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
begin{eqnarray*}
_k V&=& A_{x+k} – P_x^{(m)} ddot{a}_{x+k}^{(m)} .
end{eqnarray*}
Let us denote the policy value for a traditional whole life policy (with annual premiums) as
begin{eqnarray*}
~_k V_x &=& A_{x+k} – P_x ddot{a}_{x+k} = 1 – dddot{a}_{x+k}- P_x
ddot{a}_{x+k} \
&=& 1- frac{ddot{a}_{x+k}}{ddot{a}_x}.
end{eqnarray*}
Then, we may write the difference between these as
begin{eqnarray*}
_k V- ~_k V_x &=& P_x ddot{a}_{x+k} – P_x^{(m)}
ddot{a}_{x+k}^{(m)} \
&=& frac{A_x}{ddot{a}_x^{(m)}} frac{ddot{a}_x^{(m)}}{ddot{a}_x}
ddot{a}_{x+k} – P_x^{(m)} ddot{a}_{x+k}^{(m)} \
&=& P_x^{(m)} frac{alpha(m)ddot{a}_x – beta(m)}{ddot{a}_x}
ddot{a}_{x+k} – P_x^{(m)} (alpha(m)ddot{a}_{x+k}-beta(m)) \
&=& P_x^{(m)} left( frac{ – beta(m)}{ddot{a}_x}
ddot{a}_{x+k} – (-beta(m)) right) \
&=& P_x^{(m)}beta(m) left(
1- frac{ddot{a}_{x+k}}{ddot{a}_x} right) = P_x^{(m)}beta(m) ~_k V_x .\
end{eqnarray*}
We 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.
Example. Whole Life Policy
You 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.
Solution
The policy value is
begin{eqnarray*}
_5 V&=& A_{70} – P_{65}^{(12)} ddot{a}_{70}^{(12)} .
end{eqnarray*}
The whole life insurance net single premium is
begin{eqnarray*}
A_{70} = 1-dddot{a}_{70} = 1- frac{0.06}{1.06} 8.8 = 0.501886792.
end{eqnarray*}
To calculate the monthly annuities, we first need the interest factors.
With (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
begin{eqnarray*}
alpha(12) = frac{id}{i^{(12)} d^{(12)}}=frac{0.06 frac{0.06}{1.06}}{(0.058410607) (0.058127667)} =1.000281005,
end{eqnarray*}
and
begin{eqnarray*}
beta(12) = frac{i-d}{i^{(12)} d^{(12)}}= frac{0.06-frac{0.06}{1.06}}{(0.058410607) (0.058127667)}=0.46811951.
end{eqnarray*}
Thus,
begin{eqnarray*}
ddot{a}_{70}^{(12)} = alpha(12) ddot{a}_{70} – beta(12) = (1.000281005)8.8-0.46811951=8.334353338.
end{eqnarray*}
and
begin{eqnarray*}
P_{65}^{(12)} &=& frac{A_{65}}{ddot{a}_{65}^{(12)}} =
frac{1-dddot{a}_{65}}
{ alpha(12) ddot{a}_{65} – beta(12)}\
&=&
frac{1-frac{0.06}{1.06} 9.9}{
(1.000281005)9.9-0.46811951}=0.046596542.
end{eqnarray*}
With this, the policy value is
begin{eqnarray*}
_5 V&=& A_{70} – P_{65}^{(12)} ddot{a}_{70}^{(12)}\
& =& 0.501886792 – (0.046596542) 8.334353338= 0.11353475.
end{eqnarray*}