One way to build a multi-decrement table from the associated single decrement tables (and vice-versa) is to constant forces of transition.
Under this assumption, transition forces are constant within the year,
begin{eqnarray*}
mu_{x+t}^{(j)} = mu_{x}^{(j)} , text{ for } 0 leq t leq 1 .
end{eqnarray*}
Thus,
begin{eqnarray*}
~ _t q_x^{prime(j)} = 1- exp left{ – t mu_{x}^{(j)}right} .
end{eqnarray*}
and, with (t rightarrow 1), we have (1- q_x^{prime(j)} = exp(-mu_{x}^{(j)})). Now
begin{eqnarray*}
~ q_x^{(j)} &=& int_0^1 ~ _t p_x^{(tau)} mu_{x+t}^{(j)}~dt
~=~ frac{mu_{x}^{(j)}}{mu_{x}^{(tau)}} int_0^1 ~ _t p_x^{(tau)} mu_{x+t}^{(tau)} ~dt
\
&=& frac{mu_{x}^{(j)}}{mu_{x}^{(tau)}} ~ q_x^{(tau)} ~=~ frac{ln (1- q_x^{prime(j)})}{ln (1- q_x^{(tau)})} ~ q_x^{(tau)} .
end{eqnarray*}
Thus, we have
begin{eqnarray*}
1- q_x^{prime(j)} &=& left( p_x^{(tau)}right)^{q_x^{(j)}/q_x^{(tau)}} .
end{eqnarray*}
From this, we can use multi-decrement probabilities to calculate the associated single decrement tables, and vice-versa.
Example
In a triple decrement table, suppose that (q_{65}^1 = 0.02), (q_{65}^2 = 0.02), and (q_{65}^3 = 0.04). Calculate (p_{65}^{01}).
Solution
First, (p_{65}^{00} = prod_{j=1}^3 ~ p_x^j = (1-0.02)(1-0.02)(1-0.04) = 0.921984) so that ( p_{65}^{0bullet}= 1-0.921984 = 0.078016).
Then, for (x=65)
begin{eqnarray*}
ln (1- q_{65}^1) = ln(0.98) &=& frac{p_{65}^{01}}{p_{65}^{0bullet}}ln left( p_{65}^{00}right) \
&=& frac{p_{65}^{03}}{0.078016}ln left( 0.921984right),
end{eqnarray*}
which implies that (p_{65}^{01} = 0.019404).