Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Permodelan dan Teori Risiko |
| Periode Ujian |
: |
Mei 2018 |
| Nomor Soal |
: |
17 |
SOAL
Seorang aktuaris mengamati lima buah besaran klaim : 11,0; 15,2; 18,0; 21,0; dan 25,8. Tentukan parameter \(\mu \) dari fungsi kepadatan dibawah ini:
\(f(x) = \frac{1}{{\sqrt {2\pi x} }}\exp \left[ { – \frac{1}{{2x}}{{(x – \mu )}^2}} \right],\) \(\mu \, > \,0\)
- 12,64
- 17,64
- 14,54
- 12,85
- 16,74
| Step 1 |
Koefisien \(\frac{1}{{\sqrt {2\pi x} }}\) tidak memuat \(\mu \), sehingga dapat diabaikan.
\(L = f(11)f(15,2)f(18)f(21)f(25,8)\)
\(L = \left[ {{e^{ – \frac{1}{{2(11)}}{{(11 – \mu )}^2}}}} \right]\left[ {{e^{ – \frac{1}{{2(15,2)}}{{(15,2 – \mu )}^2}}}} \right]\left[ {{e^{ – \frac{1}{{2(18)}}{{(18 – \mu )}^2}}}} \right]\left[ {{e^{ – \frac{1}{{2(21)}}{{(21 – \mu )}^2}}}} \right]\left[ {{e^{ – \frac{1}{{2(25,8)}}{{(25,8 – \mu )}^2}}}} \right]\)
\(L = \left[ {{e^{ – \frac{1}{{2(11)}}{{(11 – \mu )}^2} – \frac{1}{{2(15,2)}}{{(15,2 – \mu )}^2} – \frac{1}{{2(18)}}{{(18 – \mu )}^2} – \frac{1}{{2(21)}}{{(21 – \mu )}^2} – \frac{1}{{2(25,8)}}{{(25,8 – \mu )}^2}}}} \right]\) |
| Step 2 |
\(\ln (L) = – \frac{1}{{2(11)}}{(11 – \mu )^2} – \frac{1}{{2(15,2)}}{(15,2 – \mu )^2} – \frac{1}{{2(18)}}{(18 – \mu )^2} – \)
\(\frac{1}{{2(21)}}{(21 – \mu )^2} – \frac{1}{{2(25,8)}}{(25,8 – \mu )^2}\)
\(\frac{{d\ln (L)}}{{d\mu }} = \frac{{(11 – \mu )}}{{(11)}} + \frac{{(15,2 – \mu )}}{{(15,2)}} + \frac{{(18 – \mu )}}{{(18)}} + \frac{{(21 – \mu )}}{{(21)}} + \frac{{(25,8 – \mu )}}{{(25,8)}}\)
- \(\frac{{d\ln (L)}}{{d\mu }} = 0\)
\(0 = 1 – \frac{1}{{11}}\mu + 1 – \frac{1}{{15,2}}\mu + 1 – \frac{1}{{18}}\mu + 1 – \frac{1}{{21}}\mu + 1 – \frac{1}{{25,8}}\mu \)
\(0 = 5 – 0,29863\mu \)
\(\hat \mu = 16,74312695\)
\(\hat \mu = 16,74\) |
| Jawaban |
e. 16,17 |
