General Discrete Policy. For a general discrete policy, we have (known) premiums \\(P_h\\) payable at time \\(h\\) and benefits payable at time \\(b_h\\). As we have seen, the policy value can be expressed recursively as
\n\\begin{eqnarray*}
\n_h V &=& v q_{x+h} b_{h+1} – P_h + v p_{x+h} ~_{h+1} V.
\n\\end{eqnarray*}
\nNow, multiply each side by \\(v^h ~_h p_x\\) to get
\n\\begin{eqnarray*}
\nv^h ~_h p_x ~_h V – v^{h+1} ~_h p_x ~p_{x+h} ~_{h+1} V&=& v^{h+1} ~_h p_x q_{x+h} b_{h+1} – P_h v^h ~_h p_x
\n\\end{eqnarray*}
\nNote the relations \\(~_{h+1} p_x=~_h p_x ~ p_{x+h}\\) and \\( ~_{h|} q_x=~_h p_x ~ q_{x+h} \\). Sum each side of the equation over \\(h=0, \\ldots, k-1\\). On the left-hand side, we have
\n\\begin{eqnarray*}
\n\\sum_{h=0}^{k-1} \\left\\{v^h ~_h p_x ~_h V – v^{h+1} ~_{h+1} p_x ~_{h+1} V \\right\\} &=&
\nv^0 ~_0 p_x ~_0 V – v^k ~_k p_x ~_k V = – ~_k E_x ~_k V ,
\n\\end{eqnarray*}
\nrecalling the relation \\(~_k E_x = v^k ~_k p_x\\) and assuming that \\(_0 V=0\\). On the right-hand side, we have
\n\\begin{eqnarray*}
\n\\sum_{h=0}^{k-1} \\left\\{v^{h+1} ~_{h|} q_x b_{h+1} – P_h v^h ~_h p_x \\right\\} .
\n\\end{eqnarray*}
\nThus, we may write
\n\\begin{eqnarray*}
\nV_k &=& \\frac{\\sum_{h=0}^{k-1} P_h v^h ~_h p_x }{ ~_k E_x}-
\n\\frac{\\sum_{h=0}^{k-1} v^{h+1} ~_{h|} q_x b_{h+1} }{ ~_k E_x} \\\\
\n&=& \\textrm{Accumulated Value of Premium} – \\textrm{Accumulated Cost of Insurance} .
\n\\end{eqnarray*}<\/p>\n
Example<\/strong> Solution<\/em>
\nFor a fully discrete policy for \\(x\\), the first year benefit is 10,000 and the first year premium payable at the beginning of the year is 500. Calculate the policy value at duration 1, assuming \\(i=5\\%\\) and \\(q_x = 0.03\\).<\/p>\n
\nFrom the retrospective formula, we have
\n\\begin{eqnarray*}
\n_1 V &=& \\frac{\\sum_{h=0}^0 P_h v^h ~_h p_x }{ ~_1 E_x}-
\n\\frac{\\sum_{h=0}^0 v^{h+1} ~_{h|} q_x b_{h+1} }{ ~_1 E_x} =
\n\\frac{P_0}{ ~_1 E_x}-
\n\\frac{ v q_x b_1 }{ ~_1 E_x} \\\\
\n&=& \\frac{500}{ v p_x}-
\n\\frac{ v q_x (10000)}{ v p_x} = \\frac{500-10000\\frac{1}{1.05}(0.03)}{\\frac{1}{1.05} (1-0.03)} = 231.96 .
\n\\end{eqnarray*}<\/p>\n