Pembahasan Ujian PAI: A50 - No. 4 - November 2016

Pembahasan Soal Ujian Profesi Aktuaris

Institusi	:	Persatuan Aktuaris Indonesia (PAI)
Mata Ujian	:	Metoda Statistika
Periode Ujian	:	November 2016
Nomor Soal	:	4

SOAL

Jika diketahui force of mortality adalah \(\mu _x^{\left( d \right)} = \frac{4}{{5\left( {100 – x} \right)}}\) dan force of withdrawal adalah \(\mu _x^{\left( w \right)} = \frac{6}{{5\left( {100 – x} \right)}}\), hitunglah conditional density function untuk kematian seseorang pada umur \(70 + t\), jika orang tersebut hidup pada umur 70.

\(\frac{{30 – t}}{{600}}\)
\(\frac{{30 – t}}{{1200}}\)
\(\frac{{30 – t}}{{1125}}\)
\(\frac{{70 – t}}{{1200}}\)
\(\frac{{70 – t}}{{1800}}\)

[showhide type more_text=”Kunci Jawaban & Pembahasan” less_text=”Sembunyikan Kunci Jawaban & Pembahasan”]

Diketahui	\(\mu _x^{\left( d \right)} = \frac{4}{{5\left( {100 – x} \right)}}\) \(\mu _x^{\left( w \right)} = \frac{6}{{5\left( {100 – x} \right)}}\) 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	\(\mu _x^{\left( \tau \right)} = \mu _x^{\left( d \right)} + \mu _x^{\left( w \right)}\) \(\mu _x^{\left( \tau \right)} = \frac{4}{{5\left( {100 – x} \right)}} + \frac{6}{{5\left( {100 – x} \right)}}\) \(\mu _x^{\left( \tau \right)} = \frac{{10}}{{5\left( {100 – x} \right)}}\)\({}_tp_x^{\left( \tau \right)} = \exp \left( { – \int\limits_0^t {\frac{2}{{\left( {100 – s} \right)}}ds} } \right),{\rm{ Misal }}u = 100 – s \to du = – ds\) \({}_tp_x^{\left( \tau \right)} = \exp \left( {2\int\limits_{100}^{100 – t} {\frac{1}{u}du} } \right)\) \({}_tp_x^{\left( \tau \right)} = \exp \left( {2\ln \left( {100 – t} \right) – 2\ln \left( {100} \right)} \right)\) \({}_tp_x^{\left( \tau \right)} = \frac{{{{\left( {100 – t} \right)}^2}}}{{{{100}^2}}}\) \(f\left( {t,j} \right) = \frac{{{{\left( {100 – t} \right)}^2}}}{{{{100}^2}}} \cdot \frac{4}{{5\left( {100 – t} \right)}}\) \(f\left( {t,j} \right) = \frac{{4\left( {100 – t} \right)}}{{5 \cdot {{100}^2}}}\) \(Peluangnya = \frac{{f\left( {t,j} \right)}}{{S\left( x \right)}}\) \(Peluangnya = \frac{{{}_tp_{70}^{\left( \tau \right)} \cdot \mu _{70}^{\left( d \right)}}}{{S\left( {70} \right)}}\) \(Peluangnya = \frac{{\frac{{4\left( {100 – 70 – t} \right)}}{{5 \cdot {{100}^2}}}}}{{\frac{{{{\left( {100 – 70} \right)}^2}}}{{{{100}^2}}}}}\) \(Peluangnya = \frac{{4\left( {30 – t} \right)}}{5} \cdot \frac{1}{{900}}\) \(Peluangnya = \frac{{30 – t}}{{1125}}\)
Jawaban	C. \(\frac{{30 – t}}{{1125}}\)