For a double-decrement model, you are given<\/p>\n
Determine the probability of eventually exiting due to cause 1.
\nSolution.<\/em>
\nThe probability of eventually exiting due to cause 1 may be \\expressed as
\n\\begin{eqnarray*}
\n_{\\infty} p_x^{01}=
\n\\int_0^{\\infty} ~ _s p_x^{00} \\mu_{x+s}^{01} ~ ds = \\frac{r_1}{c} \\int_0^{\\infty} ~ _s p_x^{00} s ~ ds .
\n\\end{eqnarray*}
\nThe probability of surviving is
\n\\begin{eqnarray*}
\n~ _t p_x^{00} &=& \\exp \\left\\{- \\int_0^t \\mu_{x+s}^{0\\bullet} ~ds \\right\\} \\\\
\n&=& \\exp \\left\\{- \\int_0^t
\n\\frac{r_1+r_2}{c} s ~ds \\right\\} =
\n\\exp \\left\\{-\\frac{r_1+r_2}{2c} t^2 \\right\\} .
\n\\end{eqnarray*}
\nWith this, the probability of eventually exiting due to cause 1 is
\n\\begin{eqnarray*}
\n_{\\infty} p_x^{01} &=& \\frac{r_1}{c} \\int_0^{\\infty} ~ \\exp \\left\\{-\\frac{r_1+r_2}{c} \\frac{s^2}{2} \\right\\} s ~ ds \\\\
\n&=& \\left. -\\frac{r_1}{c} \\frac{1}{\\frac{r_1+r_2}{c}}\\exp \\left\\{-\\frac{r_1+r_2}{c}
\n\\frac{t^2}{2} \\right\\} \\right|_0^{\\infty} = \\frac{r_1}{r_1+r_2} .
\n\\end{eqnarray*}<\/p>\n