Pembahasan Ujian PAI: A60 - No. 7 - November 2016

Pembahasan Soal Ujian Profesi Aktuaris

Institusi	:	Persatuan Aktuaris Indonesia (PAI)
Mata Ujian	:	Matematika Aktuaria
Periode Ujian	:	November 2016
Nomor Soal	:	7

SOAL

Diberikan 2 “ independent lives, \((x)\) dan \((y)\), bergantung pada “identical forces of mortality”:

\({\mu _{x + t}} = {\mu _{y + t}} = 0,05\) untuk \(0 < t \le 20\)

Tentukan peluang “last survivor \(\left( {\bar x\bar y} \right)\) akan hidup 10 tahun.

Rumus	\({}_t{p_{\overline {xy} }} = {}_t{p_x} + {}_t{p_y} – {}_t{p_{xy}}\)
Step 1	\({}_t{p_x} = {e^{\int_0^t {{\mu _{x + t}}dt} }} = {e^{ – \int_0^t {0,05dt} }} = {e^{ – 0,05t}}\) \({}_t{p_y} = {e^{\int_0^t {{\mu _{y + t}}dt} }} = {e^{ – \int_0^t {0,05dt} }} = {e^{ – 0,05t}}\)
Step 2	\({}_{10}{p_{\overline {xy} }} = {}_{10}{p_x} + {}_{10}{p_y} – {}_{10}{p_{xy}} = {}_{10}{p_x} + {}_{10}{p_y} – \left( {\left( {{}_{10}{p_x}} \right)\left( {{}_{10}{p_y}} \right)} \right)\) \({}_{10}{p_{\overline {xy} }} = {e^{ – 0,05(10)}} + {e^{ – 0,05(10)}} – \left( {\left( {{e^{ – 0,05(10)}}} \right)\left( {{e^{ – 0,05(10)}}} \right)} \right)\) \({}_{10}{p_{\overline {xy} }} = 1,213061319 – 0,3678794412\) \({}_{10}{p_{\overline {xy} }} = 0,8451818778\)
Jawaban	e. Lebih besar sama dengan 0,80

Pembahasan Ujian PAI: A60 – No. 7 – November 2016