{"id":880,"date":"2015-01-17T14:21:42","date_gmt":"2015-01-17T20:21:42","guid":{"rendered":"http:\/\/www.ssc.wisc.edu\/~jfrees\/?page_id=880"},"modified":"2015-02-20T17:12:03","modified_gmt":"2015-02-20T23:12:03","slug":"2-exercise-fully-continuous-n-year-endowment-policy-eqnarray-not-working","status":"publish","type":"page","link":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/actuarial-mathematics\/policy-values\/2-exercise-fully-continuous-n-year-endowment-policy-eqnarray-not-working\/","title":{"rendered":"2. Exercise: Fully Continuous n-Year Endowment Policy"},"content":{"rendered":"

In this policy, a benefit of 1 is payable immediately upon failure. Premiums are payable at a constant annual rate of \\(P\\) per year that is determined at contract initiation by the equivalence principle. Assume a constant force of interest and no expenses. The policy valuation is at time \\(t < n\\).\n\n1. Express the (net) premium using standard actuarial notation.\n\nSolution.<\/em>
\nThe net premium is \\(P = \\frac{\\bar{A}_{x:\\overline{n}|}}{\\bar{a}_{x:\\overline{n}|}}\\).<\/p>\n

2. Express the loss random variable using standard interest theory symbols.<\/p>\n

Solution.<\/em>
\nIf \\(T< t\\), then the policy is complete and we do not consider it further. If \\(T \\ge n\\) (exceeding the endowment period), then the loss random variable is \\(L_t = v^{n-t} - P \\bar{a}_{\\overline{n-t}|}\\), where \\(P\\) is from part (1). If \\(t \\leq T < n\\), then the loss random variable is \\(L_t = v^{T-t} - P \\bar{a}_{\\overline{T-t}|}\\). We summarize this as\n\\begin{eqnarray*}\nL_t &=&\n\\left\\{\n\\begin{array}{cc}\nv^{T-t} - P \\ddot{a}_{\\overline{T-t}|} & t \\leq T < n \\\\\nv^{n-t} - P \\ddot{a}_{\\overline{n-t}|} & T \\ge n\n\\end{array}\n\\right .\n\\end{eqnarray*}\n3. Express the policy value using standard actuarial notation.\n\nSolution.<\/em>
\nRecall that the “\\(n-t\\)” single premium endowment issued to a life age \\(x+t\\) has loss
\n\\begin{eqnarray*}
\nL_t^{Insurance} &=&
\n\\left\\{
\n\\begin{array}{cc}
\nv^{T-t} & T < n \\\\\nv^{n-t} & T \\ge n\n\\end{array}\n\\right .\n\\end{eqnarray*}\nand expectation \\(\\bar{A}_{x+t:\\overline{n-t}|}\\). Similarly, an \"\\(n-t\\)\"temporary life annuity issued to a life age \\(x+t\\) has loss\n\\begin{eqnarray*}\nL_t^{Annuity} &=&\\left\\{\n\\begin{array}{cc}\n\\bar{a}_{\\overline{T-t}|} & T < n \\\\\n\\bar{a}_{\\overline{n-t}|} & T \\ge n \\end{array}\n\\right .\n\\end{eqnarray*}\nand expectation \\(\\bar{a}_{x+t:\\overline{n-t}|}\\). Thus, the loss random variable is just \\(L_t = L_t^{Insurance} -P L_t^{Annuity}\\) with expectation (policy value) \\(_t V =\\bar{A}_{x+t:\\overline{n-t}|}-P \\bar{a}_{x+t:\\overline{n-t}|}\\). \n\n4. Express the standard deviation of the loss random variable using standard actuarial notation.\n\nSolution.<\/em>
\nUsing the relation \\(\\bar{a}_{\\overline{n}|} = (1-v^{n})\/\\delta\\), we have
\n\\begin{eqnarray*}
\nL_t &=&
\n\\left\\{
\n\\begin{array}{cc}
\n\\left(1+\\frac{P}{\\delta}\\right) v^{T-t} – \\frac{P}{\\delta} & t < T < n \\\\\n\\left(1+\\frac{P}{\\delta}\\right) v^{n-t} - \\frac{P}{\\delta} & T \\ge n\n\\end{array}\n\\right .\n= \\left(1+\\frac{P}{\\delta}\\right) L_t^{Insurance} - \\frac{P}{\\delta} .\n\\end{eqnarray*}\nFrom this, we have\n\\begin{eqnarray*}\n\\textrm{Var}(L_t|T>t) &=& \\left(1+\\frac{P}{\\delta}\\right)^2 \\textrm{Var}(L_t^{Insurance}|T > t) \\\\
\n&=& \\left(1+\\frac{P}{\\delta}\\right)^2
\n\\left(~^2 \\bar{A}_{x+t:\\overline{n-t}|}- \\bar{A}_{x+t:\\overline{n-t}|}^2 \\right)
\n\\end{eqnarray*}<\/p>\n

5. Suppose that \\(x=35\\), \\(n=20\\), \\(i=6\\%\\), \\(t=5\\), and mortality follows deMoivre’s law with limiting age \\(w=100\\). Calculate the policy value and standard deviation.<\/p>\n

Solution.<\/em>
\nWe first compute the premium
\n\\begin{eqnarray*}
\nP &=& \\frac{\\bar{A}_{35:\\overline{20}|}}{\\bar{a}_{35:\\overline{20}|}} = \\frac{\\delta \\bar{A}_{35:\\overline{20}|}}{1- \\bar{A}_{35:\\overline{20}|}}
\n\\end{eqnarray*}
\nFor our problem, we have
\n\\begin{eqnarray*}
\n\\bar{A}_{35:\\overline{20}|} &=& \\int_0^{20} v^t ~_t p_{35} \\mu_{35+t} dt + v^{20} ~_{20} p_{35} \\\\
\n&=& \\frac{1}{65} \\bar{a}_{\\overline{20|}} + v^{20} \\frac{45}{65} = 0.39756 .
\n\\end{eqnarray*}
\nthat yields
\n\\begin{eqnarray*}
\nP &=&\\frac{\\ln(1.06) (0.39756)}{1- 0.39756} = 0.03845.
\n\\end{eqnarray*}
\nWe can express the policy value as
\n\\begin{eqnarray*}
\nV_5 =\\bar{A}_{40:\\overline{15}|}-P \\bar{a}_{40:\\overline{15}|}
\n\\end{eqnarray*}
\nAs before,
\n\\begin{eqnarray*}
\n\\bar{A}_{40:\\overline{15}|} &=& \\frac{1}{60} \\bar{a}_{\\overline{15|}} + v^{15} \\frac{45}{60} = 0.479628 .
\n\\end{eqnarray*}
\nand
\n\\begin{eqnarray*}
\n\\bar{a}_{40:\\overline{15}|} &=& \\frac{1-\\bar{A}_{\\overline{40:15}|}}{\\delta}= 8.930516 ,
\n\\end{eqnarray*}
\nso
\n\\begin{eqnarray*}
\nV_5 =0.479628-(0.03845 ) (8.930516) = 0.13625 .
\n\\end{eqnarray*}
\nFor the variance, we have
\n\\begin{eqnarray*}
\n\\textrm{Var}(L_5|T>5)
\n&=& \\left(1+\\frac{P}{\\delta}\\right)^2
\n\\left(~^2 \\bar{A}_{40:\\overline{15}|}- \\bar{A}_{40:\\overline{15}|}^2 \\right) \\\\
\n&=& \\left(1+\\frac{0.03845}{\\ln(1.06)}\\right)^2
\n\\left(0.24869- 0.479628^2 \\right) =0.051378,
\n\\end{eqnarray*}
\nusing \\(~^2 \\bar{A}_{40:\\overline{15}|}=\\frac{1}{60} ~^2 \\bar{a}_{\\overline{15|}} + v^{2\\times15} \\frac{45}{60}=0.24869\\). Thus, the standard deviation is \\(\\sqrt{0.051378}=0.22667\\). <\/p>\n

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In this policy, a benefit of 1 is payable immediately upon failure. Premiums are payable at a constant annual rate of \\(P\\) per year that is determined at contract initiation by the equivalence principle. Assume …<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":280,"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-ec","acf":[],"_links":{"self":[{"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/880"}],"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=880"}],"version-history":[{"count":8,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/880\/revisions"}],"predecessor-version":[{"id":1560,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/880\/revisions\/1560"}],"up":[{"embeddable":true,"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/pages\/280"}],"wp:attachment":[{"href":"https:\/\/users.ssc.wisc.edu\/~ewfrees\/wp-json\/wp\/v2\/media?parent=880"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}