;\\quad \\quad m_j=\\text{exposure} \\quad \\quad (1)<br \/>\n$$<\/p>\n<p>This is because the variance of the sample mean is the distribution variance divided by the number of observations. It follows that the credibility factor \\(Z\\) is given by ;<\/p>\n<p>$$\\begin{array}{ll}<br \/>\nZ &#038; =\\frac{\\sum m_j}{\\sum m_j+k} &#038; &#038;  \\quad \\quad (2)\\\\<br \/>\n\\end{array}$$<\/p>\n<p>When \\(m_j=1\\) it implies that \\(\\sum m_j =n\\) and the formulae is the same as that of the B\u00fchlmann model. As before <em>k<\/em> is the ratio of &#8220;expectation of the process variance&#8221;  to the &#8220;variance of the hypothetical mean&#8221;. <\/p>\n<p>$$ k = \\frac{\\mathrm{E~} v(\\theta)}{\\mathrm{Var~}\\mu(\\theta)}<br \/>\n= \\frac{\\mathrm{E}\\left(\\mathrm{Var}(X_i| \\theta)\\right)}{\\mathrm{Var}\\left(\\mathrm{E}(X_i| \\theta)\\right)} \\quad \\quad (3)<br \/>\n$$<br \/>\nand the <\/p>\n<p>To apply these rules, <\/p>\n<ul>\n<li> Use the problem to identify the conditional <em>model distribution<\/em> of each observation given \\(\\theta\\), \\(f_{X|\\theta}(x|\\theta)\\) and the <em>prior distribution<\/em> of \\(\\theta\\), \\(\\pi(\\theta)\\).<\/li>\n<li> Using the conditional <em>model distribution<\/em> \\(f_{X|\\theta}(x|\\theta)\\), find the hypothetical mean \\(E(X|\\theta)\\) and process variance \\(Var(X|\\theta)\\). <\/li>\n<li> Using the results above and the prior distribution \\(\\pi(\\theta)\\), Calculate \\(k\\) using equation (3) above.  <\/li>\n<li> To get credibility factor \\(Z\\) use equation (2) above.<\/li>\n<li> If the question requires you to calculate the credibility premium, it is the same as that for the B\u00fchlmann model.<\/li>\n<\/ul>\n<hr \/>\n<\/div>\n<p><div class=\"scbb-content-box scbb-content-box-gray\"><br \/>\n<strong>B\u00fchlmann-Straub Credibility SOA #21<\/strong>\n<p>You are given:<br \/>\n(i)  The number of claims incurred in a month by any insured has a Poisson distribution with mean \\(\\lambda\\)<br \/>\n(ii)  The claim frequencies of different insureds are independent.<br \/>\n(iii)  The prior distribution is gamma with probability density function: <\/p>\n<p>$$ f(\\lambda)=\\frac{(100\\lambda)^6 e^{-100\\lambda}}{120\\lambda}$$<\/p>\n<p>(iv)  <\/p>\n<p>$$<br \/>\n{\\scriptsize<br \/>\n\\begin{matrix}<br \/>\n\\begin{array}{ccc}<br \/>\n    \\hline<br \/>\n    \\text{Month} &#038; \\text{Number of Insureds}  &#038; \\text{Number of Claims} \\\\<br \/>\n    \\hline<br \/>\n    1     &#038; 100  &#038; 6 \\\\<br \/>\n    2     &#038; 150  &#038; 8 \\\\<br \/>\n    3     &#038; 200  &#038; 11 \\\\<br \/>\n    4     &#038; 300  &#038; ? \\\\<br \/>\n    \\hline \\\\<br \/>\n  \\end{array}<br \/>\n\\end{matrix}<br \/>\n}<br \/>\n$$<\/p>\n<p>Calculate the B\u00fchlmann-Straub credibility estimate of the number of claims in Month 4. <\/p>\n<p>[WpProQuiz 78]<br \/>\n<\/p><\/div><br \/>\n\r\n\r\n<hr \/>\r\n\r\n<\/p>\n<p><a id=\"displayText45\" href=\"javascript:toggleStrat('toggleText45','displayText45');\"><strong>Strategy for Solving Bayesian Premium Problems <\/strong><\/a> <\/p>\n<div id=\"toggleText45\" style=\"display: none\">\n<hr \/>\n<p>For Bayesian Premium problems:<\/p>\n<ul>\n<li> Use the problem to identify the conditional <em>model distribution<\/em> of each observation given \\(\\theta\\), \\(f_{X|\\theta}(x|\\theta)\\). Then get the likelihood of the model distribution, which is typically the product of the model distributions for all observations,<br \/>\n$$f_{X|\\theta}(x_1&#8230;x_n|\\theta)=\\prod_{i=1}^N f_{X|\\theta}(x_i|\\theta) .$$\n<\/li>\n<li> From the question identify the <em>prior distribution<\/em> of \\(\\theta\\), \\(\\pi(\\theta)\\).<\/li>\n<li> Now obtain the <em>posterior distribution<\/em> using the Bayes theorem;<br \/>\n$$\\pi(\\theta|x_1&#8230;x_n)=\\frac{f_{X|\\theta}(x_1&#8230;x_n|\\theta)\\pi(\\theta)}{\\int f_{X|\\theta}(x_1&#8230;x_n|\\theta)\\pi(\\theta) d\\theta}$$Note that the denominator of this expression \\( f(x) = \\int f_{X|\\theta}(x_1&#8230;x_n|\\theta)\\pi(\\theta) d\\theta\\) is the <em>marginal distribution<\/em> of \\(x\\) and does not involve \\( \\theta \\). <\/li>\n<li> With this, to calculate the <em>posterior probability<\/em> that \\(a \\lt \\theta \\lt b\\);<br \/>\n$$\\pi(a \\lt \\theta \\lt b|x_1&#8230;x_n)=\\int_{a}^{b} \\pi(\\theta|x_1&#8230;x_n) d\\theta$$\n<\/li>\n<li>For a new data \\(y\\) , the<em> predictive distribution<\/em> is<br \/>\n$$f(y|x_1&#8230;x_n)=\\int f(y|\\theta)\\pi(\\theta|x_1&#8230;x_n)d\\theta .$$Here, \\(f(y|\\theta)\\) is obtained from the conditional <em>model distribution<\/em>\n<\/li>\n<\/ul>\n<p>There are several ways that one can use the predictive distribution. For some problems, you may wish to calculate Bayesian prediction for a new data \\(y\\). Specifically, given the past observations \\(x_1&#8230;x_n\\), you can use<br \/>\n$$E(y|x_1&#8230;x_n)=\\int E(y|\\theta)\\pi(\\theta|x_1&#8230;x_n)d\\theta. $$Here, \\(E(y|\\theta)\\) is the expected value of the conditional model distribution.<\/p>\n<hr \/>\n<\/div>\n<p><div class=\"scbb-content-box scbb-content-box-gray\"><br \/>\n<strong>Bayesian Premium SOA #45<\/strong><br \/>\nYou are given:<br \/>\n(i)  The amount of a claim, \\(X\\), is uniformly distributed on the interval \\([0,\\theta]\\).<br \/>\n(ii) The prior density of \\(\\theta\\) is \\(\\pi(\\theta)=\\frac{500}{\\theta^2},  \\quad \\quad \\theta \\gt 500 \\). \n<p>Two claims, \\(x_1=400\\) and \\(x_2=600\\) are observed. You calculate the posterior distribution as: <\/p>\n<p>$$f(\\theta|x_1, x_2)=3\\left(\\frac{600^3}{\\theta^4}\\right), \\theta \\gt 600$$  <\/p>\n<p>Calculate the Bayesian premium , \\(E(X_3|x_1,x_2)\\). <\/p>\n<hr \/>\n<p>[WpProQuiz 79]<\/p><\/div><br \/>\n\r\n\r\n<hr \/>\r\n\r\n<br \/>\n<div class=\"scbb-content-box scbb-content-box-gray\"><br \/>\n<strong>Bayesian Estimation SOA #64\/11<\/strong><br \/>\nFor a group of insureds, you are given;<br \/>\n(i) The amount of a claim in uniformly distributed but will not exceed a certain unknown limit \\(\\theta\\).<br \/>\n(ii) The prior distribution of \\(\\theta\\) is  \\(\\pi(\\theta)=\\frac{500}{{\\theta}^2}, \\theta&gt;500\\).<br \/>\n(iii) Two claims of 400 and 600 are observed. \n<p>(a) Calculate the posterior probability that \\(700 \\lt \\theta \\lt 900\\) .<br \/>\n(b) Calculate the probability that the next claim will exceed 550.<br \/>\n(c) Calculate the Bayesian premium, that is, the expected value of the next claim \\(y\\) given two claims of 400 and 600 are observed, \\(E(y|x_1=400,x_2=600)\\).<\/p>\n<hr \/>\n<p>[WpProQuiz 80]<\/p><\/div><br \/>\n<hr \/><\/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":12,"featured_media":0,"parent":5824,"menu_order":4,"comment_status":"closed","ping_status":"closed","template":"","meta":{"jetpack_post_was_ever_published":false},"jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/P8cLPd-1Bj","acf":[],"_links":{"self":[{"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/6157"}],"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=6157"}],"version-history":[{"count":94,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/6157\/revisions"}],"predecessor-version":[{"id":6258,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/6157\/revisions\/6258"}],"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=6157"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}