A third way to build a multi-decrement table from the associated single decrement tables (and vice-versa) is to assume uniform distribution of decrements (UDD) in each of the associated single decrement tables.
\nUnder this assumption, for integer age \\(x\\) and fraction \\(t\\) \\(( 0 \\leq t \\lt 1)\\), we have
\n\\begin{eqnarray*}
\n_t p_x^{\\prime(j)} = 1- ~_t q_x^{\\prime(j)} = 1- t \\times q_x^{\\prime(j)} .
\n\\end{eqnarray*}
\nNow, recall that \\( ~_t p_x^{\\prime(j)} = \\exp \\left\\{ – \\int_0^t \\mu_{x+t}^{(j)} ~ du
\n\\right\\} \\). Taking derivatives of each sides, it is easy to check that
\n\\begin{eqnarray*}
\n_t p_x^{\\prime(j)} \\mu_{x+t}^{(j)} = q_x^{\\prime(j)} .
\n\\end{eqnarray*}
\nThus,
\n\\begin{eqnarray*}
\n~ q_x^{(1)} &=& \\int_0^1 ~ _t p_x^{(\\tau)} \\mu_{x+t}^{(1)} ~dt ~=~ \\int_0^1 ~ \\left(\\prod_{j=1}^n ~_t p_x^{\\prime(j)} \\right) \\mu_{x+t}^{(1)} ~dt \\\\
\n&=& q_x^{\\prime(1)} \\int_0^1 ~ \\left(\\prod_{j=2}^n (1- t \\times q_x^{\\prime(j)}) \\right) ~dt .
\n\\end{eqnarray*}
\nSo, for example, with \\(n=2\\), we have
\n\\begin{eqnarray*}
\n~ q_x^{(1)} &=& q_x^{\\prime(1)} \\left(1- \\frac{1}{2} q_x^{\\prime(2)} \\right) .
\n\\end{eqnarray*}
\nand, by symmetry,\\(~ q_x^{(2)} = q_x^{\\prime(2)} \\left(1- \\frac{1}{2} q_x^{\\prime(1)} \\right) .\\) With \\(n=3\\), we have
\n\\begin{eqnarray*}
\n~ q_x^{(1)} &=& q_x^{\\prime(1)} \\left(1- \\frac{1}{2} (q_x^{\\prime(2)}+q_x^{\\prime(3)}) + \\frac{1}{3} q_x^{\\prime(2)} \\times
\nq_x^{\\prime(3)} \\right) ,
\n\\end{eqnarray*}
\nand similarly for \\(q_x^{(2)}\\) and \\(q_x^{(3)}\\).<\/p>\n