Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Matematika Aktuaria |
| Periode Ujian |
: |
November 2014 |
| Nomor Soal |
: |
5 |
SOAL
Sebuah anuitas menaik (temporary annuity – due) membayarkan 2 pada tahun pertama, 3 di tahun kedua dan 4 di tahun ketiga. Diketahui nilai berikut:
\({p_x} = 0,80\)
\({p_{x + 1}} = 0,75\)
\({p_{x + 2}} = 0,50\)
\(v = 0,90\)
Hitunglah variance terhadap nilai sekarang dari variabel acak anuitas ini (present value random variable)
- 3,59
- 4,79
- 5,79
- 7,59
- 8,79
| Diketahui |
Anuitas menaik (temporary annuity – due) membayarkan 2 pada tahun pertama, 3 di tahun kedua dan 4 di tahun ketiga. Diketahui nilai berikut:
\({p_x} = 0,80\)
\({p_{x + 1}} = 0,75\)
\({p_{x + 2}} = 0,50\)
\(v = 0,90\) |
| Rumus yang digunakan |
Rumus rekursif
\(E\left[ Z \right] = A_{x:\left. {\overline {\, n \,}}\! \right| }^1 = {b_k}v{q_x} + {b_k}v{p_x}A_{x + 1:\left. {\overline {\, {n – 1} \,}}\! \right| }^1\) dan \(E\left[ {{Z^2}} \right] = {}^2A_{x:\left. {\overline {\, n \,}}\! \right| }^1 = {b_k}{v^2}{q_x} + {b_k}{v^2}{p_x}{}^2A_{x + 1:\left. {\overline {\, {n – 1} \,}}\! \right| }^1\)
\(Var\left( X \right) = E\left[ {{Z^2}} \right] – {\left( {E\left[ Z \right]} \right)^2}\)
\({}_t{p_x} = \prod\limits_{k = 0}^{t – 1} {{p_{x + k}}} \) |
| Proses pengerjaan |
\(E\left[ Z \right] = {b_1} \cdot {v^0} \cdot {q_x} + \left[ {{b_1} \cdot {v^0} + {b_2} \cdot v} \right] \cdot {p_x} \cdot {q_{x + 1}} + \left[ {{b_1} \cdot {v^0} + {b_2} \cdot v + {b_3} \cdot {v^2}} \right] \cdot {}_2{p_x}\)
\(E\left[ Z \right] = \left[ {2\left( 1 \right)} \right]\left( {1 – 0.8} \right) + \left[ {2\left( 1 \right) + 3\left( {0.9} \right)} \right]\left( {0.8} \right)\left( {1 – 0.75} \right) + \)
\(\left[ {2\left( 1 \right) + 3\left( {0.9} \right) + 4{{\left( {0.9} \right)}^2}} \right]\left( {0.8} \right)\left( {0.75} \right)\)
\(E\left[ Z \right] = 6.1042\) |
|
\(E\left[ {{Z^2}} \right] = {\left( {{b_1} \cdot {v^0}} \right)^2} \cdot {q_x} + {\left( {{b_1} \cdot {v^0} + {b_2} \cdot v} \right)^2} \cdot {p_x} \cdot {q_{x + 1}} + {\left( {{b_1} \cdot {v^0} + {b_2} \cdot v + {b_3} \cdot {v^2}} \right)^2} \cdot {}_2{p_x}\)
\(E\left[ {{Z^2}} \right] = {\left[ {2\left( 1 \right)} \right]^2}\left( {1 – 0.8} \right) + {\left[ {2\left( 1 \right) + 3\left( {0.9} \right)} \right]^2}\left( {0.8} \right)\left( {1 – 0.75} \right) + \) \({\left[ {2\left( 1 \right) + 3\left( {0.9} \right) + 4{{\left( {0.9} \right)}^2}} \right]^2}\left( {0.8} \right)\left( {0.75} \right)\)
\(E\left[ {{Z^2}} \right] = 43.0442\) |
|
\(Var\left( X \right) = E\left[ {{Z^2}} \right] – {\left( {E\left[ Z \right]} \right)^2} = 43.0442 – {6.1042^2} = 5.79\) |
| Jawaban |
c. 5,79 |
