Pembahasan Ujian PAI: A60 - No. 1 - November 2018

Pembahasan Soal Ujian Profesi Aktuaris

Institusi	:	Persatuan Aktuaris Indonesia (PAI)
Mata Ujian	:	Matematika Aktuaria
Periode Ujian	:	November 2018
Nomor Soal	:	1

SOAL

Diketahui

\(k\)	\({\ddot a_{\left. {\overline {\, k \,}}\! \right\| }}\)	\({}_{\left. {k – 1} \right\|}{q_x}\)
1	1,00	0,33
2	1,93	0,24
3	2,80	0,16
4	3,62	0,11

Tentukan nilai \({\ddot a_{x:\left. {\overline {\, 4 \,}}\! \right| }}\) !

2,22
4,44
6,67
8,89
11,11

Kunci Jawaban & Pembahasan

Diketahui

\(k\)	\({\ddot a_{\left. {\overline {\, k \,}}\! \right\| }}\)	\({}_{\left. {k – 1} \right\|}{q_x}\)
1	1,00	0,33
2	1,93	0,24
3	2,80	0,16
4	3,62	0,11

Rumus yang digunakan

n-year endowment: \({\ddot a_{x:\left. {\overline {\, n \,}}\! \right| }} = \ddot a_{x:\left. {\overline {\, n \,}}\! \right| }^1 + {\ddot a_{x:\mathop {\left. {\overline {\, n \,}}\! \right| }\limits^1 }}\)
\(\ddot a_{x:\left. {\overline {\, n \,}}\! \right| }^1 = \sum\limits_{k = 0}^{n – 1} {{{\ddot a}_{\left.{\overline {\, {k + 1} \,}}\! \right| }} \cdot {}_k{p_x} \cdot {q_{x + k}}} = \sum\limits_{k = 0}^{n – 1} {{{\ddot a}_{\left. {\overline {\, {k + 1} \,}}\! \right| }} \cdot {}_{\left. k \right|}{q_x}} \)
\({\ddot a_{x:\mathop {\left. {\overline {\, n \,}}\! \right| }\limits^1 }} = {\ddot a_{\left. {\overline {\, n \,}}\! \right| }} \cdot {}_n{p_x}\)
\({}_n{p_x} = 1 – {}_n{q_x} = 1 – \left( {\prod\limits_{k = 0}^{n – 1} {{q_{x + k}}} } \right) = 1 – \sum\limits_{k = 0}^{n – 1} {{}_{\left. k \right|}{q_x}} \)

Proses pengerjaan

\({{\ddot a}_{x:\left. {\overline {\, 4 \,}}\! \right| }} = \ddot a_{x:\left. {\overline {\, 4 \,}}\! \right| }^1 + {{\ddot a}_{x:\mathop {\left. {\overline {\, 4 \,}}\! \right| }\limits^1 }}\) \({{\ddot a}_{x:\left. {\overline {\, 4 \,}}\! \right| }} = \sum\limits_{k = 0}^3 {{{\ddot a}_{\left. {\overline {\, {k + 1} \,}}\! \right| }} \cdot {}_{\left. k \right|}{q_x}} + {{\ddot a}_{\left. {\overline {\, 4 \,}}\! \right| }} \cdot {}_4{p_x} = \sum\limits_{k = 0}^3 {{{\ddot a}_{\left. {\overline {\, {k + 1} \,}}\! \right| }} \cdot {}_{\left. k \right|}{q_x}} + {{\ddot a}_{\left. {\overline {\, 4 \,}}\! \right| }} \cdot \left( {1 – \sum\limits_{k = 0}^{n – 1} {{}_{\left. k \right|}{q_x}} } \right)\) \({{\ddot a}_{x:\left. {\overline {\, 4 \,}}\! \right| }} = \left[ {{{\ddot a}_{\left. {\overline {\, 1 \,}}\! \right| }} \cdot {}_{\left. 0 \right|}{q_x} + {{\ddot a}_{\left. {\overline {\, 2 \,}}\! \right| }} \cdot {}_{\left. 1 \right|}{q_x} + {{\ddot a}_{\left. {\overline {\, 3 \,}}\! \right| }} \cdot {}_{\left. 2 \right|}{q_x} + {{\ddot a}_{\left. {\overline {\, 4 \,}}\! \right| }} \cdot {}_{\left. 3 \right|}{q_x}} \right]\) \(+ {{\ddot a}_{\left. {\overline {\, 4 \,}}\! \right| }} \cdot \left( {1 – \left[{{}_{\left. 0 \right|}{q_x} + {}_{\left. 1 \right|}{q_x} + {}_{\left. 2 \right|}{q_x} + {}_{\left. 3\right|}{q_x}} \right]} \right)\) \({{\ddot a}_{x:\left. {\overline {\, 4 \,}}\! \right| }} = \left[ {1\left( {0.33} \right) + 1.93\left( {0.24} \right) + 2.8\left( {0.16} \right) + 3.62\left( {0.11} \right)}\right]\) \(+ 3.62\left( {1 – \left[ {0.33 + 0.24 + 0.16 + 0.11} \right]} \right)\) \({{\ddot a}_{x:\left. {\overline {\, 4 \,}}\! \right| }} = 2.2186\)

Jawaban

A. 2,22