Benefits are payable at the end of year of failure and premiums are payable at the beginning of the \\(m\\)thly period. In this case, the policy value at duration \\(k\\) is
\n\\begin{eqnarray*}
\n_k V&=& A_{x+k} – P_x^{(m)} \\ddot{a}_{x+k}^{(m)} .
\n\\end{eqnarray*}
\nLet us denote the policy value for a traditional whole life policy (with annual premiums) as
\n\\begin{eqnarray*}
\n~_k V_x &=& A_{x+k} – P_x \\ddot{a}_{x+k} = 1 – d\\ddot{a}_{x+k}- P_x
\n\\ddot{a}_{x+k} \\\\
\n&=& 1- \\frac{\\ddot{a}_{x+k}}{\\ddot{a}_x}.
\n\\end{eqnarray*}
\nThen, we may write the difference between these as
\n\\begin{eqnarray*}
\n_k V- ~_k V_x &=& P_x \\ddot{a}_{x+k} – P_x^{(m)}
\n\\ddot{a}_{x+k}^{(m)} \\\\
\n&=& \\frac{A_x}{\\ddot{a}_x^{(m)}} \\frac{\\ddot{a}_x^{(m)}}{\\ddot{a}_x}
\n\\ddot{a}_{x+k} – P_x^{(m)} \\ddot{a}_{x+k}^{(m)} \\\\
\n&=& P_x^{(m)} \\frac{\\alpha(m)\\ddot{a}_x – \\beta(m)}{\\ddot{a}_x}
\n\\ddot{a}_{x+k} – P_x^{(m)} (\\alpha(m)\\ddot{a}_{x+k}-\\beta(m)) \\\\
\n&=& P_x^{(m)} \\left( \\frac{ – \\beta(m)}{\\ddot{a}_x}
\n\\ddot{a}_{x+k} – (-\\beta(m)) \\right) \\\\
\n&=& P_x^{(m)}\\beta(m) \\left(
\n1- \\frac{\\ddot{a}_{x+k}}{\\ddot{a}_x} \\right) = P_x^{(m)}\\beta(m) ~_k V_x .\\\\
\n\\end{eqnarray*}
\nWe think about the \\(m\\)thly premium policy value as equal to the policy value for a traditional (annual premium) policy plus a reserve on an auxiliary policy to cover the premium loss in the year of death.<\/p>\n
Example. Whole Life Policy<\/strong> Solution<\/em>
\nYou are given \\(i=6\\%\\), \\(\\ddot{a}_{65} = 9.9,\\) and \\(\\ddot{a}_{70} = 8.8.\\) Assuming UDD, calculate the policy value for a whole life policy issued to a life (65) at duration \\(k=5\\), assuming that premiums are payable at the beginning of each month. <\/p>\n
\nThe policy value is
\n\\begin{eqnarray*}
\n_5 V&=& A_{70} – P_{65}^{(12)} \\ddot{a}_{70}^{(12)} .
\n\\end{eqnarray*}
\nThe whole life insurance net single premium is
\n\\begin{eqnarray*}
\nA_{70} = 1-d\\ddot{a}_{70} = 1- \\frac{0.06}{1.06} 8.8 = 0.501886792.
\n\\end{eqnarray*}
\nTo calculate the monthly annuities, we first need the interest factors.
\nWith \\(1.06 = (1+\\frac{i^{(12)}}{12})^{12},\\) we have \\(i^{(12)}= 0.058410607.\\) With \\(1-\\frac{0.06}{1.06}=1-d = (1-\\frac{d^{(12)}}{12})^{12},\\) we have \\(d^{(12)}= 0.058127667.\\) This yields
\n\\begin{eqnarray*}
\n\\alpha(12) = \\frac{id}{i^{(12)} d^{(12)}}=\\frac{0.06 \\frac{0.06}{1.06}}{(0.058410607) (0.058127667)} =1.000281005,
\n\\end{eqnarray*}
\nand
\n\\begin{eqnarray*}
\n\\beta(12) = \\frac{i-d}{i^{(12)} d^{(12)}}= \\frac{0.06-\\frac{0.06}{1.06}}{(0.058410607) (0.058127667)}=0.46811951.
\n\\end{eqnarray*}
\nThus,
\n\\begin{eqnarray*}
\n\\ddot{a}_{70}^{(12)} = \\alpha(12) \\ddot{a}_{70} – \\beta(12) = (1.000281005)8.8-0.46811951=8.334353338.
\n\\end{eqnarray*}
\nand
\n\\begin{eqnarray*}
\nP_{65}^{(12)} &=& \\frac{A_{65}}{\\ddot{a}_{65}^{(12)}} =
\n\\frac{1-d\\ddot{a}_{65}}
\n{ \\alpha(12) \\ddot{a}_{65} – \\beta(12)}\\\\
\n&=&
\n\\frac{1-\\frac{0.06}{1.06} 9.9}{
\n(1.000281005)9.9-0.46811951}=0.046596542.
\n\\end{eqnarray*}
\nWith this, the policy value is
\n\\begin{eqnarray*}
\n_5 V&=& A_{70} – P_{65}^{(12)} \\ddot{a}_{70}^{(12)}\\\\
\n& =& 0.501886792 – (0.046596542) 8.334353338= 0.11353475.
\n\\end{eqnarray*}<\/p>\n