Pembahasan Soal Ujian Profesi Aktuaris
Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
Mata Ujian |
: |
Matematika Aktuaria |
Periode Ujian |
: |
November 2018 |
Nomor Soal |
: |
27 |
SOAL
(x) memiliki tiga produk asuransi yang sepenuhnya diskrit
- Asuransi berjangka 20 tahun dengan uang pertanggungan sebesar 50
- Asuransi seumur hidup yang ditunda 20 tahun dengan uang pertanggungan sebesar 100
- Asuransi seumur hidup denga uang pertanggungan sebesar 100
\({Z_i}\) adalah present value random variable untuk asuransi-asuransi di atas, dengan \(i = 1,2,3.\)
Diketahui,
i |
\(E({Z_i})\) |
\(Var({Z_i})\) |
1 |
1,65 |
46,75 |
2 |
10,75 |
50,78 |
3 |
|
|
Berapakah \(Var({Z_3})?\)
- 113
- 133
- 167
- 233
- 267
Step 1 |
\(E[{Z_1}] = 50{A_{\mathop x\limits^| :\left. {\overline {\, {20} \,}}\! \right| }} = 1,65\)
\(E[{Z_2}] = 100{}_{20|}{A_x} = 10,75\) |
|
\(E[{Z_3}] = 100{A_x}\)
\(E[{Z_3}] = 100({A_{\mathop x\limits^| :\left. {\overline {\, {20} \,}}\! \right| }} +{}_{20|}{A_x})\)
\(E[{Z_3}] = 100\left( {\frac{{1,65}}{{50}} + \frac{{10,75}}{{100}}} \right)\)
\(E[{Z_3}] = 14,05\) |
Step 2 |
\(E[{Z_1}^2] = 50{}^2\left( {{}^2{A_{\mathop x\limits^| :\left. {\overline {\, {20} \,}}\! \right| }}} \right)\)
\(E[{Z_1}^2] = Var[{Z_1}] + E{[{Z_1}]^2}\)
\(E[{Z_1}^2] = 46,75 + {1,65^2}\, = 49,47\) |
|
\(E[{Z_2}^2] = 100{}^2\,\left( {{}^2{}_{20|}{A_x}} \right)\)
\(E[{Z_2}^2] = Var[{Z_2}] – E{[{Z_2}]^2}\)
\(E[{Z_2}^2] = 50,78 + {10,75^2}\, = 166,34\) |
|
\(E[{Z_3}^2] = {100^2}\,\left( {{}^2{A_x}} \right)\)
\(E[{Z_3}^2] = 100{}^2({}^2{A_{\mathop x\limits^| :\left. {\overline {\, {20} \,}}\! \right| }} + {}^2{}_{20|}{A_x})\)
\(E[{Z_3}^2] = {100^2}\left( {\frac{{49,47}}{{50{}^2}} + \frac{{166,34}}{{100{}^2}}} \right)\)
\(E[{Z_3}^2] = 364,22\) |
Step 3 |
\(Var[{Z_3}] = E[{Z_3}^2] – E{[{Z_3}]^2}\)
\(Var[{Z_3}] = 364,22 – {14,05^2}\)
\(Var[{Z_3}] = 166,8175\)
\(Var[{Z_3}] \simeq \,167\) |
Jawaban |
c. 167 |