Pembahasan Soal Ujian Profesi Aktuaris
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 | \(\ln (L) = – \frac{1}{{2(11)}}{(11 – \mu )^2} – \frac{1}{{2(15,2)}}{(15,2 – \mu )^2} – \frac{1}{{2(18)}}{(18 – \mu )^2} – \) \(\frac{1}{{2(21)}}{(21 – \mu )^2} – \frac{1}{{2(25,8)}}{(25,8 – \mu )^2}\) \(\frac{{d\ln (L)}}{{d\mu }} = \frac{{(11 – \mu )}}{{(11)}} + \frac{{(15,2 – \mu )}}{{(15,2)}} + \frac{{(18 – \mu )}}{{(18)}} + \frac{{(21 – \mu )}}{{(21)}} + \frac{{(25,8 – \mu )}}{{(25,8)}}\)
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| Jawaban | e. 16,17 |