{"id":97,"date":"2014-12-26T13:45:05","date_gmt":"2014-12-26T13:45:05","guid":{"rendered":"http:\/\/frees.pajarel.net\/?page_id=97"},"modified":"2015-02-20T16:12:22","modified_gmt":"2015-02-20T22:12:22","slug":"special-case-premiums-for-dependent-lives","status":"publish","type":"page","link":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/actuarial-mathematics\/classic-joint-life-models\/2-joint-life-and-last-survivor-annuities-and-insurances-continuous\/special-case-premiums-for-dependent-lives\/","title":{"rendered":"Special Case: Premiums For Dependent Lives"},"content":{"rendered":"

Consider a fully continuous life insurance for \\(x\\) and \\(y\\). Suppose that
\n1) premiums are payable until the first death
\n2) a benefit of 1 is payable at the second death.<\/p>\n

The joint mortality for \\(T(x)\\) and \\(T(y)\\) is not independent but given by
\n\\begin{eqnarray*}
\n~_t p_{xy} = r ~_t p_x + (1-r) ~_t p_x ~_t p_y.
\n\\end{eqnarray*}
\nFor simplicity, assume that the mortality for \\(T(x)\\) and \\(T(y)\\) follow a common exponential distribution,
\n\\begin{eqnarray*}
\n~_t p_x = e^{-\\mu t} = ~_t p_y .
\n\\end{eqnarray*}
\nShow that the annual premium is \\(\\frac{\\mu(2\\mu+\\delta r)}{\\delta + (1+r)\\mu}.\\)<\/p>\n

\n
– Soln<\/div>\n
+ Soln<\/div>\n
\nSolution. <\/em>The premium (P) is the solution of \\(\\bar{A}_{\\overline{xy}} = P \\bar{a}_{xy}\\). Now
\n\\begin{eqnarray*}
\n\\bar{a}_{xy} &=& \\int_0^{\\infty} e^{-\\delta t} ~_t p_{xy} ~ dt \\\\
\n&=& \\int_0^{\\infty} e^{-\\delta t} \\left( r ~_t p_x + (1-r) ~_t p_x ~_t p_y \\right) ~ dt \\\\
\n&=& \\int_0^{\\infty} e^{-\\delta t} \\left( r e^{-\\mu t} + (1-r) e^{-2\\mu t} \\right) ~ dt \\\\
\n&=& \\frac{r}{\\mu+\\delta} + \\frac{1-r}{2\\mu+\\delta} = \\frac{\\delta+(1+r)\\mu}{(\\mu+\\delta)(2\\mu+\\delta)}
\n\\end{eqnarray*}
\nFurther
\n\\begin{eqnarray*}
\n\\bar{A}_x &=& \\int_0^{\\infty} e^{-\\delta t} ~_t p_{xy} \\mu_{x+t}~ dt \\\\
\n&=& \\mu \\int_0^{\\infty} e^{-\\delta t} e^{-\\mu t} ~ dt = \\frac{\\mu}{\\mu+\\delta} = \\bar{A}_y
\n\\end{eqnarray*}and
\n\\begin{eqnarray*}
\n\\bar{A}_{\\overline{xy}} = \\bar{A}_x + \\bar{A}_y – \\bar{A}_{xy} =2 \\frac{\\mu}{\\mu+\\delta} –
\n\\left( 1 – \\delta \\bar{a}_{xy} \\right) .
\n\\end{eqnarray*}
\nThus
\n\\begin{eqnarray*}
\nP &=& \\frac{\\bar{A}_{\\overline{xy}}}{\\bar{a}_{xy}} = \\frac{2 \\frac{\\mu}{\\mu+\\delta} -1}{\\bar{a}_{xy}} + \\delta \\\\
\n&=& \\frac{\\mu(2\\mu+\\delta r)}{\\delta + (1+r)\\mu}.
\n\\end{eqnarray*}\n<\/div>\n<\/div>\n

<\/br><\/br><\/p>\n

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Consider a fully continuous life insurance for \\(x\\) and \\(y\\). Suppose that 1) premiums are payable until the first death 2) a benefit of 1 is payable at the second death. The joint mortality for …<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":90,"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-1z","acf":[],"_links":{"self":[{"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/97"}],"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=97"}],"version-history":[{"count":5,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/97\/revisions"}],"predecessor-version":[{"id":1525,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/97\/revisions\/1525"}],"up":[{"embeddable":true,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/90"}],"wp:attachment":[{"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/media?parent=97"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}