How do the general multiple state probability properties reduce to the multiple decrement case?<\/p>\n
1. Just to reinforce the fundamentals, in a multiple decrement probability model, we have
\n\\begin{eqnarray*}
\n_t p_x^{j0}=0, ~ j=1,\\ldots,n \\text{ and } _t p_x^{jj}=1, ~ j=1,\\ldots,n.
\n\\end{eqnarray*}
\n2. For state (i=0) (alive or active), transition probabilities are equal to occupancy probabilities. That is,
\n\\begin{eqnarray*}
\n~ _t p_x^{00} = ~ _t p_x^{\\overline{00}} .
\n\\end{eqnarray*}
\nThis is because it is not possible to return to state 0 after having left it (all other states are absorbing).<\/p>\n
As a consequence, we may express
\n\\begin{eqnarray*}
\n~ _t p_x^{00} &=& \\exp \\left\\{- \\int_0^t \\sum_{j=1}^n \\mu_{x+s}^{0j} ~ds
\n\\right\\}
\n= \\exp \\left\\{- \\int_0^t mu_{x+s}^{0\\bullet} ~ds \\right\\},
\n\\end{eqnarray*}
\nwhere \\( \\mu_{x}^{0\\bullet}= \\sum_{j=1}^n \\mu_{x}^{0j}\\) is the total force of transition out of state 0 at age \\(x\\).<\/p>\n
3. The Chapman-Kolmogorov equations reduce to
\n\\begin{eqnarray*}
\n_{m+n} p_x^{0j} &=& \\sum_{k=0}^n ~ _m p_x^{0k} \\times ~_n p_{m+x}^{kj}
\n= ~ _m p_x^{00} \\times ~ _n p_{m+x}^{0j} + ~ _m p_x^{0j} .
\n\\end{eqnarray*}
\n4. Kolmogorov’s forward equation reduces to
\n\\begin{eqnarray*}
\n\\frac{\\partial}{\\partial t} ~ _t p_x^{0j}=
\n\\sum_{k=0, k ne j}^n
\n\\left(
\n_t p_x^{0k} \\mu_{x+t}^{kj}
\n– ~ _t p_x^{0j} \\mu_{x+t}^{jk}
\n\\right)
\n= ~ _t p_x^{00} \\mu_{x+t}^{0j}.
\n\\end{eqnarray*} Thus, we may write
\n\\begin{eqnarray*}
\n_t p_x^{0j}=
\n\\int_0^t ~ _s p_x^{00} \\mu_{x+s}^{0j} ~ ds .
\n\\end{eqnarray*}<\/p>\n