Pembahasan Soal Ujian Profesi Aktuaris
Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
Mata Ujian |
: |
Matematika Aktuaria |
Periode Ujian |
: |
November 2017 |
Nomor Soal |
: |
7 |
SOAL
Untuk suatu model double decrement, diketahui sebagai berikut:
(i) T adalah variabel acak dari time-until-death
(ii) J adalah variabel acak dari cause-of-decrement
(iii) \({f_{T,J}}\) adalah joint p.d.f dari T dan J
(iv) \({f_{T,J}}(t,j) = \left\{ \begin{array}{l} 0,6k\,{e^{ – 0,8t}} + 0,9(1 – k){e^{ – 1,5t}},\,\,t \ge 0\,and\,J = 1\\ 0,2k\,{e^{ – 0,8t}} + 0,6(1 – k){e^{ – 1,5t}},\,\,t \ge 0\,and\,J = 2 \end{array} \right.\)
(v) \({}_\infty {q_x}^{(1)} = 3{}_\infty {q_x}^{(2)}\)
Hitunglah k.
- 3/8
- 4/9
- 1/2
- 2/3
- 1
Rumus |
\({}_\infty {q_x}^{(j)} = \int\limits_0^\infty {{}_t{p_x}^{(\tau )}{\mu _{x + t}}^{(j)}\,dt} \)
\({}_t{p_x}^{(\tau )}{\mu _{x + t}}^{(j)}\,mirip\,dengan\,{f_{T,J}}(t,j)\) |
Step 1 |
\({}_\infty {q_x}^{(1)} = \int\limits_0^\infty {{f_{T,J}}(t,1)\,dt} \)
\({}_\infty {q_x}^{(1)} = \int\limits_0^\infty {0,6k\,{e^{ – 0,8t}} + 0,9(1 – k){e^{ – 1,5t}}\,\,dt} \)
\({}_\infty {q_x}^{(1)} = \frac{{0,6k}}{{ – 0,8}}(0 – 1) + \frac{{0,9(1 – k)}}{{ – 1,5}}(0 – 1)\)
\({}_\infty {q_x}^{(1)} = \frac{3}{4}k + \frac{3}{5}(1 – k)\)
\({}_\infty {q_x}^{(1)} = \frac{3}{{20}}k + \frac{3}{5}\) |
Step 2 |
\({}_\infty {q_x}^{(2)} = \int\limits_0^\infty {{f_{T,J}}(t,2)\,dt} \)
\({}_\infty {q_x}^{(2)} = \int\limits_0^\infty {0,2k\,{e^{ – 0,8t}} + 0,6(1 – k){e^{ – 1,5t}}\,\,dt} \)
\({}_\infty {q_x}^{(2)} = \frac{{0,2k}}{{ – 0,8}}(0 – 1) + \frac{{0,6(1 – k)}}{{ – 1,5}}(0 – 1)\)
\({}_\infty {q_x}^{(2)} = \frac{1}{4}k + \frac{2}{5}(1 – k)\)
\({}_\infty {q_x}^{(2)} = \frac{2}{5} – \frac{3}{{20}}k\) |
Step 3 |
\({}_\infty {q_x}^{(1)} = 3{}_\infty {q_x}^{(2)}\)
\(\frac{3}{{20}}k + \frac{3}{5} = 3\left( {\frac{2}{5} – \frac{3}{{20}}k} \right)\)
\(\frac{3}{{20}}k + \frac{3}{5} = \left( {\frac{6}{5} – \frac{9}{{20}}k} \right)\)
\(\frac{{12}}{{20}}k = \frac{3}{5}\)
\(k = \frac{3}{5}\left( {\frac{{20}}{{12}}} \right)\)
\(k = 1\) |
Jawaban |
e. 1 |