{"id":411,"date":"2015-01-05T19:40:03","date_gmt":"2015-01-05T19:40:03","guid":{"rendered":"http:\/\/frees.pajarel.net\/?page_id=411"},"modified":"2015-02-20T18:57:33","modified_gmt":"2015-02-21T00:57:33","slug":"example-continued","status":"publish","type":"page","link":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/actuarial-mathematics\/interest-rate-risks-and-simulation\/3-diversifiable-risks\/example-whole-life-policy-values\/example-continued\/","title":{"rendered":"Example &#8211; Continued"},"content":{"rendered":"<p> Instead, suppose that beginning at time 1, the insurer faces one of three interest rate scenarios given by<br \/>\n\\begin{eqnarray*}<br \/>\n\\textbf{i} = \\left\\{<br \/>\n\\begin{array}{cr}<br \/>\ni_1 = 4  &#038; \\textrm{with probability 0.25} \\\\<br \/>\ni_2 = 5  &#038; \\textrm{with probability 0.50} \\\\<br \/>\ni_3 = 6  &#038; \\textrm{with probability 0.25} \\\\<br \/>\n\\end{array} \\right.<br \/>\n\\end{eqnarray*}<br \/>\nFor a fixed interest scenario \\(i_s\\), we have that the random future loss variable at policy duration \\(k\\) is<br \/>\n\\begin{eqnarray*}<br \/>\n_k L(i_s) = v_s^{K+1-k} &#8211; P_{40} \\ddot{a}_{\\overline{K+1-k|}} = \\left(1+\\frac{P_{40}}{1-v_s}\\right) v_s^{K+1-k} &#8211; \\frac{P_{40}}{1-v_s} ,<br \/>\n\\end{eqnarray*}<br \/>\nwhere \\(v_s=1\/(1+i_s)\\). This has conditional (on the interest scenario) mean<br \/>\n\\begin{eqnarray*}<br \/>\n\\textrm{E}(_k L(i_s)|i_s) = A_{40+k}@i_s &#8211; P_{40} \\ddot{a}_{40+k}@i_s,<br \/>\n\\end{eqnarray*} and variance<br \/>\n\\begin{eqnarray*}<br \/>\n\\textrm{Var}(_k L(i_s)|i_s) = \\left(1+\\frac{P_{40}}{1-v_s}\\right)^2 \\left( ^2A_{40+k}@i_s &#8211; A_{40+k}^2 @i_s\\right).<br \/>\n\\end{eqnarray*}<br \/>\nThe following table summarizes the calculations at duration \\(k=1\\) based on the Illustrative Life Table (recall \\(P_{40} = 0.125021\\) was set at contract initiation and so is constant across scenarios).<\/p>\n<table>\n<tr>\n<th><\/th>\n<th><\/th>\n<th><\/th>\n<th><\/th>\n<th>Conditional<\/th>\n<th><\/th>\n<th>Conditional<\/th>\n<\/tr>\n<tr>\n<td>\\(i\\)<\/td>\n<td>prob<\/td>\n<td>\\(A_{41}\\)<\/td>\n<td>\\(\\ddot{a}_{41}\\)<\/td>\n<td>Exp Loss<\/td>\n<td>\\(^2 A_{41}\\)<\/td>\n<td>Var Loss<\/td>\n<\/tr>\n<tr>\n<td>0.04<\/td>\n<td>0.25<\/td>\n<td>0.282<\/td>\n<td>18.658<\/td>\n<td>0.049<\/td>\n<td>0.105<\/td>\n<td>0.042<\/td>\n<\/tr>\n<tr>\n<td>0.05<\/td>\n<td>0.50<\/td>\n<td>0.216<\/td>\n<td>16.461<\/td>\n<td>0.010<\/td>\n<td>0.072<\/td>\n<td>0.034<\/td>\n<\/tr>\n<tr>\n<td>0.06<\/td>\n<td>0.25<\/td>\n<td>0.169<\/td>\n<td>14.686<\/td>\n<td>-0.015<\/td>\n<td>0.052<\/td>\n<td><\/td>\n<\/tr>\n<\/table>\n<p>For example, at \\(i=0.04\\), the conditional mean loss is (0.282 &#8211; 0.012502 (18.658) = 0.049). The conditional variance is<br \/>\n\\begin{eqnarray*}<br \/>\n\\left(1+\\frac{0.012502}{1-1\/(1.04)}\\right)^2 \\left( 0.105 &#8211; (282)^2\\right) = 0.042.<br \/>\n\\end{eqnarray*}<br \/>\nComputing quantities over interest scenarios, we may write the expected loss as<br \/>\n\\begin{eqnarray*}<br \/>\n_1 V = \\textrm{E}(_1 L(\\textbf{i})) &#038;=&#038; \\sum_{j=1}^3 \\textrm{E}(_1 L(i_s)|i_s) \\times Pr(\\textbf{i}=i_s) \\\\<br \/>\n&#038;=&#038; (0.25) (0.049) + (0.50) (0.010) + (0.25) (-0.015) = 0.014 .<br \/>\n\\end{eqnarray*}<br \/>\nIn this case, the expected fund is<br \/>\n\\begin{eqnarray*}<br \/>\n\\textrm{E}~FUND_1 &#038;=&#038; N P_{40}(1.05) &#8211; N q_{40} &#8211; N (1-q_{40}) ~_1 V \\\\<br \/>\n&#038;=&#038; N \\left\\{ (0.012502)(1.05) &#8211; 0.0027812 &#8211; (1-0.0027812) (0.014) \\right\\}\\\\<br \/>\n&#038;=&#038; -0.0033 N,<br \/>\n\\end{eqnarray*}<br \/>\na sad situation for the insurer.<\/p>\n<p><div class=\"alignleft\"><a href=\"https:\/\/users.ssc.wisc.edu\/~ewfrees\/actuarial-mathematics\/interest-rate-risks-and-simulation\/3-diversifiable-risks\/example-whole-life-policy-values\/\" title=\"Example. Whole Life Policy Values\">&#9668 Previous page<\/a><\/div><div class=\"alignright\"><a href=\"https:\/\/users.ssc.wisc.edu\/~ewfrees\/actuarial-mathematics\/interest-rate-risks-and-simulation\/3-diversifiable-risks\/example-whole-life-policy-values\/example-continued-2\/\" title=\"Example &#8211; Continued\">Next page &#9658<\/a><\/div><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Instead, suppose that beginning at time 1, the insurer faces one of three interest rate scenarios given by \\begin{eqnarray*} \\textbf{i} = \\left\\{ \\begin{array}{cr} i_1 = 4 &#038; \\textrm{with probability 0.25} \\\\ i_2 = 5 &#038; &hellip;<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":407,"menu_order":0,"comment_status":"closed","ping_status":"open","template":"","meta":{"jetpack_post_was_ever_published":false},"jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/P8cLPd-6D","acf":[],"_links":{"self":[{"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/411"}],"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=411"}],"version-history":[{"count":3,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/411\/revisions"}],"predecessor-version":[{"id":811,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/411\/revisions\/811"}],"up":[{"embeddable":true,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/407"}],"wp:attachment":[{"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/media?parent=411"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}