Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Jika diketahui force of mortality adalah \(\mu _x^{\left( d \right)} = \frac{3}{{4\left( {100 – x} \right)}}\) dan force of withdrawal adalah \(\mu _x^{\left( w \right)} = \frac{5}{{4\left( {100 – x} \right)}}\), hitunglah conditional density function untuk kematian seseorang pada umur \(70 + t\), jika orang tersebut hidup pada umur 70.
- \(\frac{{30 – t}}{{600}}\)
- \(\frac{{30 – t}}{{1200}}\)
- \(\frac{{70 – t}}{{600}}\)
- \(\frac{{70 – t}}{{1200}}\)
- \(\frac{{30 – t}}{{600 + t}}\)
| Diketahui | \(\mu _x^{\left( d \right)} = \frac{3}{{4\left( {100 – x} \right)}}\) dan \(\mu _x^{\left( w \right)} = \frac{5}{{4\left( {100 – x} \right)}}\) Kondisional jika orang tersebut hidup pada umur 70 |
| Rumus yang digunakan | \(\mu _x^{\left( \tau \right)} = \mu _x^{\left( d \right)} + \mu _x^{\left( w \right)}\) \({}_tp_x^{\left( \tau \right)} = \exp \left( { – \int\limits_0^t {\mu _x^{\left( \tau \right)}\left( s \right)ds} } \right)\) \(S\left( x \right) = {}_x{p_0}\) |
| Proses pengerjaan | \(\mu _x^{\left( \tau \right)} = \mu _x^{\left( d \right)} + \mu _x^{\left( w \right)}\) \(= \frac{3}{{4\left( {100 – x} \right)}} + \frac{5}{{4\left( {100 – x} \right)}}\) \(= \frac{8}{{4\left( {100 – x} \right)}}\) \({}_tp_x^{\left( \tau \right)} = \exp \left( { – \int\limits_0^t {\frac{8}{{4\left( {100 – s} \right)}}ds} } \right),\_{\rm{Misal\_}}u = 400 – 4s \to du = – 4ds\) \(= \exp \left( {8\int\limits_{400}^{4\left( {100 – t} \right)} {\frac{1}{{4u}}du} } \right)\) \(= \exp \left( {2\ln \left( {4\left( {100 – t} \right)} \right) – 2\ln \left( {400} \right)} \right)\) \(= \frac{{16{{\left( {100 – t} \right)}^2}}}{{{{400}^2}}}\) \(= \frac{{16{{\left( {100 – t} \right)}^2}}}{{160000}} = \frac{{{{\left( {100 – t} \right)}^2}}}{{10000}}\) \(f\left( {t,j} \right) = \frac{{{{\left( {100 – t} \right)}^2}}}{{10000}} \cdot \frac{3}{{4\left( {100 – t} \right)}}\) \(= \frac{{3\left( {100 – t} \right)}}{{40000}}\) \(Peluangnya = \frac{{f\left( {t,j} \right)}}{{S\left( x \right)}}\) \(= \frac{{{}_tp_{70}^{\left( \tau \right)} \cdot \mu _{70}^{\left( d \right)}}}{{S\left( {70} \right)}}\) \(= \frac{{\frac{{3\left( {100 – \left( {70 + t} \right)} \right)}}{{40000}}}}{{\frac{{{{\left( {100 – 70} \right)}^2}}}{{10000}}}}\) \(= \frac{{3\left( {30 – t} \right)}}{{40000}} \cdot \frac{{10000}}{{900}}\) \(= \frac{{30 – t}}{{1200}}\) |
| Jawaban | b. \(\frac{{30 – t}}{{1200}}\) |