Pembahasan Soal Ujian Profesi Aktuaris
Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
Mata Ujian | : | Pemodelan dan Teori Risiko |
Periode Ujian | : | November 2018 |
Nomor Soal | : | 18 |
SOAL
Diberikan
Banyaknya Klaim | Peluang | Besar Klaim | Peluang |
0 | \(\frac{1}{5}\) | | |
1 | \(\frac{3}{5}\) | 25 | \(\frac{1}{3}\) |
150 | \(\frac{2}{3}\) |
2 | \(\frac{1}{5}\) | 50 | \(\frac{2}{3}\) |
200 | \(\frac{1}{3}\) |
Masing-masing besar klaim ialah saling bebas. Hitung variansi dari kerugian gabungan (aggregate loss)
- 4.050
- 8.100
- 10.500
- 12.100
- 15.930
Diketahui | Banyaknya Klaim | Peluang | Besar Klaim | Peluang | 0 | \(\frac{1}{5}\) | | | 1 | \(\frac{3}{5}\) | 25 | \(\frac{1}{3}\) | 150 | \(\frac{2}{3}\) | 2 | \(\frac{1}{5}\) | 50 | \(\frac{2}{3}\) | 200 | \(\frac{1}{3}\) | |
Rumus yang digunakan | Bernoulli: \(Var\left[ S \right] = {\left( {b – a} \right)^2}q\left( {1 – q} \right)\)
\({E\left[ S \right] = E\left[ N \right]E\left[ X \right],}\) \({\mu = \sum\limits_{i = 1}^n {n \cdot \Pr \left( S \right)} ,}\) \({{\sigma ^2} = \sum\limits_{i = 1}^n {{n^2} \cdot \Pr \left( S \right)} – {\mu ^2}}\) |
Proses pengerjaan | Varians dari aggregate klaim berdasarkan besar klaim (menggunakan distribusi Bernoulli karena hnaya ada 2 peluang) - Untuk 0 klaim : 0
- Untuk 1 klaim : \({\left( {150 – 25} \right)^2}\left( {\frac{1}{3}} \right)\left( {\frac{2}{3}} \right) = \frac{{2\left( {{{125}^2}} \right)}}{9}\)
- Untuk 2 klaim : \(2\left[ {{{\left( {200 – 50} \right)}^2}\left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right)} \right] = 10,000\)
Diperoleh nilai ekspektasi dari varians
\(E\left[ {Var\left( {\left. S \right|N} \right)} \right] = \left( {\frac{3}{5}} \right)\left( {\frac{{2\left( {{{125}^2}} \right)}}{9}} \right) + \left( {\frac{1}{5}} \right)\left( {10,000} \right) = \frac{{12250}}{3} = 4,083\frac{1}{3}\) |
Rata-rata dari aggregate klaim - Untuk 0 klaim : 0
- Untuk 1 klaim : \(25\left( {\frac{1}{3}} \right) + 150\left( {\frac{2}{3}} \right) = \frac{{325}}{3}\)
- Untuk 2 klaim : \(2\left[ {50\left( {\frac{2}{3}} \right) + 200\left( {\frac{1}{3}} \right)} \right] = 200\)
Diperoleh nilai varians dari mean
\(Var\left[ {E\left( {\left. S \right|N} \right)} \right] = \left[ {\left( {\frac{3}{5}} \right){{\left( {\frac{{325}}{3}} \right)}^2} + \left( {\frac{1}{5}} \right){{\left( {200} \right)}^2}} \right] – {\left[ {\left( {\frac{3}{5}} \right)\left( {\frac{{325}}{3}} \right) + \left( {\frac{1}{5}} \right)\left( {200} \right)} \right]^2}\)
\(Var\left[ {E\left( {\left. S \right|N} \right)} \right] = \frac{{45,125}}{3} – {105^2} = \frac{{12,050}}{3} = 4,016\frac{2}{3}\) |
Jadi varians dari aggregate loss
\(Var\left[ S \right] = E\left[ {Var\left( {\left. S \right|N} \right)} \right] + Var\left[ {E\left( {\left. S \right|N} \right)} \right] = 4,083\frac{1}{3} + 4,016\frac{2}{3} = 8,100\) |
Jawaban | b. 8.100 |