Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Matematika Aktuaria |
| Periode Ujian |
: |
November 2018 |
| Nomor Soal |
: |
11 |
SOAL
Sebuah asuransi diskrit berjangka 2 tahun diterbitkan untuk (x) dengan \(i = 0\).
Diketahui \({q_x} = 0,50\) dan \(Var[{Z_{\mathop x\limits^1 :\left. {\overline {\, 2 \,}}\! \right| }}] = 0,1771\). Hitunglah \({q_{x + 1}}!\)
- 0,54
- 0,65
- 0,76
- 0,87
- 0,98
| Step 1 |
\({A_{\mathop x\limits^1 :\left. {\overline {\, 2 \,}}\! \right| }} = \sum\limits_0^1 {{V^{K(x) + 1}}\Pr (K(x) = k)} \)
\({A_{\mathop x\limits^1 :\left. {\overline {\, 2 \,}}\! \right| }} = v{q_x} + {v^2}{p_x}{q_{x + 1}}\)
\({A_{\mathop x\limits^1 :\left. {\overline {\, 2 \,}}\! \right| }} = {q_x} + {p_x}{q_{x + 1}}\)
\({A_{\mathop x\limits^1 :\left. {\overline {\, 2 \,}}\! \right| }} = 0,5 + (1 – 0,5){q_{x + 1}}\)
\({A_{\mathop x\limits^1 :\left. {\overline {\, 2 \,}}\! \right| }} = 0,5 + 0,5{q_{x + 1}}\) |
| Step 2 |
\({}^2{A_{\mathop x\limits^1 :\left. {\overline {\, 2 \,}}\! \right| }} = \sum\limits_0^1 {{V^{2(K(x) + 1)}}\Pr (K(x) = k)} \)
\({}^2{A_{\mathop x\limits^1 :\left. {\overline {\, 2 \,}}\! \right| }} = v{}^2{q_x} + {v^4}{p_x}{q_{x + 1}}\)
\({}^2{A_{\mathop x\limits^1 :\left. {\overline {\, 2 \,}}\! \right| }} = {q_x} + {p_x}{q_{x + 1}}\)
\({}^2{A_{\mathop x\limits^1 :\left. {\overline {\, 2 \,}}\! \right| }} = 0,5 + (1 – 0,5){q_{x + 1}}\)
\({}^2{A_{\mathop x\limits^1 :\left. {\overline {\, 2 \,}}\! \right| }} = 0,5 + 0,5{q_{x + 1}}\) |
| Maka |
\(Var[{Z_{\mathop x\limits^1 :\left. {\overline {\, 2 \,}}\! \right| }}] = {}^2{A_{\mathop x\limits^1 :\left. {\overline {\, 2 \,}}\! \right| }} – {A_{\mathop x\limits^1 :\left. {\overline {\, 2 \,}}\! \right| }}{}^2\)
\(Var[{Z_{\mathop x\limits^1 :\left. {\overline {\, 2 \,}}\! \right| }}] = 0,5 + 0,5{q_{x + 1}} – \left( {0,5 + 0,5{q_{x + 1}}} \right){}^2\)
\(0,1771 = 0,5 + 0,5{q_{x + 1}} – 0,5{}^2 – (2){\left( {0,5} \right)^2}{q_{x + 1}} – 0,5{}^2\left( {{q_{x + 1}}} \right){}^2\)
\(0,1771 = 0,25 – 0,5{}^2\left( {{q_{x + 1}}} \right){}^2\)
\(0 = 0,5{}^2\left( {{q_{x + 1}}} \right){}^2 – 0,0729\)
\(0,0729 = 0,5{}^2\left( {{q_{x + 1}}} \right){}^2\)
\({q_{x + 1}} = 0,54\) |
| Jawaban |
a. 0,54 |
