Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Permodelan dan Teori Risiko |
| Periode Ujian |
: |
Mei 2018 |
| Nomor Soal |
: |
14 |
SOAL
Random acak sebanyak 5 klaim dari suatu distribusi log-normal diberikan sebagai berikut :
| 500 |
1.000
|
1.500 |
2.500 |
4.500 |
Hitung estimasi parameter \(\hat \mu \) menggunakan metode momen
- 7,398
- 2,458
- 3,984
- 3,832
- 7,148
| Rumus |
\(Lognormal\,\,(\mu ,\sigma )\)
\(E[{X^k}] = \exp \left( {k\mu + \frac{{{k^2}{\sigma ^2}}}{2}} \right)\) |
| Step 1 |
Momen pertama,
- Lognormal Distribution
\(E[X] = \exp \left( {\mu + \frac{{{\sigma ^2}}}{2}} \right)\)
- Rumus Umum
\(E[X] = \frac{1}{5}\sum\limits_{t = 1}^5 {{x_t}} \)
\(\exp \left( {\mu + \frac{{{\sigma ^2}}}{2}} \right) = \frac{1}{5}\sum\limits_{t = 1}^5 {{x_t}} \)
\(\exp \left( {\mu + \frac{{{\sigma ^2}}}{2}} \right) = \frac{1}{5}\left( {500 + 1.000 + 1.500 + 2.500 + 4.500} \right)\)
\(\exp \left( {\mu + \frac{{{\sigma ^2}}}{2}} \right) = 2.000\) |
| Step 2 |
Momen kedua,
- Lognormal Distribution
\(E[{X^2}] = \exp \left( {2\mu + \frac{{{2^2}{\sigma ^2}}}{2}} \right)\)
- Rumus Umum
\(E[{X^2}] = \frac{1}{5}\sum\limits_{t = 1}^5 {{x_t}^2} \)
\(\exp \left( {2\mu + \frac{{{2^2}{\sigma ^2}}}{2}} \right) = \frac{1}{5}\sum\limits_{t = 1}^5 {{x_t}^2} \)
\(\exp \left( {2\mu + \frac{{{2^2}{\sigma ^2}}}{2}} \right) = \frac{1}{5}\left( {{{500}^2} + {{1.000}^2} + {{1.500}^2} + {{2.500}^2} + {{4.500}^2}} \right)\)
\(\exp \left( {2\mu + \frac{{{2^2}{\sigma ^2}}}{2}} \right) = 6.000.000\) |
| Maka |
\(\frac{{\exp \left( {2\mu + \frac{{{2^2}{\sigma ^2}}}{2}} \right)}}{{\exp \left( {\mu + \frac{{{\sigma ^2}}}{2}} \right)}} = \frac{{6.000.000}}{{2.000}}\)
\(\frac{{2\mu + \frac{{{2^2}{\sigma ^2}}}{2}}}{{\mu + \frac{{{\sigma ^2}}}{2}}} = \frac{{\ln 6.000.000}}{{\ln 2.000}}\)
\(\left( {2\mu + \frac{{{2^2}{\sigma ^2}}}{2}} \right)\ln 2.000 = \ln 6.000.000\left( {\mu + \frac{{{\sigma ^2}}}{2}} \right)\)
\(2\left( {\ln 2.000} \right)\mu + 2\left( {\ln 2.000} \right){\sigma ^2} = \left( {\ln 6.000.000} \right)\mu + \left( {\ln 6.000.000} \right)\frac{{{\sigma ^2}}}{2}\)
\(2\left( {\ln 2.000} \right){\sigma ^2} – \left( {\ln 6.000.000} \right)\frac{{{\sigma ^2}}}{2} = \left( {\ln 6.000.000} \right)\mu – 2\left( {\ln 2.000} \right)\mu \)
\(0,4054651081\mu = 7,398169905{\sigma ^2}\)
Solusi,
- \(\hat \mu = 7,398169905\)
\(\hat \mu \cong 7,398\)
- \({\hat \sigma ^2} = 0,4054651081\)
|
| Jawaban |
a. 7, 398 |
