Pembahasan Soal Ujian Profesi Aktuaris
Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
Mata Ujian | : | Matematika Aktuaria |
Periode Ujian | : | November 2017 |
Nomor Soal | : | 9 |
SOAL
Suatu pembayaran dilakukan sebesar 10 di akhir minggu untuk memenuhi kebutuhan pembelian detergen. Kegunaan detergen adalah variabel, “the week of exhaustion of supply” adalah variabel acak K
k | Pr(K=k) |
1 | 0,20 |
2 | 0,30 |
3 | 0,20 |
4 | 0,15 |
5 | 0,15 |
Misalkan \(Z = 10{v^K}\) menyatakan “present value” dari pembayaran variabel acak. Dengan asumsi bunga \(i = 0,01\), “effective per week”
Hitunglah “variansi” dari Z
- 0,01663
- 0,02663
- 0,03663
- 0,04663
- 0,05663
Rumus | \(E(Z) = 10\sum\limits_{k = 1}^5 {{v^K}P({K_x} = k)} \)
\(E({Z^2}) = 10\sum\limits_{k = 1}^5 {{v^2}^KP({K_x} = k)} \) |
Step 1 | \(E(Z) = 10\left( {{v^1}\,P({K_x} = 1) + {v^2}\,P({K_x} = 2) + {v^3}\,P({K_x} = 3) + {v^4}\,P({K_x} = 4) + {v^5}\,P({K_x} = 5)} \right)\)
\(E(Z) = 10\left( {\frac{{0,20}}{{1,01}} + \frac{{0,30}}{{{{1,01}^2}}} + \frac{{0,20}}{{{{1,01}^3}}} + \frac{{0,15}}{{{{1,01}^4}}} + \frac{{0,15}}{{{{1,01}^5}}}} \right)\)
\(E(Z) = 10\left( {0,9730936} \right)\)
\(E(Z) = 9,730936\) |
Step 2 | \(E({Z^2}) = 100\left( {{v^2}\,P({K_x} = 1) + {v^4}\,P({K_x} = 2) + {v^6}\,P({K_x} = 3) + {v^8}\,P({K_x} = 4) + {v^{10}}\,P({K_x} = 5)} \right)\)
\(E({Z^2}) = 100\left( {\frac{{0,20}}{{{{1,01}^2}}} + \frac{{0,30}}{{{{1,01}^4}}} + \frac{{0,20}}{{{{1,01}^6}}} + \frac{{0,15}}{{{{1,01}^8}}} + \frac{{0,15}}{{{{1,01}^{10}}}}} \right)\)
\(E({Z^2}) = 100\left( {0,94707789} \right)\)
\(E({Z^2}) = 94,707789\) |
Step 3 | \(Var(Z) = E({Z^2}) – {(E(Z))^2}\)
\(Var(Z) = 94,707789 – {(9,730936)^2}\)
\(Var(Z) = 94,707789 – 94,691115\)
\(Var(Z) = 0,016674\)Jawaban paling mendekati ialah
\(Var(Z) \cong 0,01663\) |
Jawaban | a. 0,01663 |