Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Metoda Statistika |
| Periode Ujian |
: |
November 2018 |
| Nomor Soal |
: |
23 |
SOAL
Sebuah studi pada interval \(\left( {x,x + 1} \right)\)
Diketahui:
\({}_s{p_x} = 1 – \frac{3}{2}{s^2}{\left( {{q_x}} \right)^2}\) untuk \(0 \le s \le \frac{2}{3}\)
\({}_s{p_x} = 1 – s{\left( {{q_x}} \right)^2}\) untuk \(\frac{2}{3} < s < 1\)
Jika \({n_x} = 300\), dan 1 kematian terjadi di usia \(x + 0,45\) dan 1 kematian lagi pada usia \(x + 0,85\)
Tentukanlah MLE dari \({q_x}\)
- 0,013
- 0,018
- 0,020
- 0,022
- 0,024
| Diketahui |
\({}_s{p_x} = 1 – \frac{3}{2}{s^2}{\left( {{q_x}} \right)^2}\) atau \(0 \le s \le \frac{2}{3}\)
\({}_s{p_x} = 1 – s{\left( {{q_x}} \right)^2}\) atau \(\frac{2}{3} < s < 1\)
\({n_x} = 300\)
1 kematian terjadi di usia \(x + 0,45\) dan 1 kematian lagi pada usia \(x + 0,85\) |
| Rumus yang digunakan |
\(L = {\left( {1 – {q_x}} \right)^{{n_x} – {d_x}}} \cdot \prod\limits_{i = 1}^d {{}_{{s_i}}{p_x}{\mu _{x + {s_i}}}} \)
Untuk asumsi linear
\({}_s{p_x} = 1 – s \cdot {q_x}\) dan \({\mu _{x + s}} = \frac{{{q_x}}}{{1 – s \cdot {q_x}}}\) |
| Proses Pengerjaan |
Untuk \(0 \le s \le \frac{2}{3}\)
\({}_{{s_i}}{p_x}{\mu _{x + {s_i}}} = \left[ {1 – \frac{3}{2}{s^2}{{\left( {{q_x}} \right)}^2}} \right] \cdot \left[ {\frac{{{{\left( {{q_x}} \right)}^2}}}{{1 – \frac{3}{2}{s^2}{{\left( {{q_x}} \right)}^2}}}} \right]\)
\(= {\left( {{q_x}} \right)^2}\) |
|
Untuk \(\frac{2}{3} < s < 1\)
\({}_{{s_i}}{p_x}{\mu _{x + {s_i}}} = \left[ {1 – s{{\left( {{q_x}} \right)}^2}} \right] \cdot \left[ {\frac{{{{\left( {{q_x}} \right)}^2}}}{{1 – s{{\left( {{q_x}} \right)}^2}}}} \right]\)
\(= {\left( {{q_x}} \right)^2}\) |
|
\(L = {\left( {1 – {q_x}} \right)^{{n_x} – {d_x}}} \cdot \prod\limits_{i = 1}^d {{}_{{s_i}}{p_x}{\mu _{x + {s_i}}}} \)
\(= {\left( {1 – {q_x}} \right)^{300 – 2}} \cdot {\left( {{q_x}} \right)^2} \cdot {\left( {{q_x}} \right)^2}\)
\(l = \ln \left[ {{{\left( {1 – {q_x}} \right)}^{300 – 2}} \cdot {{\left( {{q_x}} \right)}^2} \cdot {{\left( {{q_x}} \right)}^2}} \right]\)
\(= 288\ln \left( {1 – {q_x}} \right) + 2\ln {q_x} + 2\ln {q_x}\)
Dilakukan proses differential
\(\frac{{dl}}{{d{q_x}}} = 288\ln \left( {1 – {q_x}} \right) + 2\ln {q_x} + 2\ln {q_x}\)
\(0 = – \frac{{288}}{{1 – {q_x}}} + \frac{2}{{{q_x}}} + \frac{2}{{{q_x}}}\)
\(\frac{{288}}{{1 – {q_x}}} = \frac{4}{{{q_x}}}\)
\(4 – 4{q_x} = 288{q_x}\)
\({q_x} = \frac{4}{{302}} = 0,01324\) |
| Jawaban |
a. 0,013 |
