{"id":247,"date":"2014-12-27T22:23:30","date_gmt":"2014-12-27T22:23:30","guid":{"rendered":"http:\/\/frees.pajarel.net\/?page_id=247"},"modified":"2015-02-20T16:55:05","modified_gmt":"2015-02-20T22:55:05","slug":"common-shock-exercise","status":"publish","type":"page","link":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/actuarial-mathematics\/classic-joint-life-models\/3-special-case-common-shock-model-for-dependent-lives\/common-shock-exercise\/","title":{"rendered":"Common Shock Exercise"},"content":{"rendered":"<p>You have calculated the expected present value of a last survivor life insurance of 1 on \\(x\\) and \\(y\\). You assumed:<\/p>\n<ul>\n<li>The death benefit is payable immediately on the second death.<\/li>\n<li>The future lifetimes of \\(x\\) and \\(y\\) are independent, and each life has a constant force of mortality with \\(\\mu=0.06\\).<\/li>\n<li>\\(\\delta = 0.05\\).<\/li>\n<\/ul>\n<p>Your supervisor points out that these are not independent future lifetimes. Each mortality assumption is correct, but each includes a common shock component with constant force 0.02.<\/p>\n<p>Calculate the increase in the expected present value over what you originally calculated.<\/p>\n<div class=\"contingut_complert\">\n<div class=\"itm_simple_hide\">&#8211; Soln<\/div>\n<div class=\"itm_simple_show\">+ Soln<\/div>\n<div class=\"itm_simple_hidden\"><i>Solution.<\/i>We wish to calculate \\(\\bar{A}_{\\overline{xy}}^{CS} &#8211; \\bar{A}_{\\overline{xy}}^{IND}\\). Under the independence model, we have<br \/>\n\\begin{eqnarray*}<br \/>\n\\bar{A}_x &amp;=&amp; \\int_0^{\\infty} e^{-\\delta t} ~_t p_x \\mu_{x+t} dt =\\int_0^{\\infty} e^{-\\delta t} e^{-\\mu t} \\mu dt = \\frac{\\mu}{\\mu+\\delta} .<br \/>\n\\end{eqnarray*}<br \/>\nThus,<br \/>\n\\begin{eqnarray*}<br \/>\n\\bar{A}_x = \\frac{0.06}{0.06+0.05} = \\frac{6}{11} = \\bar{A}_y .<br \/>\n\\end{eqnarray*}<br \/>\nFurther,<br \/>\n\\begin{eqnarray*}<br \/>\n\\bar{A}_{xy} &amp;=&amp; \\int_0^{\\infty} e^{-\\delta t} ~_t p_{xy} \\mu_{xy}(t) dt =\\int_0^{\\infty} e^{-\\delta t} e^{-2\\mu t} 2\\mu dt = \\frac{2\\mu}{2\\mu+\\delta} ,<br \/>\n\\end{eqnarray*}<br \/>\nso that \\( \\bar{A}_{xy} = \\frac{0.12}{0.17} = \\frac{12}{17} \\) and<br \/>\n\\begin{eqnarray*}<br \/>\n\\bar{A}_{\\overline{xy}}^{IND} = \\bar{A}_x +\\bar{A}_y-\\bar{A}_{xy} = 2 \\frac{6}{11} -\\frac{12}{17} = 0.385 .<br \/>\n\\end{eqnarray*}<br \/>\nFor the common shock model, the original force of mortality was 0.06, so the new force of mortality is 0.04. Thus,<br \/>\n\\begin{eqnarray*}<br \/>\n\\bar{A}_x &amp;=&amp; \\int_0^{\\infty} e^{-\\delta t} ~_t p_x \\mu_{x+t} dt =\\int_0^{\\infty} e^{-0.05t} e^{-0.04 t} e^{-0.02 t}\\mu dt = \\frac{6}{11},<br \/>\n\\end{eqnarray*}<br \/>\nas before. The same is true for \\(\\bar{A}_y \\). The joint force of mortality is \\(\\mu_{xy}(t) = 0.04+0.04+0.02 = 0.10\\). Thus, the joint survival function is<br \/>\n\\(~_t p_{xy} e^{-0.10 t}\\). This yields<br \/>\n\\begin{eqnarray*}<br \/>\n\\bar{A}_{xy}^{CS} = \\frac{\\mu_{xy}}{\\mu_{xy}+\\delta} =\\frac{0.10}{0.10+0.05} = \\frac{2}{3},<br \/>\n\\end{eqnarray*}<br \/>\nand<br \/>\n\\begin{eqnarray*}<br \/>\n\\bar{A}_{\\overline{xy}}^{CS} = \\bar{A}_x +\\bar{A}_y-\\bar{A}_{xy}^{CS} = 2 \\frac{6}{11} -\\frac{2}{3} = 0.42423 .<br \/>\n\\end{eqnarray*}<br \/>\nThus, the increase is<br \/>\n\\begin{eqnarray*}<br \/>\n\\bar{A}_{\\overline{xy}}^{CS} &#8211; \\bar{A}_{\\overline{xy}}^{IND} = 0.42423 &#8211; 0.385 = 0.0392.<br \/>\n\\end{eqnarray*}<\/div>\n<\/div>\n<p><\/br><br \/>\n<\/br><br \/>\n<div class=\"alignleft\"><a href=\"https:\/\/users.ssc.wisc.edu\/~ewfrees\/actuarial-mathematics\/classic-joint-life-models\/3-special-case-common-shock-model-for-dependent-lives\/\" title=\"3. Special Case: &#8220;Common Shock&#8221; Model for Dependent Lives\">&#9668 Previous page<\/a><\/div><div class=\"alignright\"><a href=\"https:\/\/users.ssc.wisc.edu\/~ewfrees\/actuarial-mathematics\/policy-values\/\" title=\"Policy Values\">Next page &#9658<\/a><\/div><\/p>\n","protected":false},"excerpt":{"rendered":"<p>You have calculated the expected present value of a last survivor life insurance of 1 on \\(x\\) and \\(y\\). You assumed: The death benefit is payable immediately on the second death. The future lifetimes of &hellip;<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":113,"menu_order":1,"comment_status":"closed","ping_status":"open","template":"","meta":{"jetpack_post_was_ever_published":false},"jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/P8cLPd-3Z","acf":[],"_links":{"self":[{"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/247"}],"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=247"}],"version-history":[{"count":5,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/247\/revisions"}],"predecessor-version":[{"id":1548,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/247\/revisions\/1548"}],"up":[{"embeddable":true,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/113"}],"wp:attachment":[{"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/media?parent=247"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}